Chapter 4 Continuous Variables and Their Probability ... › wp-content › uploads › 2020 › 03 › Lecture-Notes-6.pdfContinuous Variables and Their Probability Distributions

Chapter 4

Continuous Variables and TheirProbability Distributions

4.1 Introduction

We now look at probability distributions for continuous random variables.

4.2 The Probability Distribution for a Continuous

Random Variable

The (cumulative) distribution function for (either discrete or continuous) randomvariable Y is

F (y) = P (Y ≤ y) , −∞ < y < ∞,

and has properties1

(1) limy→−∞ F (y) = 0,(2) limy→∞ F (y) = 1,(3) if y1 < y2, then F (y1) ≤ F (y2); that is, F is nondecreasing.

A random variable Y with distribution function F (y) is continuous if F (y) iscontinuous2. The (probability) density function, f(y), is given by:

f(y) =dF (y)

dy= F ′(y),

and so, also, F (y) =∫ y

−∞f(t) dt. Properties of the density function for a continuous

random variable are:

1Also, limn→∞ F (yn) = F (y); that is, F must be right continuous (which determines where thesolid and empty endpoints are on the graph of a distribution function).

2And the first derivative of F (y) exists and continuous except, possibly, at a finite number ofpoints in a finite interval.

93

94Chapter 4. Continuous Variables and Their Probability Distributions (ATTENDANCE 6)

(1) f(y) ≥ 0, for all y,−∞ < y < ∞,(2)

∫∞

−∞f(y) dy = 1.

Also, the pth quantile (or pth percentile) of random variable Y is denoted φp andis the smallest value where P (Y ≤ φp) = F (φp) = p. And,

P (a ≤ Y ≤ b) = P (Y ≤ b)− P (Y ≤ a) = F (b)− F (a) =

∫ b

a

f(y) dy.

Quantile is defined as φp in equation F (φp) = p, where 0 ≤ p ≤ 1. In special case p happens to equal r

100, r = 1, . . . , 99,

quantile can also be called a percentile. Having said this, quantile and percentiles are used fairly interchangeably.

Exercise 4.2(The Probability Distribution for a Continuous Random Vari-able)

1. Discrete distribution function: flipping a coin twice.Let the number of heads flipped in two flips of a coin be a binomial randomvariable Y where n = 2 and p = 0.5, in other words, since the sample spacein this case is TT (Y = 0), HT or TH (Y = 1) and HH (Y = 2), then p(0) =P (Y = 0) = 0.25, p(1) = P (Y = 1) = 0.50 and p(2) = P (Y = 2) = 0.25.Density p(y) is given in table and Figure 4.1(a), and distribution F (y) = P (Y ≤y is given in Figure 4.1(b).

y 0 1 2p(y) = P (Y = y) 1

424

14

0 1 2

0.25

0.50

0.75

1

0 1 2

0.25

0.50

0.75

1

(a) density (b) distribution

Figure 4.1: Density and distribution function: flipping a coin twice

2nd DISTR binompdf(2,0.5).

(a) F (0) = P (Y ≤ 0) = P (Y = 0) = (choose one)(i) 0 (ii) 0.25 (iii) 0.75 (iv) 1.

(b) F (1) = P (Y ≤ 1) = P (Y = 0) + P (Y = 1) = (choose one)(i) 0 (ii) 0.25 (iii) 0.75 (iv) 1.

Section 2. The Probability Distribution for a Continuous Random Variable (ATTENDANCE 6)95

(c) F (2) = P (Y ≤ 2) = P (Y = 0) + P (Y = 1) + P (Y = 2) = (choose one)(i) 0 (ii) 0.25 (iii) 0.75 (iv) 1.

(d) Also, if y < 0, F (y) = P (Y ≤ y) = (choose one)(i) 0 (ii) 0.25 (iii) 0.75 (iv) 1.

(e) And, if y ≥ 2, F (y) = P (Y ≤ y) = (choose one)(i) 0 (ii) 0.25 (iii) 0.75 (iv) 1.

(f) So, in this case,

F (y) =

0, y < 00.25, 0 ≤ y < 10.75, 1 ≤ y < 21, y ≥ 2.

(i) True (ii) False

(g) The discontinuous “step function” graph of this F (y) is given in Fig-ure 4.1(b). Notice that F (y) is right continuous, which is indicated bythe solid and empty endpoints on the graph of this distribution function.The height of a “step” in distribution F (y) is equal to the height of thecorresponding “stick” in density p(y).

