MMCAS34SB 2ed 13 3pp - St Leonard's College · the characteristics of the question. μ – 3σ μ μ +3σ 178 99.7% 167 193 x y 0 μ – σ μ μ+σ 178 183 68% 16% 16% 173 x y 0

As seen in chapter 12 there are many different continuous random variables. One of the mostimportant of these is the normal distribution. Examples that are normally distributed include height,weight, IQ and study scores in VCE subjects, as well as an enormous range of other variables andphysical characteristics such as the volume of venom milked from a taipan or the wingspan of a typeof fly or the weight of a species of adult kangaroo.

The normaldistributionThe normaldistribution

1313

MMCAS34SB_2ed_13_3pp.fm Page 505 Thursday, June 18, 2009 11:53 AM

MathsWorld

Mathematical Methods CAS Units 3 & 4

506

The normal distribution

Continuous random variables, such as height, weight and time, involve characteristics that can be

measured

and can

take any value

in a given range. As there are infinitely many values that a characteristic such as height can take, we group the values into class intervals to enable us to construct frequency tables and histograms. For example, the frequency table and histogram for the heights of 100 adult males are shown below.

(Note that [165, 170) means 165

≤

height

<

170.)

Clearly we are unable to determine any information about the heights of individuals from this data. Nor can we determine how many people are between 183 cm and 188 cm, as these values are not at the ends of the given class intervals. We can, however, deduce that there are 17 people who are at least 185 cm tall by adding the 15 people in the [185, 190) intervalto the two in the [190, 195) interval.

If the number of males whose heights were measured was increased to a very large number and the class intervals were made very small, the histogram would approach a smooth frequency curve as shown opposite. We use

μ

to represent the mean. The smooth frequency curve can be scaled so that it has the properties of a

probability density function

or

pdf

for some random variable

X

. In many such cases, the pdf of the variable can be modelled by the

normal distribution

.

The graph of the

normal pdf

is the familiar bell-shaped curve and is referred to as the

normal curve

. It is symmetrical about its mean.

If

X

is a normally distributed random variable with mean

μ

and variance

σ

2

, we write

X

∼

N(

μ

,

σ

2

). A graph representing Pr(

a

≤

X

≤

b

) as the area under the curve between

x

=

a

and

x

=

b

is shown.

Height (cm)

[165, 170) [170, 175) [175, 180) [180, 185) [185, 190) [190, 195)

Frequency (

f

)

4 18 32 29 15 2

Height (cm)165 170 175 180 185 190 195

Frequency (f)

20

25

15

10

5

0

30

35

165 195180

f

0

a smoothfrequency curve

h

μ

y

a xb0

Area = Pr(a ≤ X ≤ b)

13.1


13 The normal distribution

507

Properties of the normal distribution

The normal distribution pdf satisfies the usual properties of any probability density function:

f

(

x

)

≥

0 and

f

(

x

)

dx

=

1. In addition, it has the following properties.

Probability intervals of the normal distribution

We are often concerned with questions that require us to find, for example, the proportion of the population in a particular interval. Due to the symmetry of the normal distribution about the mean, and the characteristics of the normal curve, we are able to establish the following

probability intervals

relating to the normal distribution:

.

Approximately 68% of the population will fall within one standard deviation of the mean. So Pr(

μ

−

σ

≤

X

≤

μ

+

σ

)

0.68. This is sometimes referred to as the one sigma limits, as the values in the interval are within one standard deviation (i.e.

σ

) of the mean.

Properties of the normal distribution

.

The equation of the probabilitydensity function or pdf for the normal curve is given by

.

.

The normal curve is bell-shaped and is symmetrical.

.

x

=

μ

is not only the mean of the distribution, but also the median and the mode.

.

The maximum value of , occurs when

x

=

μ

.

.

The curve continues infinitely in both directions with the

x

-axis as an asymptote of the curve.

Warning Don’t forget the square!A common mistake is, given the mean μ and standard deviation σ of a distribution, to incorrectly write X ∼ N(μ, σ), forgetting to square the standard deviation to get the variance. The standard deviation is often needed to answer questions and in these cases we must remember to take the square root of the variance.

⌠⌡−∞

∞

μ

y

y = f(x)

x0

σ π12

μ( , )σ π

12

f x( ) 1

σ 2π----------------e

12--- x μ–

σ------------⎝ ⎠

⎛ ⎞ 2–

x R∈,=

f x( ) 1

σ 2π----------------,

~

13.1


MathsWorld Mathematical Methods CAS Units 3 & 4

508

. Approximately 95% of the population will fall within two standard deviations of the mean. So Pr(μ − 2σ ≤ X ≤ μ + 2σ) 0.95. This is sometimes referred to as two sigma limits. It can be stated that a randomly chosen member of the population will most probably be, or is highly likely to be, within two standard deviations of the mean.

. Approximately 99.7% of the population will fall within three standard deviations of the mean. So Pr(μ − 3σ ≤ X ≤ μ + 3σ) 0.997. This is sometimes referred to as three sigma limits. It can be stated that a randomly chosen member of the population will almost certainly be within three standard deviations of the mean.

Example 1If X is a normally distributed variable with μ = 80 and σ = 10, sketch the normal curve for this variable and shade three standard deviations above and below μ.

SolutionIt is useful to first get an idea about the spread. We know thatPr(μ − 3σ ≤ X ≤ μ + 3σ) 0.997 and so we deduce that 99.7%of values will be between 80 − 3 × 10 = 50 and 80 + 3 × 10 = 110.

~

~

μμ σ– μ σ+ x

‘about two-thirds’

μ − σ ≤ X ≤ μ + σ) ≈ 0.68Pr(y

0

μμ σ x– 2 μ σ+ 2

‘most probably’

μ − 2σ ≤ X ≤ μ + 2σ) ≈ 0.95Pr(y

0

μμ σ x– 3 μ σ+3

‘almost certainly’

μ − 3σ ≤ X ≤ μ + 3σ ) ≈ 0.997Pr(y

0

50 80 90 1007060 110 xμ σ– 3 μ σ– 2 μ σ– μ σ+ μ μ σ+2 μ σ+3

y

0

~



509

Example 2The heights of adult males in a certain city are normally distributed with mean 178 cm and standard deviation 5 cm. Find the approximate percentage of the adult males who are:

a between 168 cm and 188 cm. b taller than 193 cm.

c taller than 173 cm. d shorter than 183 cm.

Solutiona 168 cm is two standard deviations below the

mean and 188 cm is two standard deviations above the mean. So approximately 95% of the adult males are between 168 cm and 188 cm.

b 193 cm is three standard deviations above the mean. We know that approximately 99.7% of values arein the interval (μ − 3σ, μ + 3σ). Itfollows that 0.3% of values are notin that interval. As the curve is symmetrical, 0.15% of values must lie in each tail and so approximately0.15% of the adult males are taller than 193 cm.

c 173 cm is one standard deviation below the mean. We know that approximately 68% of values are in the interval (μ − σ, μ + σ). It follows that 32% of values are not in that interval. As the curve is symmetrical, 16% of values must lie in each tail and so approximately 68% + 16% = 84% of the adult males are taller than 173 cm.

d 183 cm is one standard deviation above the mean. We know that approximately 68% of values are in the interval (μ − σ, μ + σ). It follows that 32% of values are not in that interval. As the curve is symmetrical, 16% of values must lie in each tail and so approximately 68% + 16% = 84% of the adult males are shorter than 183 cm. Note that using symmetry, this area is the same as the area considered in part c.

TipA rough sketch of a normal curve can be extremely useful, especially in examples that are not straightforward and in which the symmetry of the normal distribution is to be used to simplify, clarify, or visualise the characteristics of the question.

μ σ– 3 μμ σ+3178

99.7%

167 193 x

y

0

μμ σ– μ σ+178 183

68%16%16%

173 x

y

0

μμ σ– μ σ+178 183

68%16%16%

173 x

y

0

13.1



510

Example 3A continuous random variable X is normally distributed with μ = 54 and σ = 3. Find an interval which will contain approximately:

a 95% of all possible values. b 99.7% of all possible values.

c the highest 16% of all possible values.