(i) True (ii) False

(h) P (Y < 1) = P (Y = 0) = 0.25 = (choose one)(i) F (0) (ii) F (1) (iii) F (2) (iv) F (3).

(i) P (Y < 2) = P (Y = 0) + P (Y = 1) = 0.75 = (choose one)(i) F (0) (ii) F (1) (iii) F (2) (iv) F (3).

(j) P (Y > 1) = 1− P (Y ≤ 1) = 1− F (1) = 1− 0.75 = (choose one)(i) 0 (ii) 0.25 (iii) 0.75 (iv) 1.

2. Another discrete distribution function.Let random variable Y have distribution,

F (y) =

0, y < 013, 0 ≤ y < 1

12, 1 ≤ y < 2

1, y ≥ 2.

(a) F (1) = P (Y ≤ 1) = P (Y = 0) + P (Y = 1) = (choose one)(i) 0 (ii) 1

6(iii) 1

3(iv) 1

2.

(b) P (Y < 2) = P (Y = 0) + P (Y = 1) = (choose one)(i) 0 (ii) 1

6(iii) 1

3(iv) 1

2.

(c) P (Y ≤ 1.5) = P (Y = 0) + P (Y = 1) = (choose one)(i) 0 (ii) 1

6(iii) 1

3(iv) 1

2.


0 1 2

1

0 1 2

1

(a) density (b) distribution

1/3

1/2

1/6

1/3

1/2

Figure 4.2: Density and distribution function

(d) p(1) = P (Y = 1) = P (Y ≤ 1)−P (Y ≤ 0) = F (1)−F (0) = (choose one)(i) 0 (ii) 1

6(iii) 1

3(iv) 1

2.

(e) p(2) = P (Y = 2) = P (Y ≤ 2)−P (Y ≤ 1) = F (2)−F (1) = (choose one)(i) 0 (ii) 1

6(iii) 1

3(iv) 1

2.

(f) Random variable Y is discrete, not continuous, because the associatedF (y) is a discontinuous (“step”, in this case) function.(i) True (ii) False

(g) Notice that(1) limy→−∞ F (y) = 0,(2) limy→∞ F (y) = 1,(3) if y1 < y2, then F (y1) ≤ F (y2); that is, F is nondecreasing.(i) True (ii) False

3. Continuous distribution function: uniform.Let the time waiting in line, in minutes, be described by the random variableY which has the following probability density (not distribution),

f(y) =

0, y < 2,12, 2 ≤ y < 4,

0, y ≥ 4.

(a) The chance of waiting at most y = 1 minute is

F (1) = P (Y ≤ 1) =

∫ 1

−∞

f(y) dy =

∫ 1

−∞

0 dy =

(choose one) (i) 0 (ii) 0.1 (iii) 0.5 (iv) 1.


0 1 2 3

0.25

0.50

0.75

1

4

density f(y)

0 1 2 3

0.25

0.50

0.75

1

4

distribution F(y)

Figure 4.3: Distribution function: continuous uniform

(b) The chance of waiting at most y = 3 minutes is

F (3) = P (Y ≤ 3) =

∫ 3

−∞

f(y) dy

=

∫ 2

−∞

0 dy +

∫ 3

2

1

2dy

= 0 +y

2

]3

2=

3

2− 2

2=


(c) The chance of waiting at most y = 5 minutes is3

F (5) = P (Y ≤ 5) =

∫ 5

−∞

f(y) dy

=

∫ 2

−∞

0 dy +

∫ 4

2

1

2dy +

∫ 5

4

0 dy

= 0 +y

2

]4

2+ 0 =

4

2− 2

2=


(d) The chance of waiting any time less than 2 minutes, y < 2,F (y) = P (Y ≤ y) =

∫ y

−∞0 dy = (choose one)

(i) 0 (ii) 0.1 (iii) 0.5 (iv) 1.

(e) The chance of waiting any time more than 4 minutes, y ≥ 4,

F (y) = P (Y ≤ y) =∫ y

−∞f(y) dy =

∫ 4

212dy = (choose one)

(i) 0 (ii) 0.1 (iii) 0.5 (iv) 1.

3The integrals involving zero will not always be explicitly stated. In other words, instead of∫ 5

−∞f(y) dy =

∫ 2

−∞0 dy +

∫ 4

2

1

2dy +

∫ 5

40 dy, the integral

∫ 5

−∞f(y) dy =

∫ 4

2

1

2dy is used instead.


(f) In general, the distribution function is

F (y) =

0, y < 2,y2− 1, 2 ≤ y < 4,

1, y ≥ 4.

Random variable Y is continuous because, as shown in Figure 4.2, eventhough the density, f(y), is a discontinuous function, the associated dis-tribution, F (y), is a continuous function in this case.(i) True (ii) False

(g) F (3) = 12is both the (positive) area under density f(y) from −∞ up to 3

and also the point on distribution F (y) at y = 3.(i) True (ii) False

(h) P (Y < 3) =∫ 3

212dt = y

2

]3

2= 3

2− 2

2= 1

2= (choose one)

(i) P (Y > 3) (ii) P (Y < 4) (iii) P (Y ≤ 3) (iv) P (Y ≤ 4).since Y is a continuous random variable and, consequently, the chance ofwaiting exactly y = 3 minutes is zero,

P (Y = 3) =

∫ 3

3

f(y) dy = 0.

(i) P (2.5 < Y < 3) =∫ 3

2.512dy = y

2

]3

2.5= 3

2− 2.5

2= 0.5

2= 1

4=

(choose one or more)(i) P (2.5 ≤ Y < 3) (ii) P (2.5 ≤ Y ≤ 3)(iii) P (2.5 < Y < 3) (iv) P (2.5 < Y ≤ 3).because the chance of waiting exactly y minutes is zero,

P (Y = y) =

∫ y

y

f(t) dt = 0.

(j) Percentile. If P (Y ≤ φ0.95) = 0.95, what is the 95th percentile, φ0.95,waiting time? In other words, what is that time where 95% of people waitless than this time? Since

P (Y ≤ φ0.95) = F (φ0.95) =

∫ φ0.95

2

1

2dy =

y

2

]φ0.95

2=

φ0.95

2− 2

2= 0.95,

then, rearranging, φ0.95 = (choose one)(i) 3.2 (ii) 3.5 (iii) 3.9 (iv) 4.1.

4. Continuous distribution function: triangle.Let random variable Y have the following probability density,

f(y) =

0, y < 2,16y, 2 ≤ y < 4,

0, y ≥ 4.


(a) In general, the distribution function is

F (y) =

∫ y

−∞0 dt = 0, y < 2,

∫ 2

−∞0 dt +

∫ y

2t6dt = 0 + t2

12

]y

2= y2

12− 4

12, 2 ≤ y < 4,

∫ 2

−∞0 dt +

∫ 4

2t6dt +

∫∞

40 dt = 0 + y2

12

]4

2+ 0 = 1, y ≥ 4.

Both density and distribution are given in Figure 4.3.(i) True (ii) False

0 1 2 3

0.25

0.50

0.75

1

4

density f(y)

0 1 2 3

0.25

0.50

0.75

1

4

distribution F(y)


(b) F (1) = (choose one) (i) 0 (ii) 5

12(iii) 7

12(iv) 1.

(c) F (2) = 22

12− 4

12= (choose one) (i) 0 (ii) 5

12(iii) 9

12(iv) 1.

(d) F (3) = 32

12− 4

12= (choose one) (i) 0 (ii) 5

12(iii) 9

12(iv) 1.

(e) F (5) = (choose one) (i) 0 (ii) 5

12(iii) 9

12(iv) 1.

(f) Also,

P (1 < Y < 3) = P (Y < 3)− P (Y < 1)

= P (Y ≤ 3)− P (Y ≤ 1)

= F (3)− F (1) =

(

32

12− 4

12

)

− 0 =

(choose one) (i) 0 (ii) 5

12(iii) 9

12(iv) 1.

(g) P (2.5 < Y < 3.5) = F (3.5)− F (2.5) =(

3.52

12− 4

12

)

−(

2.52

12− 4

12

)

=


(h) P (Y > 3.5) = 1− P (Y ≤ 3.5) = 1− F (3.5) = 1−(

3.52

12− 4

12

)

=

(choose one) (i) 0.3125 (ii) 0.4425 (iii) 0.7650 (iv) 1.