Solutiona We know that approximately 95% of values are in the interval μ − 2σ ≤ X ≤ μ + 2σ,

so 54 − 2 × 3 ≤ X ≤ 54 + 2 × 3, which gives 48 ≤ X ≤ 60.

b We know that approximately 99.7% of values are in the interval μ − 3σ ≤ X ≤ μ + 3σ, so 54 − 3 × 3 ≤ X ≤ 54 + 3 × 3, which gives 45 ≤ X ≤ 63.

c We know that approximately 68% of values are in the interval μ − σ ≤ X ≤ μ + σ.It follows that 32% of values are not in that interval. As the curve is symmetrical, 16% of values must lie in each tail, and so approximately 16% of values must be at least μ + σ. So, X ≥ 54 + 3, which gives X ≥ 57.

Example 4Find the value of μ and the (approximate) value of σ if the graph of the pdf of a normal random variable X has a symmetrical region that contains:

a 68% of its values between 12 and 22. b 95% of its values between 48 and 72.

c 99.7% of its values between 23 and 35.

Solutiona We know that the region is symmetrical and so the mean must be at its midpoint. So

. We also know that approximately 68% of values are in the interval

μ − σ ≤ X ≤ μ + σ, so 12 and 22 are each one standard deviation from the mean. So σ = 5.

Alternatively, there are two standard deviations between 12 and 22 and so .

b We have . We know that approximately 95% of values are in the

interval μ − 2σ ≤ X ≤ μ + 2σ, so 48 and 72 are each two standard deviations from the

mean. Therefore . Alternatively, there are four standard deviations

between 48 and 72 and so .

c We have . We know that approximately 99.7% of values are in the

interval μ − 3σ ≤ X ≤ μ + 3σ, so 23 and 35 are each three standard deviations from

the mean. Therefore . Alternatively, there are six standard deviations

between 23 and 35 and so .

μ 12 22+2

------------------ 17= =

σ 22 12–2

------------------ 5= =

μ 48 72+2

------------------ 60= =

σ 72 60–2

------------------ 6= =

σ 72 48–4

------------------ 6= =

μ 23 35+2

------------------ 29= =

σ 35 29–3

------------------ 2= =

σ 35 23–6

------------------ 2= =



511

exercise 13.11 Sketch a normal curve for which:

a μ = 12 and σ = 2. b μ = 0 and σ = 1.

2 For each of the graphs at theright:

i find the mean.

ii estimate the standard deviation.

3 a Sketch, using the same scale in each case, a normal curve for which:

i μ = 16 and σ = 3. ii μ = 16 and σ = 1.

b What effect does decreasing σ have?

4 a Sketch, using the same scale in each case, a normal curve for which:

i μ = 20 and σ = 2. ii μ = 24 and σ = 2.

b What effect does increasing μ have?

5 Sketch a graph for each of the following distributions.

a X ∼ N(14, 4) b X ∼ N(80, 100)

6 For each of the three graphs of normal pdfs:

i find the mean. ii estimate the standard deviation.

y

0 2–2–4–6 4 6 x

19.514.5 17129.50

22 24.5 x

y

a

b

a

b

c

y

2 4 6 8 10 12 14 16 18 20 24 26 28220

13.1



512

7 Which of the graphs of normal pdfs have the same:

a means?

b standard deviations?

8 A random variable X is normally distributed and has a mean of 16 and a standard deviation of 3.

a Find the percentage of values that are:

i between 10 and 22. ii greater than 19.

b Find the probability that X is:

i between 13 and 19. ii less than 10.

9 A normally distributed variable X has μ = 28 and σ = 4.

a Find the percentage of values that are:


b Find the probability that X is


10 A normally distributed variable X has μ = 63 and σ = 5.

a Find an interval within which:

i 68% of values lie. ii 95% of values lie. iii 99.7% of values lie.

b Find the number x such that:

i 16% of values are greater than x. ii 2.5% of values are less than x.

iii 84% of values are greater than x.

11 If X ∼ N(86, 81), find:

a an interval within which the following percentages lie.

i 68% ii 95% iii 99.7%

b the number x such that

i 0.15% of values are less than x. ii 16% of values are greater than x.

iii 97.5% of values are less than x.

12 Find the values of μ and σ, if a normal distribution has a symmetrical region with:

a 68% of values between 28 and 62. b 95% of values between 11 and 47.

c 99.7% of values between −3 and 25.

13 Find the values of μ and σ, if a normal distribution has a symmetrical region with:

a 68% of values between −7 and 5. b 95% of values between 57 and 86.

c 99.7% of values between 86 and 99.

AB

EC

0

D

x

y



513

14 The number of apples harvested from each tree in a large orchard is normally distributed with a mean of 220 and a standard deviation of 15.

a Find the percentage of trees for which the number of apples harvested is expected to be:


b Find the probability that the number of apples harvested from a randomly selected tree is:


15 The marks on a Maths Methods test are normally distributed and have a mean of 67 and a standard deviation of 8.5. To pass the test a student must achieve a score of 50 or more.

a What percentage of students who sat the test are expected to achieve a score:

i between 41.5 and 92.5? ii less than 58.5?

b If a student was randomly selected from those who sat the test, what is the probability that the student achieved a score:

i between 58.5 and 75.5? ii greater than 41.5?

c What percentage of the students who sat the test are expected to pass?

16 A manufacturer makes washers to be used in the production of a large machine. The diameters of the washers are normally distributed and have a mean of 12 mm and a standard deviation of 0.4 mm.

a If a washer is randomly selected, find the probability that the washer has a diameter that is:

i between 11.2 mm and 12.8 mm. ii greater than 11.2 mm.

b What percentage of washers are expected to have a diameter

i between 11.6 mm and 12.4 mm? ii less than 13.2 mm?

c The washers must have a diameter of at least 11.6 mm to be suitable for use in the machine. What percentage of the washers are expected to be suitable?

Example 5The heights of adult males in a certain city are normally distributed with mean 178 cm and standard deviation 5 cm. Find the approximate percentage of adult males who are

a between 168 cm and 183 cm. b between 183 cm and 188 cm.

Solutiona 168 cm is two standard deviations

below the mean, 183 cm is one standard deviation above the mean. We know that approximately 95% of values are in the interval μ − 2σ ≤ X ≤ μ + 2σ and that approximately 68% of values arein the interval μ − σ ≤ X ≤ μ + σ. It follows that 27% (95% − 68%) of values are between these intervals. As the curve is symmetrical, 13.5% of values mustlie in each of the left and right regions between the two intervals and so approximately 68% + 13.5% = 81.5% of these adult males are between 168 cm and 183 cm.

μ σ– 2 μ σ+ 2178 183 188

68%

13.5% 13.5%173168 x0

y

13.1

continued �



514

exercise 13.117 There is a flea infestation at K9 Kennels. The number of fleas per dog can be modelled by

a normal distribution with mean 240 and standard deviation 20. Find the approximate percentage of dogs that have:

a between 200 and 300 fleas.

b between 180 and 220 fleas.

18 The mass of fully grown fish found in Lake Pescatore is normally distributed, with mean 500 g and standard deviation 25 g.

a Find the approximate percentage of these fish that have a mass between 425 g and 525 g.

b What is the probability that a fully grown fish caught from the lake has a mass between 475 g and 575 g?

c Comment on your answers to parts a and b.

19 The wingspan of the blue emperor butterfly is normally distributed with mean 77 mm and variance 9 mm.

a Find the approximate percentage of butterflies that have a wingspan between 77 mm and 83 mm.

b A butterfly is selected at random. What is the probability that it has a wingspan between 83 mm and 86 mm?

20 A professional tennis player took part in a study to measure the speed of his serve. He made numerous serves and the study found that the speeds of the serves could be modelled by a normally distributed random variable with mean 190 km/h and variance 81 km/h. Find the approximate percentage of serves that have a speed between:

a 181 km/h and 208 km/h.

b 172 km/h and 217 km/h.

b From the answer to part a, the graph on the right has already been deduced. So approximately 13.5% of these adult males are between 183 cm and 188 cm.