(i) P (Y ≤ 3|Y < 3.5) = P (Y≤3∩Y <3.5)P (Y <3.5)

= P (Y≤3)P (Y <3.5)

= F (3)F (3.5)

=32

12− 4

12

3.52

12− 4

12

=

(choose one) (i) 0.32 (ii) 0.45 (iii) 0.57 (iv) 0.61.


5. Continuous distribution function: another triangle.Let random variable Y have the following probability density,

f(y) =

0, y < 1,ky, 1 ≤ y < 5,0, y ≥ 5.

(a) The distribution function is

F (y) =

0, y < 1,∫ y

1kt dt = k t2

2

]y

1= ky2

2− k

2= k(y2−1)

2, 1 ≤ y < 5,

1, y ≥ 5.

(i) True (ii) False

(b) What is k? Since

F (5) =

∫ 5

1

kt dt = kt2

2

]5

1

=k(52 − 1)

2=

24k

2= 1,

then k = (choose one) (i) 1

11(ii) 1

12(iii) 1

13(iv) 1

14.

(c) In other words,

f(y) =

0, y < 1,y12, 1 ≤ y < 5,

0, y ≥ 5.

and

F (y) =

0, y < 1,(y2−1)

24, 1 ≤ y < 5,

1, y ≥ 5.

(i) True (ii) False

(d) P (2.5 < Y < 3.5) = F (3.5)− F (2.5) =(

3.52−124

)

−(

2.52−124

)

=


(e) Percentile. What is the 78th percentile, φ0.78? Since

P (Y ≤ φ0.78) = F (φ0.78) =(φ2

0.78 − 1)

24= 0.78,

φ0.78 =√19.72 ≈ (choose one) (i) 3.9 (ii) 4.1 (iii) 4.4 (iv) 4.9.

6. One last continuous distribution function.Let random variable Y have the following probability density,

f(y) =

y, 0 < y < 1,2− ky, 1 ≤ y < 2,0, elsewhere.


(a) For 0 < y < 1, the distribution function is

F (y) =

∫ y

0

t dt =t2

2

]y

0

=y2

2− 02

2=

y2

2

and for 1 ≤ y < 2,

F (y) =

∫ 1

0

t dt+

∫ y

1

(2− kt) dt

=t2

2

]1

0

+

[

2t− kt2

2

]y

1

=

(

1

2− 0

2

)

+

((

2y − ky2

2

)

−(

2(1)− k12

2

))

= 2y − ky2

2+

k

2− 3

2

and for y ≥ 2, F (y) = 1; in other words,

F (y) =

y2

2, 0 < y < 1,

2y − ky2

2+ k

2− 3

2, 1 ≤ y < 2,

1, y ≥ 2.

(i) True (ii) False

(b) What is k? Since

F (2) =

∫ 2

0

f(y) dy = 2(2)− k22

2+

k

2− 3

2= 1,

then k = (choose one) (i) 1 (ii) 2 (iii) 3 (iv) 4.

(c) In other words,

f(y) =

y, 0 < y < 1,2− y, 1 ≤ y < 2,0, elsewhere.

and

F (y) =

y2

2, 0 < y < 1,

2y − y2

2− 1, 1 ≤ y < 2,

1, y ≥ 2.

(i) True (ii) False

(d) Check if F (y) correct. Since it should be that F ′(y) = f(y), notice, for0 < y < 1,

F ′(y) =d

dy

(

y2

2

)

= y = f(y),


and for y ≥ 1,

F ′(y) =d

dy(1) = 0 = f(y),

and also for 1 ≤ y < 2,

F ′(y) =d

dy

(

2y − y2

2− 1

)

=

(choose one) (i) 2 − 2y (ii) 2 − y2 (iii) 2y − y2 − 1 (iv) 2 − y

which equals f(y) and, together, this indicates F (y) is correct.

4.3 Expected Values for Continuous Random

Variables

The expected value, E(Y ), of a continuous random variable, Y is given by4

E(Y ) =

∫ ∞

−∞

yf(y) dy.

The expected value of a function g of the random variable Y , E[g(Y )], is given by

E[g(Y )] =

∫ ∞

−∞

g(y)f(y) dy.

The variance, σ2 = V (Y ), and standard deviation, σ, are defined exactly as they werein the discrete case. Other properties are

E(c) = c,

E[cg(Y )] = cE[g(Y )],

E[g1(Y ) + g2(Y ) + · · ·+ gk(Y )] = E[g1(Y )] + E[g2(Y )] + · · ·+ E[gk(Y )],

where c is a constant (number, not a random variable).