μ σ– 2 μ σ+2178 183 188

68%

13.5% 13.5%173168 x

0

y

continued



515

The standard normal distribution

One of the difficulties associated with the normal distribution, particularly before the advent of calculators, was the fact that a normal distribution and therefore its related probabilities are dependent on both μ and σ. A consequence of this is that if we were to attempt to construct a normal table to calculate probabilities of various intervals for every possible normal distribution, we would need infinitely many tables. Clearly this is absurd and makes the extensive use of tables impractical. This problem is overcome by the use of a transformed or ‘standardised’ form of normal distribution called the standard normal distribution.

The variable in a standard normal distribution is denoted by Z, to distinguish it from X as used in other normal distributions. The standard normal distribution has μ = 0 and σ = 1, and the z value indicates how many standard deviations the corresponding x value is from the mean. To find the value of z we first find the difference between the x value and the mean by finding x − μ. To find how many standard deviations this equals, we divide by σ. This gives

us , which is an important result that we will use many times.

For the standard normal distribution, we write Z ∼ N(0, 1). After we have standardised a normal distribution in this way, we require only one table and have only one curve, the curve opposite, to deal with.

All of the properties that applied tothe normal distribution also apply tothe standard normal distribution. Somestatements, while equivalent, look somewhat different. For example, we know that Pr(μ − σ ≤ X ≤ μ + σ) 0.68. To convert this to an equivalent statement for a standard normal distribution, we substitute μ = 0 and σ = 1 to get Pr(−1 ≤ Z ≤ 1) 0.68. Similarly Pr(μ − 2σ ≤ X ≤ μ + 2σ) 0.95 can be converted to Pr(−2 ≤ Z ≤ 2) 0.95 and Pr(μ − 3σ ≤ X ≤ μ + 3σ) 0.997 can be converted to Pr(−3 ≤ Z ≤ 3) 0.997.

The connection between X ∼ N(μ, σ2) and Z ∼ N(0, 1) can be symbolised by . This is known as standardising.

The standard normal distributionIf we substitute μ = 0, σ = 1 and into the equation for a normal curve, that is

, we have the equation for a standard normal curve:

. Note that g(z) is a transformed version of f (x).

zx μ–

σ------------=

zx μ–

σ------------=

f x( ) 1

σ 2π----------------e

12--- x μ–

σ------------⎝ ⎠

⎛ ⎞ 2–

x R∈,=

g z( ) 1

2π------------e

12---z2–

z R∈,=

y = g(z)

y

0 1–1–2–3 2 3 zμ σ– 3 μ σ– 2 μ σ– μ σ+μ μ σ+2 μ σ+3

~~

~ ~~ ~

ZX μ–

σ-------------=

13.213.2



516

There are other features of the standard normal distribution that we will use extensively.

. The probability that Z is less than some value z plus the probability that Z is greater than the same value z is equal to 1, i.e. Pr(Z < z) + Pr(Z > z) = 1.

This equation can be transformed to give Pr(Z > z) = 1 − Pr(Z < z). This is an important result, especially if a standard normal table is to be used, as the table is constructed to give Pr(Z < z). That is, it gives the probability that Z < z.

. The symmetry of the distribution shows that Pr(Z < −z) = Pr(Z > z).

By considering the unshaded areas, we can also see that Pr(Z > −z) = Pr(Z < z).

Symmetry properties of the standard normal distribution. Pr(Z > z) = 1 − Pr(Z < z)

. Pr(Z < −z) = Pr(Z > z)

. Pr(Z > −z) = Pr(Z < z)

Example 1If Z ∼ N(0, 1) and Pr(Z < a) = 0.6, find:

a Pr(Z > a) b Pr(Z < −a) c Pr(Z > −a)

Solutiona We know that Pr(Z < a) + Pr(Z > a) = 1

so Pr(Z > a) = 1 − Pr(Z < a) = 0.4.

b We know that Pr(Z < −a) = Pr(Z > a) so Pr(Z < −a) = 0.4.

c We know that Pr(Z > −a) = Pr(Z < a) so Pr(Z > −a) = 0.6.

y

z0 1–1–2–3 2 3 z

y

z0 1–1–2–3 2 3 z

y

–z0 1–1–2–3 2 3 z

y

z0 1–1–2–3 2 3 z

TipWe could use graphs similar to those above to answer these questions, but it is not necessary to sketch graphs for such simple cases. We can use the equations from above instead. Note that if we did sketch the graphs, a would be to the right of the mean as half of the values are to the left of the mean and Pr(Z < a) = 0.6.



517

exercise 13.21 For a standard normal variable Z, find:

a the percentage of values for which Z is

i between −3 and 3. ii less than −1.

b the probability that Z is

i between −2 and 2. ii greater than −2. iii less than zero.

2 For a standard normal variable, find:

a an interval within which the following percentages of values lie.

i 68% ii 95% iii 99.7%

b the number such z that

i 0.15% of values are greater than z.

ii 84% of values are less than z.

iii 97.5% of values are greater than z.

3 If Z ∼ N(0, 1) and Pr(Z < a) = 0.2, find:

a Pr(Z > a). b Pr(Z < −a). c Pr(Z > −a).

4 If Z ∼ N(0, 1) and Pr(Z > a) = 0.75, find:

a Pr(Z > −a). b Pr(Z < −a). c Pr(Z < a).

5 Convert the variable in each of the following expressions to a standard normal variable Z and use it to write an equivalent expression. Do not attempt to evaluate the expression.

a Pr(X < 63); μ = 66, σ = 2

b Pr(X ≥ 117); μ = 95, σ = 11

c Pr(−15 < X ≤ 0); μ = −6, σ = 3

Example 2Convert the variable in each of the following expressions to a standard normal variable Zthen use it to write an equivalent expression involving Z. Do not attempt to evaluate the expression.

a Pr(X < 45); μ = 37, σ = 4 b Pr(X ≥ 12); μ = 15, σ = 2 c Pr(23 < X ≤ 29); μ = 27, σ = 5

Solutiona We have , so . The equivalent expression is Pr(Z < 2).

b . The equivalent expression is Pr(Z ≥ −1.5).

c If x = 23, . If x = 29, . The equivalent expression is

Pr(−0.8 < Z ≤ 0.4).

zx μ–

σ------------= z

45 37–4

------------------ 2= =

zx μ–

σ------------ 12 15–

2------------------ 1.5–= = =

z23 27–

5------------------ 0.8–= = z

29 27–5

------------------ 0.4= =

13.2



518

6 Write an equivalent expression to the one given by converting the variables in each of the following questions to a standard normal variable Z. Do not attempt to evaluate the expression.

a Pr(X > 47); μ = 53, σ = 12

b Pr(X ≤ 4.26); μ = 5, σ = 0.37

c Pr(−32 < X ≤ 28); μ = −4, σ = 20

7 Give an approximate value for Pr(X < 72), if μ = 69, σ = 3, by first converting the variable to a standard normal variable Z.

8 Give an approximate value for Pr(X ≥ 45), if μ = 37, σ = 4, by first converting the variable to a standard normal variable Z.

9 If X ∼ N(114, 121), find an approximate value for Pr(X ≤ 81) by first converting the variable to a standard normal variable Z.

10 If X ∼ N(4.50, 0.0036), find an approximate value for Pr(4.32 ≤ X < 4.68) by first converting the variable to a standard normal variable Z.

A study of a large number of adults finds that their resting heart rate, X beats per minute, is normally distributed such that X ∼ N(68, 25).

a Find the probability that a randomly selected person from this group will have a heart rate


b Find an interval within which the following percentages of values for the heart rate lie.

i 68%

ii 95%

iii 99.7%

c Find k such that:

i 16% of heart rates are greater than k.

ii 2.5% of heart rates are lower than k.

iii 84% of heart rates are greater than k.

d Given that when the variable is converted to a standard normal variable Z, Pr(Z > a) = 0.3, find:

i Pr(Z > − a).

ii Pr(Z < − a).

iii Pr(Z < a).

e Find the probability that a randomly selected person from this group will have a heart rate greater than 78, given that it is greater than 73.

Analysis task 1—have a rest

SAC

Have a rest

SAC analysis task

13.1



519

Calculation of probabilitiesA wide range of random variables follow a normal distribution. Some of these are found in nature and include the weights, lengths, areas, volumes, and times taken, for a huge variety of natural phenomena, and will be used in the examples in this chapter. A random variable can also be designed to have a normal distribution. An example of this that is of great interest to students studying for their VCE is the way in which study scores are calculated.

In all VCE subjects, the study scores are normally distributed, with a mean of 30 and a standard deviation of 7. From what we have learned about probability intervals, we can determine that approximately 68% of study scores are between 23 and 37, 95% of study scores are between 16 and 44, approximately 2.5% of study scores are above 44, and so on.

In this section we will calculate probabilities associated with normal distributions, first, by using some of the properties of normal distributions and then by using a CAS as an aid.

Using properties of normal distributionsWe have previously considered probabilities that can be calculated using some of the symmetry properties of normal distributions, as well as using probability intervals. Now we examine other probabilities that are associated with more general regions.

Often we wish to find the probability that an object, sampled randomly from a variable that is normally distributed, lies between two specified values. That is, we are required to find the probability that a < X < b. This probability is represented by the shaded area in the graph shown below left. Its area is equal to the shaded area in the middle graph minus the shaded area in the right graph. That is Pr(a < X < b) = Pr(X < b) − Pr(X < a).

Calculating and visualising normal probabilities

Example 1If X is normally distributed with Pr(X < a) = 0.2 and Pr(X < b) = 0.7, find:

a Pr(a < X < b). b Pr(X < a | X < b).

Solutiona By constructing graphs similar to those in the example above, or by simply using the result

above, Pr(a < X < b) = Pr(X < b) − Pr(X < a) = 0.7 − 0.2 = 0.5.

μ0 xb

yPr(X < b)

x

y

0 a

Pr(X < a)

μ0 xa b

yPr(a < X < b)

Pr(a < X < b) = Pr(X < b) − Pr(X < a)

= −

continued �

13.313.3



520

b Using the conditional probability formula we have:

Pr(A | B) =

Pr(X < a | X < b) =

To find where X < a and X < b on a graph, shade both regions. The area where both are true will be shaded twice. Thus:

so:

=

Example 2If X is normally distributed with Pr(X > x1) = 0.17 and Pr(X > x2) = 0.85, find:

a Pr(x2 < X < x1).

b an approximate value for c if Pr(x2 < X < x1) = Pr(−c < Z < c) where Z ∼ N(0, 1).

Solutiona If Pr(X > x1) = 0.17 then Pr(X < x1) = 0.83. If Pr(X > x2) = 0.85 then Pr(X < x2) = 0.15.

Thus Pr(x2 < X < x1) = Pr(X < x1) − Pr(X < x2) = 0.83 − 0.15 = 0.68.

b Pr(x2 < X < x1) = 0.68 and we know that Pr(−1 < Z < 1) 0.68 so c = 1.

Example 3If X is normally distributed with Pr(X > d) = 0.16 and Pr(c < X < d) = 0.59, draw diagrams to represent these probabilities and find:

a Pr(X < c).

b an approximate value for z if Pr(X > d) = Pr(Z > z) where Z ∼ N(0, 1).

Solutiona If Pr(X > d) = 0.16 then Pr(X < d) = 0.84.

We know that Pr(c < X < d) = Pr(X < d) − Pr(X < c), so

Pr(X < c) = Pr(X < d) − Pr(c < X < d)= 0.84 − 0.59 = 0.25.

b We know that Pr(−1 < Z < 1) 0.68 so Pr(Z > 1) 0.16. As Pr(X > d) = 0.16, z 1.

Pr A B∩( )Pr B( )

--------------------------

Pr X a<( ) X b<( )∩[ ]Pr X b<( )

-------------------------------------------------------

μ

y

0 xba

Pr X a<( ) X b<( )∩[ ] Pr X a<( )=

Pr X a | X b< <( ) Pr X a<( )Pr X b<( )------------------------=

27---

~

y

c0 d

Pr(c < X < d) = 0.59

Pr(X > d) = 0.16

~~

~



521

Using technologyIn most questions relating to a normal distribution we do not have the convenience of being asked about a probability or interval that involves an exact whole number of standard deviations, nor are we always given a probability that is related to the one being sought.If a question relates to finding a probability involving a standard normal distribution, we have a choice of using a normal probability table or a CAS. The normal probability table is a little tedious to use, and is limited by the fact that the probabilities are generally only shown for positive z values. To find other probabilities symmetry must be used.

To find a probability involving a normal distribution that is not a standard distribution, a normal probability table could only be used if any x values were first converted to z values.

A CAS provides us with an easy and convenient way of calculating probabilities whether or not the normal distribution is a standard normal distribution.

Calculating normal probabilities with a CASMost CAS have a command called Normal Cdf or similar that calculates normal probabilities.

On the TI-Nspire, the command and its syntax are:

• normCdf(lower bound, upper bound, µ, ), or simply

• normCdf(lower bound, upper bound) if μ = 0 and σ = 1

On the Casio ClassPad, the command and its syntax are:

• normCDf(lower bound, upper bound, , µ), or simply

• normCDf(lower bound, upper bound) if μ = 0 and σ = 1

In either case, the command can be found in the Catalog or in an appropriate menu. (Note that the values for µ and σ are entered in opposite orders in the two CAS calculators.)

If a probability such as Pr(X > a) is required, then the upper bound is entered as ∞; if a probability such as Pr(X < b) is required, then the lower bound is entered as −∞.

Example 4If Z has a standard normal distribution, find, correct to 4 decimal places:

a Pr(–1.25 < Z < −0.12)

b Pr(Z ≥ 0.56)

SolutionFrom the screenshots.

a Pr(−1.25 < Z < −0.12) = 0.3466

b Pr(Z ≥ 0.56) = 0.2877

TI-Nspire 11.4

ClassPad 11.4

13.3



522

Representing normal probabilities graphicallyMost CAS have a command called Normal Pdf or similar which can be used to plot the graph of a normal probability density function. (The command can be accessed from the Catalog.) The Integral feature can then be used to shade a given area and record its value. For example, in example 5 part a, the required probability is the area under the normal probability density curve where μ = 28 and σ = 1.7. The screenshots at right show the shaded area and the required probability.

Example 5a If X has a normal distribution with mean μ = 28 and standard deviation σ = 1.7, find,

correct to 4 decimal places:

i Pr(25 ≤ X ≤ 30) ii Pr(X < 32)

b If X ∼ N(12, 9), find, correct to 4 decimal places:

i Pr(X ≥ 14) ii Pr(|X − 12| > 5)

Solutiona The screenshot below shows the calculations for each case. Hence:

i Pr(25 ≤ X ≤ 30) = 0.8415 ii Pr(X < 32) = 0.9907

b Note that Pr(|X − 12| > 5) = 1 − Pr(|X − 12| ≤ 5) and the probability on the right is the same as Pr(−5 ≤ X − 12 ≤ 5) = Pr(7 ≤ X ≤ 17). So:

Pr(|X − 12| > 5) = 1 − Pr(7 ≤ X ≤ 17).

The screenshots at right show the calculations for each case (where σ = 3 and not 9!). Hence:

i Pr(X ≥ 14) = 0.2525

ii Pr(|X − 12| > 5) = 0.0956

TI-Nspire 11.4

ClassPad 11.4TI-N

spire 5.5, 11.4ClassPad 5.5, 11.4



523

exercise 13.3In questions 1 to 10, X is normally distributed and Z ∼ N(0, 1).

1 Given that Pr(X < a) = 0.3 and Pr(X < b) = 0.5, find:

a Pr(X > a). b Pr(a < X < b).

2 If Pr(X > x1) = 0.2 and Pr(X < x2) = 0.4, find:

a Pr(X < x1). b Pr(x1 < X < x2).

3 If Pr(X ≤ x1) = 0.3 and Pr(X ≥ x2) = 0.1, find:

a Pr(x1 ≤ X ≤ x2). b Pr(X < x1 | X < x2).

4 Given that Pr(X < c) = 0.42 and Pr(X < d) = 0.84, find:

a Pr(c < X < d). b Pr(X < c | X < d).