Exercise 4.3 (Expected Values for Continuous Random Variables)

1. Time and cost of cell phone.Let the amount of time spent (in minutes) on a cell phone call be representedby random variable Y with the following probability density,

f(y) =

{

16y, 2 ≤ y ≤ 4,

0, elsewhere.

4This assumes the expected value exists,∫

∞

−∞|y|f(y) dy < ∞.

Section 3. Expected Values for Continuous Random Variables (ATTENDANCE 6)103

(a) Expected value. The expected amount of time on the call is,

µ = E(Y ) =

∫ ∞

−∞

yf(y) dy =

∫ 4

2

y

(

1

6y

)

dy =

∫ 4

2

y2

6dy =

y3

18

]4

2

=43

18−23

18=

(choose one) (i) 23

9(ii) 28

9(iii) 31

9(iv) 35

9.

(b) E (Y 2).

E(

Y 2)

=

∫ ∞

−∞

y2f(y) dy =

∫ 4

2

y3

6dy =

y4

24

]4

2

=44

24− 24

24=

(choose one) (i) 9 (ii) 10 (iii) 11 (iv) 12.

(c) Variance. The variance in the amount of time on the call is,

σ2 = V (Y ) = E(

Y 2)

− µ2 = 10−(

28

9

)2

=

(choose one) (i) 23

81(ii) 26

81(iii) 31

81(iv) 35

81.

(d) Standard deviation. The standard deviation in the amount of time spenton the call is,

σ =√σ2 =

√

26

81≈


(e) Expected value of function g(y). The expected cost of the call, if it costs50 cents just to use the cell phone and 5 cents per minute after that; inother words, g(y) = 5y + 50, is

E (g(Y )) = E (5Y + 50) = 5E (Y ) + 50 = 5

(

28

9

)

+ 50 =


(f) Standard deviation of function g(y). The standard deviation in the cost ofthe call, if g(y) = 5y + 50, is

σ =√

V (5Y + 50) =√

52V (Y ) = 5√σ2 = 5

√

26

81=


(g) Tchebysheff. According to Tchebysheff’s theorem, there is at least a 1− 1k2

chance cost of cell phone calls should fall within the range µ ± kσ. Forexample, there is at least a probability 1− 1

22= 0.75 costs fall in the range

µ± kσ ≈ 65.56± 2(2.83) =

(i) (59.9, 68.2) (ii) (59.9, 71.2) (iii) (62.9, 71.2) (iv) (63.9, 73.2).


2. Another example.Let random variable Y have the following probability density,

f(y) =

y, 0 < y < 1,2− y, 1 ≤ y < 2,0, elsewhere.

(a) Expected value.

µ = E(Y ) =

∫ ∞

−∞

yf(y) dy

=

∫ 1

0

y (y) dy +

∫ 2

1

y (2− y) dy

=

∫ 1

0

y2 dy +

∫ 2

1

(

2y − y2)

dy

=

[

y3

3

]1

0

+

[

2y2

2− y3

3

]2

1

=

(

13

3− 03

3

)

+

(

22 − 23

3

)

−(

12 − 13

3

)

=


(b) E (Y 2).

E(

Y 2)

=

∫ 1

0

y2 (y) dy +

∫ 2

1

y2 (2− y) dy

=

∫ 1

0

y3 dy +

∫ 2

1

(

2y2 − y3)

dy

=

[

y4

4

]1

0

+

[

2y3

3− y4

4

]2

1

=

(

14

4− 04

4

)

+

(

2(2)3

3− 24

4

)

−(

2(1)3

3− 14

4

)

=

(choose one) (i) 4

6(ii) 5

6(iii) 6

6(iv) 7

6.

(c) Variance.

σ2 = V (Y ) = E(

Y 2)

− µ2 =7

6− 12 =

(choose one) (i) 1

3(ii) 1

4(iii) 1

5(iv) 1

6.

(d) Standard deviation.

σ =√σ2 =

√

1

6≈


Section 4. The Uniform Probability Distribution (ATTENDANCE 6) 105

(e) Expected value of function g(y). If g1(y) = 5y + 50 and g2(y) = −3y,

E (g1(Y ) + g2(Y )) = E (g1(Y )) + E (g2(Y ))

= E (5Y + 50) + E (−3Y )

= 5E (Y ) + 50− 3E (Y )

= 2E (Y ) + 50

= 2 (1) + 50 =


4.4 The Uniform Probability Distribution

The continuous uniform distribution of random variable Y , defined on the interval(θ1, θ2), has density

5

f(y) =

{

1θ2−θ1

, θ1 ≤ y ≤ θ2,

0, elsewhere,

distribution function,

F (y) =

0 y ≤ θ1,y−θ1θ2−θ1

θ1 < y < θ2,

1 y ≥ θ2,

and its expected value (mean), variance and standard deviation are,

µ = E(Y ) =θ1 + θ2

2, σ2 = V (Y ) =

(θ2 − θ1)2

12, σ =

√

V (Y ).

Exercise 4.4 (The Uniform Probability Distribution)

1. Potatoes. An automated process fills one bag after another with Idaho potatoes.Although each filled bag should weigh 50 pounds, in fact, because of the differingshapes and weights of each potato, each bag weighs anywhere from 49 poundsto 51 pounds, with the following uniform density:

f(y) =

{

0.5, 49 < y ≤ 51,0, elsewhere.

(a) Since θ1 = 49 and θ2 = 51, the distribution is

F (y) =

0 y < 49,y−4951−49

49 ≤ y < 51,

1 y ≥ 52,

and so graphs of density and distribution are given in Figure 4.4.(i) True (ii) False

5The constants θ1 and θ2 are called parameters.


0 49 50 51

density f(y)

0 49 50 51

distribution F(y)

0.25

0.50

0.75

1

0.25

0.50

0.75

1


(b) The chance the bags weight between 49.5 and 51 pounds isP (49.5 < Y < 51) = F (51)− F (49.5) = 51−49

51−49− 49.5−49

51−49=

(choose one) (i) 0.25 (ii) 0.50 (iii) 0.75 (iv) 1.

(c) P (Y > 49.5) = 1− P (Y ≤ 49.5) = 1− F (49.5) = 1− 49.5−4951−49

=(choose one) (i) 0.25 (ii) 0.50 (iii) 0.75 (iv) 1.

(d) P (Y ≥ 49.5) = 1− P (Y < 49.5) = 1− F (49.5) = 1− 49.5−4951−49

=(choose one) (i) 0.25 (ii) 0.50 (iii) 0.75 (iv) 1.

(e) P (Y ≤ 49.5|Y < 50.5) = P (Y≤49.5, Y <50.5)P (Y <50.5)

= P (Y≤49.5)P (Y <50.5)

= F (49.5)F (50.5)

=49.5−49

51−49

50.5−49

51−49

=

(choose one) (i) 1

2(ii) 1

3(iii) 1

4(iv) 1

5.

(f) Mean. What is the mean weight of a bag of potatoes?

µ = E(Y ) =θ1 + θ2

2=

49 + 51

2=

(i) 49 (ii) 50 (iii) 51 (iv) 52.

(g) Standard deviation. What is the standard deviation in the weight of a bagof potatoes?

σ =

√

(θ2 − θ1)2

12=

√

(51− 49)2

12=

(i) 0.44 (ii) 0.51 (iii) 0.55 (iv) 0.58.

(h) Standard deviation of cost. If bags cost $0.05 per pound to fill each bag,what is the standard deviation in cost of filling the bags?

σ =√

V (0.05Y ) =√

0.052V (Y ) = 0.05√σ2 = 0.05σ ≈ 0.05(0.58) =


Section 4. The Uniform Probability Distribution (ATTENDANCE 6) 107

(i) Percentile. What is the median, or, in other words, the 50th percentileweight of bag, P (Y ≤ φ0.50) = 0.50? Since

P (Y ≤ φ0.50) = F (φ0.50) =φ0.50 − θ1

θ2 − θ1= 0.50,

then, rearranging, φ0.50 = (choose one)(i) θ1+θ2

2(ii) θ1+θ2

3(iii) θ1+θ2

4(iv) θ1+θ2

5.

In other words, the median, φ0.50, equals the mean, µ, in this case.

2. Another example. Consider the following uniform density:

f(y) =

{

k, −4 < y ≤ 8,0, elsewhere,

where k is an unknown constant.