5 Given that Pr(X > x1) = 0.01 and Pr(X ≥ x2) = 0.96, find:

a Pr(x2 < X < x1).

b an approximate value for c if Pr(x2 < X < x1) = Pr(−c < Z < c).

6 Given that Pr(X > c) = 0.001 and Pr(X > d) = 0.998, find:

a Pr(d < X < c).

b an approximate value for z if Pr(d ≤ X ≤ c) = Pr(−z ≤ Z ≤ z).

7 Given that Pr(X < a) = 0.025 and Pr(a < X < b) = 0.345, find:

a Pr(X < b).

b an approximate value for c if Pr(X > a) = Pr(Z > c).

8 Given that Pr(X < c) = 0.0015 and Pr(c < X < d) = 0.4761, find:

a Pr(X ≤ d).

b an approximate value for z if Pr(X > c) = Pr(Z > z).

9 Given that Pr(X < a) = 0.32 and Pr(b < X < a) = 0.18, find:

a Pr(X < b). b Pr(X < a | X < b).

10 Given that Pr(X > c) = 0.45 and Pr(d ≤ X ≤ c) = 0.09, find:

a Pr(X ≥ d). b Pr(X < d | X ≤ c).

11 Find:

a Pr(X ≤ 17) if μ = 19, σ = 1.2 and X is normally distributed.

b Pr(Z > −0.43) where Z ∼ N(0, 1).

c Pr(X < 37) if X ∼ N(34, 16).

d Pr(−0.67 < Z < 1.23) where Z ∼ N(0, 1).

12 Find:

a Pr(X ≥ 61) if X ∼ N(47, 121).

b Pr(Z < −1.05) where Z ∼ N(0, 1).

c Pr(0.73 < Z ≤ 1.41) where Z ∼ N(0, 1).

d Pr(X > 13.7) if μ = 9.8, σ = 2.3 and X is normally distributed.

13.3



524

13 In the following, Z ∼ N(0, 1). Find:

a Pr(|Z| ≥ 2.11).

b Pr(X < 231.8) if μ = 193.4, σ = 37.4 and X is normally distributed.

c Pr(−1.26 ≤ Z ≤ 0.16).

d Pr(|X − μ| ≤ 0.1) if X ∼ N(μ, 0.04).

exercise 13.314 The labels on bags of flour say that the bags have a weight of 1 kg. The actual mean

weight of the bags is 1.02 kg, in order to minimise the number of bags that are underweight. If the weight of the bags is normally distributed with a standard deviation of 12 g, find the percentage of bags, correct to 1 decimal place, that would be expected to:

a weigh more than 1.035 kg.

b weigh less than 990 g, the legal meaning of ‘underweight’.

15 The life of Bright Ideas light globes is normally distributed with a mean of 960 hours and a standard deviation of 24 hours. If a light globe is purchased, what is the probability that its life is:

a greater than 1000 hours? b less than 900 hours?

c between 950 and 980 hours?

Example 6A manufacturer produces screws, the lengths of which are normally distributed with a mean of 15.7 mm and a standard deviation of 0.4 mm. If a screw is randomly selected from the production line, find correct to 4 decimal places the probability that it is:

a no longer than 16 mm. b at least 15 mm long.

c rejected, if only screws that are between 14.6 mm and 16.6 mm long are accepted.

Solutiona Let X represent the length of the screws. We want

Pr(X ≤ 16) as ‘no longer than’ is equivalent to ‘less than or equal to’. The calculation is shown in thescreenshot. Hence:

Pr(no longer than 16 mm) = 0.7734.

b We want Pr(X ≥ 15) as ‘at least’ is equivalent to ‘greater than or equal to’. The screenshot shows the result. Hence:

Pr(at least 15 mm long) = 0.9599.

c The screw is accepted if it lies between 14.6 mm and 16 mm, i.e. if 14.6 ≤ X ≤ 16.6. Then:

Pr(14.6 ≤ X ≤ 16.6) = 0.9848 (see screenshot above right)

So the probability that the screw is rejected is given by:

Pr(screw rejected) = 1 − 0.9848= 0.0152

TI-Nspire 11.4

ClassPad 11.4

continued



525

16 The Sugary Syrup Company sell bottles of maple syrup. The volume of syrup in each bottle follows a normal distribution with a mean of 500 millilitres and a standard deviation of 3.4 millilitres. If a person buys one bottle of maple syrup, find the probability that the volume of maple syrup is:

a less than 495 millilitres. b greater than 507 millilitres.

c between 496 and 505 millilitres.

17 The Choof’s Pies Company makes an exotic chicken pie that is packed full of chicken and vegetables. The chicken pies have an average weight of 750 g. Their weight is normally distributed with a standard deviation of 8 g.

a What is the probability that a randomly selected chicken pie weighs more than 762 g?

b The company offers a money-back guarantee if any chicken pie purchased weighs less than 735 g. What percentage of purchases, to the nearest per cent, are expected to be eligible for a refund?

Example 7The intelligence quotient or IQ as measured by IQ tests is a normally distributed random variable with mean 100 and standard deviation 15.

a If a person is randomly selected from the population, what is the probability that their IQ is:

i over 120? ii less than 90?

b If four people are randomly selected from the population, what is the probability that exactly one of them has an IQ over 120?

c If a person is randomly selected from the people who have an IQ that is greater than 90, what is the probability that the person selected has an IQ over 120?

Solutiona The screenshot on the right shows the results.

i Pr(IQ over 120) = 0.0912

ii Pr(IQ less than 90) = 0.2525

b From part a, the probability that a person has an IQ over 120 is 0.0912. We now need to determine the probability that exactly one of the four people who were randomly selected has an IQ that is over 120. The probability of randomly selecting one person from the population with an IQ that is over 120 does not change with successive selections—we are now dealing with a binomial distribution with four trials and the probability of success on one trial is 0.0912.

Let S = number of successes (in this case having an IQ that is over 120). Then:

=

= 0.2738

TI-Nspire 11.4

ClassPad 11.4

Pr S 1=( ) 41⎝ ⎠

⎛ ⎞ 0.0912( )1 1 0.0912–( )3

13.3

continued �



526

exercise 13.318 Perfect Pictures develops photographic film and prints the photographs on paper that is

advertised to be 18 cm long and 13 cm wide. The area of the developed photographs is normally distributed with a mean of 234 cm2 and a standard deviation of 1.1 cm2.

a If a photograph is randomly selected from those that have been printed, what is the probability that it has an area greater than 235.5 cm2?

b If a photograph is randomly selected from those that are printed, what is the probability that it has an area within 3 cm2 of the mean area?

c If six photographs are randomly selected from those that have been printed, what is the probability that two of them have an area greater than 235.5 cm2?

d If all photographs with an area of less than 232 cm2 are discarded and one photograph is randomly selected from those that remain, what is the probability that it has an area greater than 235.5 cm2?

19 Ball bearings are manufactured for use in large machines. Their diameters are normally distributed with a mean of 2.8 mm and a standard deviation of 0.1 mm.

a To the nearest tenth of a per cent, what is the percentage of ball bearings with a diameter greater than 3 mm?

b The size of the ball bearings is critical. The diameters must be between 2.72 mm and 2.88 mm or the ball bearings cannot be used in the machine. Find the probability that:

i if one ball bearing is chosen randomly, it can be used in the machine?

ii if three ball bearings are chosen randomly, they can all be used in the machine?

c This is a conditional probability question. The sample space has been restricted by the fact that the person being selected cannot have an IQ less than 90. Thus:

Pr(IQ > 120 | IQ > 90) = (using )

= (since (IQ > 120) ⊂ (IQ > 90))

= = 0.1220

Pr IQ 120>( ) IQ 90>( )∩[ ]Pr IQ 90>( )

-------------------------------------------------------------------- Pr A B( ) Pr A B∩( )Pr B( )

--------------------------=

Pr IQ 120>( )Pr IQ 90>( )

--------------------------------

0.09121 0.2525–--------------------------

TI-Nspire 1.2, 11.3, 11.4

ClassPad 1.2, 11.3, 11.4

TipIn parts b and c, it is helpful to store the answers obtained with your CAS in part a so that calculator accuracy rather than rounded answers are used. The calculations in parts b and c using this method are shown in the screenshot at right.

continued



527

20 A national study was commissioned to learn how far elite Australian Rules footballers could kick a football. Each player was given several kicks in an attempt to produce their longest kick. It was found that the lengths of their best kicks were normally distributed with a mean of 55.3 m and a standard deviation of 3.2 m.

a To the nearest per cent, what percentage of footballers were able to kick a football further than 55 m?

b If five footballers were randomly chosen, what was the probability that at least four of them were able to kick a football further than 55 m?

c If a footballer was randomly chosen from those who were able to kick a football further than 55 m, what is the probability that he could kick a football further than 60 m?

Ahab is going on a fishing trip to Lake Lotsafish. The lake is well stocked with large edible fish. The lengths of the fish are normally distributed, with a mean of 45.6 cm and a standard deviation of 4.2 cm.

a If Ahab catches one fish, find the probability that it is:

i less than 43 cm long.

ii longer than 48 cm.

iii between 46 cm and 50 cm long.

b If Ahab fishes each day for seven consecutive days, what percentage of the fish he catches, to the nearest tenth of a per cent, are likely to be longer than 52 cm?

c If Ahab catches 50 fish, how many would be expected to be longer than 52.9 cm?

The law states that fish less than 40 cm long must be thrown back in the water and only fish over 40 cm long may be taken to be eaten.

d What is the probability that a fish must be thrown back?

e If Ahab catches three fish, what is the probability that he is allowed to keep all three?

f If Ahab only catches one fish that he is allowed to keep, what is the probability that it is over 50 cm long?

Analysis task 2—a fishy story

SAC

A fishy story

SAC analysis task

13.2

13.3



528

The inverse normal distribution

We have calculated probabilities for a standard normal distribution given an interval of Z values, and for a non-standard normal distribution given an interval of X values. Sometimes we are given a probability and wish to determine a value c such that Pr(Z < c) is equal to the given probability. Alternatively, we may need to find an x value for given μ and σ; that is, to find the value b such that Pr(X < b) is equal to the given probability, by first finding c such

that Pr(Z < c) is equal to the given probability and then using to find the value of b.

Finding z or x values given a probabilityTo find a z value, given a probability for a standard normal distribution, we would have the choice of using a normal probability table (in reverse), an inverse normal probability table, or a calculator. The normal probability tables are a little tedious to use and limited by the fact that the probabilities are generally only shown for positive z values and vice versa. To find an x value given a probability for a non-standard normal distribution, either of the normal probability tables referred to could be used, but only by first finding z values and then converting

them to x values using .

Whether or not the normal distribution is a standard normal distribution, a CAS provides an easy and convenient way of calculating such values for a given probability.

For example, suppose we want to find a person’s IQ so that only 5% of the population had a higher IQ. Recall that IQ is a normally distributed variable that has a mean of 100 and a standard deviation of 15. So we are required to find the value of x given Pr(X > x) = 0.05. This is equivalent to finding the value of x such that Pr(X < x) = 0.95. It is this latter form we must use, because the inverse normal function on a CAS (and in the normal probability tables) is cumulative.

Calculating values with the inverse normal function on a CASMost CAS have a command called Inverse Normal or similar that calculates the value of x for a given normal probability p where Pr(X < x) = p.

On the TI-Nspire, the command and its syntax are:

• invNorm(p, µ, ), or simply

• invNorm(p) if μ = 0 and σ = 1

On the Casio ClassPad, the command and its syntax are:

• invNormCDf(p, , µ), or simply

• invNormCDf(p) if μ = 0 and σ = 1

In either case, the command can be found in the Catalog or in an appropriate menu.(Note that the values for μ and σ are entered in opposite orders in the two CAS calculators.)

cb μ–

σ------------=

zx μ–

σ------------=

13.4



529

Example 1If Z ∼ N(0, 1), find, correct to 4 decimal places, the value of c given that:

a Pr(Z < c) = 0.29 b Pr(Z ≥ c) = 0.19 c Pr(−c < Z ≤ c) = 0.86

SolutionThe screenshot on the right shows the results in each case.Note that in parts b and c, the first step is to rewrite thestatement in a more convenient form.

a c = −0.5534b Pr(Z ≥ c) = 0.19 so Pr(Z ≤ c) = 0.81.

c = 0.8779

c Pr(−c < Z ≤ c) = 0.86 so 0.14 of the area under the curve is found in the two tails. Therefore, 0.07 of the area under the curve is found in each tail. It follows that Pr(Z < c) = 0.93.c = 1.4758

Example 2If X is normally distributed, find, correct to 2 decimal places:

a x given that Pr(X ≤ x) = 0.37, μ = 13.6, σ = 1.4.b x given that Pr(X > x) = 0.44, μ = 421.2, σ = 15.6.c a and b given that Pr(a ≤ X ≤ b) = 0.53, μ = 41.2, σ = 4.2 and the specified interval is

symmetrical about the mean.

SolutionThe screenshot on the right shows the results.

a x = 13.14b Pr(X > x) = 0.44 so Pr(X ≤ x) = 0.56.

x = 423.56

c Pr(a ≤ X ≤ b) = 0.53 so 0.47 of the area under the curve is found in the two tails. So 0.235 of the area under thecurve is found in each tail. Hence Pr(X ≤ b) = 0.765.Similarly Pr(X ≤ a) = 0.235. So a = 38.17 and b = 44.23.

TI-Nspire 11.4

ClassPad 11.4

TipA graph shows the symmetry properties clearly and it is often helpful to sketch a graph before attempting a calculation like that in part c.

y

zc–c 0 1–1–2–3 2 3

0.86

0.07 0.07

13.4

continued �



530

Quantiles and percentilesQuantiles and percentiles are terms that are similar in meaning and are used to convey information about a distribution. The term quantile refers to the value below which there is a specified probability (given as a decimal) that a randomly selected element will fall. So the 0.5 quantile of a standard normal distribution is equal to a where Pr(Z < a) = 0.5. Of course we know that a = 0 in this case.

The term percentile refers to the value below which there is a specified percentage that a randomly selected element will fall. So the 50th percentile of a standard normal distribution is equal to a for which 50% of values are below a. Again, we know that a = 0 in this case. It follows that the 0.5 quantile has the same meaning as the 50th percentile, and both are the same as the median. Note that the mean and the median have the same value for normal distribution.

Percentile is a term commonly used with the lengths and weights of newborn babies. Parents may be told that their child is at the 70th percentile for length and the 60th percentile for weight. This means that the baby is longer than 70% of babies and weighs more than 60% of babies.

Example 3a Find the 0.45 quantile for a standard normal distribution.

b Find the 80th percentile if X ∼ N(20.6, 4).

Solutiona We require the value of z such that Pr(Z < z) = 0.45.

z = −0.1257

b First convert to a probability. We require the value of x such that Pr(X < x) = 0.80.

x = 22.28

TipAgain a sketch of the graph is helpful,particularly in part c. Try to include a sketch like this in all of your solutions.

41.2

0.235 0.53 0.235

xba

TI-Nspire 11.4

ClassPad 11.4

Warning Variance or standard deviation!Note that σ = 2 in part b. A common mistake is to put σ = 4, but 4 is the variance σ2.



531

exercise 13.4Give answers correct to 4 significant figures unless otherwise specified.

1 If Z ∼ N(0, 1), find the value of z given that:

a Pr(Z < z) = 0.36. b Pr(Z ≥ z) = 0.28. c Pr(−z < Z ≤ z) = 0.38.

2 If Z ∼ N(0, 1), find the value of z given that:

a Pr(Z < z) = 0.94. b Pr(Z ≥ z) = 0.68. c Pr(−z < Z ≤ z) = 0.75.

3 Find the value of x if X is normally distributed and:

a Pr(X ≤ x) = 0.52, μ = 21, σ = 2.6. b Pr(X > x) = 0.73, μ = 132, σ = 9.8.

c Pr(a ≤ X ≤ x) = 0.67, μ = 31.2, σ = 5.6 and the specified interval is symmetrical about the mean.