(a) What is k? Since the uniform has nonzero probability defined in the range−4 ≤ y < 8 with length 8 − (−4) = 12, and rectangular area under anyuniform must be equal to 1, k = (choose one)(i) 1

11(ii) 1

12(iii) 1

13(iv) 1

14.

(b) In other words,

f(y) =

{

112, −4 < y ≤ 8,

0, elsewhere.

(i) True (ii) False

(c) Since θ1 = −4 and θ2 = 8, the distribution isF (y) = 0, y < −4, F (y) = 1, y ≥ 8 and for −4 ≤ y < 8, F (y) =(choose one) (i) y−4

8−4(ii) y+4

8−4(iii) y−4

8+4(iv) y+4

8+4.

(d) P (−7 < Y < 0) = F (0)− F (−7),where since −4 ≤ y = 0 < 8, F (0) = 0+4

8+4= 4

12,

and since y = −7 < −4, F (−7) = 0,and so P (−7 < Y < 0) = F (0)− F (−7) = 4

12− 0 =

(choose one) (i) 1

12(ii) 2

12(iii) 3

12(iv) 4

12.

(e) P (1 < Y < 10) = F (10)− F (1),where since y = 10 > 8, F (10) = 1,and since −4 ≤ y = 1 < 8, F (1) = 1+4

8+4= 5

12,

and so P (1 < Y < 10) = F (10)− F (1) = 1− 512

=(choose one) (i) 7

12(ii) 8

12(iii) 9

12(iv) 10

12.

(f) Mean.

µ = E(Y ) =θ1 + θ2

2=

−4 + 8

2=

(i) 1 (ii) 2 (iii) 3 (iv) 4.


(g) Standard deviation.

σ =

√

(θ2 − θ1)2

12=

√

(8− (−4))2

12=

(i) 1.93 (ii) 2.69 (iii) 3.46 (iv) 4.33.

4.5 The Normal Probability Distribution

The continuous normal distribution of random variable Y , defined on the interval(−∞,∞), has density6 with parameters µ and σ,

f(y) =1

σ√2π

e−(1/2)[(y−µ)/σ]2

and its expected value (mean), variance and standard deviation are,

E(Y ) = µ, V (Y ) = σ2, σ =√

V (Y ).

A normal random variable, Y , may be transformed to a standard normal, Z,

f(z) =1√2π

e−y2/2,

where µ = 0 and σ = 1 using the following equation,

Z =Y − µ

σ.

Exercise 4.5 (The Normal Probability Distribution)

1. Nonstandard normal: IQ scores.It has been found that IQ scores, Y , can be distributed by a normal distribution.Densities of IQ scores for 16 year olds, Y1, and 20 year olds, Y2, are given by

f(y1) =1

16√2π

e−(1/2)[(y−100)/16]2 ,

f(y2) =1

20√2π

e−(1/2)[(y−120)/20]2 .

A graph of these two densities is given in Figure 4.5.

6The distribution of this density does not have a closed–form expression and so must be solvedusing numerical integration methods. We will be using the TI–84+ calculator, rather than tableswhich give approximate numerical answers or the applet suggest in the text.

Section 5. The Normal Probability Distribution (ATTENDANCE 6) 109

f(x)

f(x)

x

20 year old IQs

16 year old IQs

µ = 100 µ = 120

σ = 20

σ = 16

Figure 4.6: Normal distributions: IQ scores

(a) Mean IQ score for 20 year olds isµ = (choose one) (i) 100 (ii) 120 (iii) 124 (iv) 136.

(b) Average (or mean) IQ scores for 16 year olds isµ = (choose one) (i) 100 (ii) 120 (iii) 124 (iv) 136.

(c) Standard deviation in IQ scores for 20 year oldsσ = (choose one) (i) 16 (ii) 20 (iii) 24 (iv) 36.

(d) Standard deviation in IQ scores for 16 year olds isσ = (choose one) (i) 16 (ii) 20 (iii) 24 (iv) 36.

(e) Normal density for 20 year old IQ scores is (choose one)(i) broader than normal density for 16 year old IQ scores.(ii) as wide as normal density for 16 year old IQ scores.(iii) narrower than normal density for 16 year old IQ scores.

(f) Normal density for the 20 year old IQ scores is (choose one)(i) shorter than normal density for 16 year old IQ scores.(ii) as tall as normal density for 16 year old IQ scores.(iii) taller than normal density for 16 year old IQ scores.