4 Find the value of x if X is normally distributed and:

a Pr(X ≤ x) = 0.21, μ = 44.2, σ = 7.6. b Pr(X > x) = 0.78, μ = 61.7, σ = 6.9.

c Pr(d ≤ X ≤ x) = 0.87, μ = 0.94, σ = 0.12 and the specified interval is symmetrical about the mean.

5 If X ∼ N(87, 169), Pr(c ≤ X ≤ d) = 0.92 and the specified interval is symmetrical about the mean, find the values of c and d.

6 X ∼ N(10.4, 1.44), Pr(b ≤ X ≤ a) = 0.55 and the specified interval is symmetrical about the mean. Find the values of a and b.

7 For a standard normal distribution find:

a the 0.45 quantile. b the 0.93 quantile.

8 For a standard normal distribution find:

a the 50th percentile. b the 20th percentile.

9 If X is normally distributed and μ = 21, σ = 2.6, find:

a the 0.37 quantile. b the 86th percentile.

10 If X ∼ N(100, 225), find:

a the 95th percentile. b the 0.4 quantile.

11 The average length of female babies at birth is 48.8 cm with a standard deviation of 3.3 cm. The lengths are normally distributed. How many centimetres long is a female baby at birth if her length is at:

a the 45th percentile? b the 90th percentile?

12 The Perfect Parmesan Cheese Company produces containers of parmesan cheese labelled to have contents of 250 g. The contents are normally distributed with a mean of 251.6 g and a standard deviation of 0.8 g. What is the minimum content, in grams, of 95% of all containers produced?

13 The time taken to drive from Carnegie to Torquay is a normally distributed random variable with a mean of 83.2 minutes and a standard deviation of 3.5 minutes.

a What is the least time taken for the slowest 5% of drives? Give your answer correct to the nearest second.

b What is the greatest time taken for the fastest 5% of drives? Give your answer correct to the nearest second.

13.4



532

Example 4a If X is normally distributed with mean 28 and Pr(X ≤ 31) = 0.64, find the standard deviation

σ correct to 3 decimal places.

b If X ∼ N(μ, 16) and Pr(X > 92.3) = 0.23, find the mean μ correct to 2 decimal places.

SolutionFor a non-standard normal distribution, we need both a mean and a standard deviation to use the inverse normal method with a CAS. So first we ‘translate’ the problem so that we can use a standard normal distribution.

Recall that if X ∼ N(μ, σ2) and Z ∼ N(0, 1), then .

We use this as follows.

a We are given that Pr(X ≤ 31) = 0.64 where X ∼ N(28, σ2).

Pr(X ≤ 31) =

= (since μ = 28)

=

Using a CAS we find that if Pr(Z ≤ z) = 0.64, then z = 0.3585. So:

= 0.3585

σ =

= 8.369 (using calculator accuracy)

b We are given Pr(X > 92.3) = 0.23, which is equivalent to Pr(X ≤ 92.3) = 0.77, whereX ∼ N(μ, 16).

Pr(X ≤ 92.3) =

= (since σ = 4)

Using a CAS we find that if Pr(Z ≤ z) = 0.77, then z = 0.7388. So:

= 0.7388

92.3 − μ = 4 × 0.7388μ = 92.3 − 4 × 0.7388

= 89.34

ZX μ–

σ-------------=

TipNote that the final answer was obtained using calculator accuracy, with the answer shown in the screenshot below. If z = 0.3585 is used, we get an answer that is incorrect in the third decimal place as shown here.

Remember that even when asked to give an answer to a certain number of decimal places, you should still use as many decimal places as possible in your working to ensure that subsequent answers are not compromised.

PrX μ–

σ------------- 31 μ–

σ---------------≤⎝ ⎠

⎛ ⎞

Pr Z31 28–

σ------------------≤⎝ ⎠

⎛ ⎞

Pr Z3σ---≤⎝ ⎠

⎛ ⎞

TI-Nspire 11.4

ClassPad 11.4

3σ---

30.3585-----------------

PrX μ–

σ------------- 92.3 μ–

σ--------------------≤⎝ ⎠

⎛ ⎞

Pr Z92.3 μ–

4--------------------≤⎝ ⎠

⎛ ⎞

92.3 μ–4

--------------------



533

exercise 13.414 If X is normally distributed with mean 37 and Pr(X > 41) = 0.27, find the standard

deviation, σ .

15 If X is normally distributed with mean 108 and Pr(X ≤ 103) = 0.43, find the standard deviation, σ .

16 If X ∼ N(μ, 121) and Pr(X < 1243) = 0.69, find the mean, μ.

17 If X ∼ N(μ, 64) and Pr(X ≥ 43.7) = 0.73, find the mean, μ.

18 Studies have shown that the IQs of students studying at a tertiary institution are normally distributed with a standard deviation of 13.6. Find the mean, μ, to the nearest whole number, given that 23% of these students have an IQ of at least 120.

19 A small town in Northern Europe has, on average, 8 hours and 30 minutes of daylight per day in December, with the amount of daylight depending partly on weather conditions. Find the standard deviation, σ, given that one in every four days has fewer than 8 hours of daylight and the number of hours of daylight is normally distributed. Give your answer correct to the nearest minute.

20 A random variable, X, is normally distributed. It is known that 40% of the values are at least 49.51 and three-quarters of the values are less than 51.03. Find the mean, μ, and the standard deviation, σ, of the distribution. Give your answers correct to 1 decimal place.

TipOn the TI-Nspire, the Numerical Solve command together with the invNorm command can be used to solve for an unknown μ or unknown σ. The screenshot shows the results for example 4. (A positive guess is required to find σ; no guess is required to find μ as shown.)

TI-Nspire 7.1, 11.4

continued

13.4



534

In all VCE subjects, the study scores are normally distributed with a mean of 30 and a standard deviation of 7.

Part 1 a What percentage, to the nearest tenth of a per cent, of Maths Methods students achieve

a study score of 40 or more?

b What is the probability that a randomly selected student achieves a study score of 25 or less?

c What is the probability that a randomly selected student achieves a study score between 28 and 38?

Part 2 d If three students were randomly selected from all of those studying Maths Methods,

what is the probability that they all achieved a study score between 28 and 38?

e If five students were randomly selected from all of those studying Maths Methods, what is the probability that three of them achieved a study score between 28 and 38?

Part 3f If a student performs better than 90% of those studying Maths Methods, what is the

least study score to the nearest whole number the student achieves?

g If 60% of those studying Maths Methods perform better than a particular student, what study score, to the nearest whole number, does the student achieve?

ExtensionA large research project involving Maths Methods students demonstrated a strong relationship between the amount of effective study done by a student and the study score achieved by that student. The research showed that students who had studied effectively throughout the year never achieved a study score below 25.

h What is the probability that a randomly selected student who has studied effectively achieves a study score of 40 or more?

i What is the probability that a randomly selected student who has studied effectively achieves a study score of 30 or less?

For students who do virtually no work and do not listen in class, the study scores are still normally distributed but have a standard deviation of 3.1.

j If 20% of these students achieve a study score of less than 16, find the mean study score, μ, for this group of students. Give your answer correct to 1 decimal place.

Analysis task 3—the more you STUDY, the more you SCORE

SAC

The more you STU

DY, the more you SCO

RESAC analysis task

13.3


Chapter review13 The normal distribution

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SummaryThe normal distribution. We write X ∼ N(μ, σ 2)to represent the normally distributed random variable X with mean,

μ, variance σ2 and hence standard deviation, σ.

. The normal distribution has the following properties:

a The equation of the probability density function or pdf for the normal curve is given by

.

a The normal curve is bell-shaped and symmetrical.

a μ is not only the mean of the distribution, it is also the median and the mode.

a The maximum value of f(x), , occurs when x = μ.

a The curve continues infinitely in both directions with the x-axis as an asymptote of the curve.

. The probability intervals of the normal distribution are:

a Pr(μ − σ ≤ X ≤ μ + σ) 0.68

a Pr(μ − 2σ ≤ X ≤ μ + 2σ) 0.95

a Pr(μ − 3σ ≤ X ≤ μ + 3σ) 0.997

. The standard normal distribution has μ = 0 and σ = 1.

. A non-standard normal distribution, X ∼ N(μ, σ 2), can be standardised using .

A value z of Z indicates how many standard deviations from the corresponding value x of X is from its mean.

. The probability intervals (transformed from those involving μ and σ) and some of the symmetries of the standard normal random variable Z are:

a Pr(−1 ≤ Z ≤ 1) 0.68 a Pr(−2 ≤ Z ≤ 2) 0.95 a Pr(−3 ≤ Z ≤ 3) 0.997

a Pr(Z > z) = 1 – Pr(Z < z) a Pr(Z < −z) = Pr(Z > z) a Pr(Z > −z) = Pr(Z < z)

Calculation of probabilities Using properties of normal distributionsWe use known probability intervals and symmetries to calculate probabilities in special cases.

Using a CAS. For a standard normal distribution, for which a and b are the lower and upper z values of

the interval respectively, to calculate Pr(a < Z < b): normCdf(a, b).

. For a non-standard normal distribution with mean μ and standard deviation σ, to calculate Pr(a < X < b): normCdf(a, b, µ, ) on a TI-Nspire; normCDf(a, b, , µ) on a ClassPad.

The inverse normal distribution . For a standard normal distribution, to find the value of z if Pr(Z < z) = p: invNorm(p).

. For a non-standard normal distribution with mean μ and standard deviation σ, to find the value of x if Pr(X < x) = p: invNorm(p, µ, ) on a TI-Nspire; invNormCDf(p, , µ) on a ClassPad.

f x( ) 1

σ 2π----------------e

12--- x μ–

σ------------⎝ ⎠

⎛ ⎞ 2–

x R∈,=

1

σ 2π----------------

~

~

~

ZX μ–

σ-------------=

~ ~ ~


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Quantiles and percentiles. The term quantile refers to the value below which there is a specified probability (given as

a decimal) that a randomly selected element will fall. For example, the 0.6 quantile of a standard normal distribution is equal to a such that Pr(Z < a) = 0.6.

. The term percentile refers to the value below which a specified percentage of the values will fall. The 60th percentile of a standard normal distribution is equal to a such that 60% of values are below a.

. The 0.6 quantile has the same meaning as the 60th percentile.

Revision questionsShort answer technology-free questions1 A normally distributed random variable has μ = 24 and σ = 5. Find:

a the approximate percentage of values that are


b the approximate probability that a randomly selected object from this distribution will be


2 A normally distributed random variable has μ = 47 and σ = 6. Find:

a an interval within which the following percentages of values lie.

i 95% ii 99.7%

b the value of c such that

i 16% of values are greater than c. ii 84% of values are greater than c.

3 Find the values of μ and σ of a normal distribution that has a symmetrical region which contains:

a 68% of its values between 14 and 26. b 95% of its values between 36 and 56.

c 99.7% of its values between −5 and 22.

4 X ∼ N(15, 25). Find:

a the approximate probability that X will be


b the approximate percentage of values that are


5 The mass of tins of sweet corn is normally distributed with mean 250 g and standard deviation 15 g.

a Find the approximate percentage of tins with a mass between 205 g and 265 g.

b A tin is chosen randomly from the production line. Determine the approximate probability that the tin has a mass between 235 g and 295 g.

6 For a standard normal random variable Z, find:

a the approximate percentage of values for which Z is

i between −1 and 1. ii less than −2.

b the approximate probability that Z is

i between −3 and 3. ii greater than −2.

7 A normal random variable X has mean 70. If Pr(X < 80) = p, find in terms of p:

a Pr(X > 60) b Pr(70 < X < 80) c Pr(X < 80 | X > 60)



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8 For a standard normal random variable, find:

a an interval symmetric about the mean within which

i 68% of values lie. ii 99.7% of values lie.

b the z value such that

i 0.15% of values are greater than z. ii 97.5% of values are greater than z.

9 Given that Pr(Z < a) = 0.37 where Z ∼ N(0, 1), find:

a Pr(Z > a) b Pr(Z < −a) c Pr(Z > −a)

10 Write an equivalent statement to the one given by converting the variable in each of the following to a standard normal variable. Do not attempt to evaluate the statement.

a Pr(X ≥ 11); μ = 13, σ = 4 b Pr(57 < X ≤ 63); μ = 59, σ = 5

11 X is a normal random variable and a, b are constants such that Pr(X < a) = 0.4 and Pr(X > b) = 0.2. Find:

a Pr(X > a). b Pr(a < X < b). c Pr(X < a | X < b).

12 X is a normally distributed random variable with mean 20 and standard deviation 2. Z is a standard normal random variable. Given that Pr(Z ≤ 1.5) = 0.9332, find:

a Pr(20 ≤ X ≤ 23)

b Pr(X > 17)

c Pr(X < 17 | X < 20)

Extended response1 a If X ∼ N(137, 196), find:

i Pr(X ≤ 120) ii Pr(130 ≤ X ≤ 140)

iii c if Pr(X > c) = 0.81 iv the 75th percentile.

v the 0.3 quantile.

b The lengths of the leaves of a particular tree are normally distributed with mean 12.4 cm and standard deviation 2.3 cm. Find, in centimetres, a leaf’s length if it is at:

i the 35th percentile. ii the 60th percentile.

2 The times taken for a large group of swimmers to swim 50 metres freestyle are normally distributed with mean 23.1 seconds and variance 0.16 seconds.

a Without a calculator find the approximate percentage of these swimmers who can swim 50 metres freestyle in a time that is between 22.7 seconds and 23.5 seconds.

b If one of these swimmers is randomly chosen, what is the probability that they can swim 50 metres freestyle in less than 22.3 seconds?

c Find the percentage of these swimmers who cannot swim 50 metres freestyle in less than 23.7 seconds.

d If two swimmers are chosen randomly, what is the probability that both can swim 50 metres freestyle in less than 23.7 seconds?

e What is the probability that one of the group can swim 50 metres freestyle in less than 23 seconds, given that the swimmer is randomly chosen from those who can swim 50 metres freestyle in less than 23.7 seconds.


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3 Dracula’s Stakes sells garden stakes that are advertised as being 1 m long. The lengths of the stakes are normally distributed with a mean of 1.015 m and a standard deviation of 0.01 m.

a If a person buys one garden stake, what is the probability that it is:

i longer than 1.025 m?

ii shorter than 1.01 m?

iii between 1.025 m and 1.04 m?

A person can get their money back if a stake is less than 1 m long.

b i What is the probability that a person who buys one stake is eligible for the money-back deal?

ii What is the number of refunds the store would expect to make if 500 stakes were sold?

c If a person buys three stakes, what is the probability that they are not eligible for a refund on any of the stakes?

4 Kate’s Candles sells a type of candle that has the characteristic that the length of time the candle burns before dying is a normally distributed random variable with a mean of 5 hours 15 minutes and a standard deviation of 12 minutes.

a Find the percentage of candles that burn for at least five and a half hours.

b Find the probability that a randomly selected candle burns for less than 295 minutes.

c If four candles are selected randomly, what is the probability that, at most, one of them burns for less than 295 minutes?

d What is the probability that a randomly selected candle burns for at least five and a half hours, given that it burns for more than 5 hours 15 minutes.

5 Jeremy’s Juice Company sells apple juice in cartons that advertise the contents as 2 litres. In practice, the volume of juice is normally distributed, with a mean of 2005 millilitres and a standard deviation of 2.8 millilitres.

a What is the probability that the volume of juice in a randomly selected carton is less than 2 litres?

b Find the percentage of cartons with a volume of juice that is at least 2008 millilitres.

c If a customer buys three cartons of apple juice, what is the probability that the volume of juice in at most two of them is greater than 2 litres?

d The company decides to introduce a quality control system so that any carton from the production line that has a volume of juice of less than 2 litres is not put out to be sold. What is the probability that a randomly selected carton, which is put out to be sold, contains at least 2008 millilitres, given the new quality control system.

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MMCAS34SB 2ed 13 3pp - St Leonard's College · the characteristics of the question. μ – 3σ μ μ +3σ 178 99.7% 167 193 x y 0 μ – σ μ μ+σ 178 183 68% 16% 16% 173 x y 0