(g) Total area (probability) under normal density for 20’s IQ scores is(i) smaller than area under normal density for 16’s IQ scores.(ii) the same as area under normal density for 16’s IQ scores.(iii) larger than area under normal density for 16’s IQ scores.

(h) Number of different normal densities: (choose one)(i) one (ii) two (iii) three (iv) infinity.

2. Percentages: IQ scores.


Densities of IQ scores for 16 year olds, Y1, and 20 year olds, Y2, are given by

f(y1) =1

16√2π

e−(1/2)[(y−100)/16]2 ,

f(y2) =1

20√2π

e−(1/2)[(y−120)/20]2 .

(a) For the sixteen year old normal distribution, where µ = 100 and σ = 16,

P (Y1 < 84) =

∫ 84

−∞

1

16√2π

e−(1/2)[(y−100)/16]2 dy1 ≈

(choose one) (i) −0.1587 (ii) 0.1587 (iii) 0.3587 (iv) 0.8413.(2nd DISTR 2:normalcdf(− 2nd EE 99, 84, 100, 16)7.

(b) P (Y1 < 100) = (choose one)(i) 0.4413 (ii) 0.5000 (iii) 0.6587 (iv) 0.8413.(2nd DISTR 2:normalcdf(− 2nd EE 99, 100, 100, 16).)

(c) P (84 < Y1 < 100) = (choose one)(i) 0.3413 (ii) 0.4901 (iii) 0.5587 (iv) 0.7413.(2nd DISTR 2:normalcdf(84, 100, 100, 16).)

(d) For the twenty year old normal distribution, where µ = 120 and σ = 20,P (84 < Y2 < 100) = (choose one)(i) 0.0413 (ii) 0.1227 (iii) 0.3597 (iv) 0.5413.(2nd DISTR 2:normalcdf(84, 100, 120, 20).)

(e) Consider the following table of probabilities and possible values of proba-bilities. Use Figure 4.6.

Column I Column II

(a) P (Y1 > 84) ≈ (a) 0.4931(2nd DISTR 2:normalcdf(84, 2nd EE 99, 100, 16))

(b) P (96 < Y1 < 120) ≈ (b) 0.9641(2nd DISTR 2:normalcdf(96, 120, 100, 16))

(c) P (Y2 > 84) ≈ (c) 0.8413(2nd DISTR 2:normalcdf(84, 2nd EE 99, 120, 20))

(d) P (96 < Y2 < 120) ≈ (d) 0.3849(2nd DISTR 2:normalcdf(96, 120, 120, 20))

Using your calculator and the figure above, match the four items in columnI with the items in column II.

7The normalcdf function has four arguments: normalcdf( low, high, µ, σ). In this case, the “low”number is “− 2nd EE 99” and approximates negative infinity. The “high” number is 84. Finally,this is a nonstandard normal, where the µ and σ are 100 and 16, respectively.


f(x)

x100 84

(a)

N(100,16 )P(X > 84) = ? f(x)

f(x)f(x)

x100 96 120

(b)

(d)

P(96 < X < 120) = ?

x12096

N(120,20 )P(96 < X < 120) = ?

(c)

x120 84

N(120,20 )P(X > 84) = ?

N(100,16 )2

2

2 2

Figure 4.7: Normal probabilities: IQ scores

Column I (a) (b) (c) (d)Column II

3. Percentiles: IQ scores.Densities of IQ scores for 16 year olds, Y1, and 20 year olds, Y2, are given by

f(y1) =1

16√2π

e−(1/2)[(y−100)/16]2 ,

f(y2) =1

20√2π

e−(1/2)[(y−120)/20]2 .

(a) The 75th percentile for 16s, where µ = 100 and σ = 16 is φ0.75 where

P (Y1 < φ0.75) =

∫ φ0.75

−∞

1

16√2π

e−(1/2)[(y−100)/16]2 dy1 = 0.75.

In other words, φ0.75 ≈ (choose one)(i) 103.5 (ii) 106.7 (iii) 110.8 (iv) 112.3.(2nd DISTR 3:invNorm(0.75, 100, 16).)

(b) If P (Y1 < φ0.35) = 0.35, then φ0.35 ≈ (circle one)(i) 83.5 (ii) 92.5 (iii) 93.8 (iv) 112.3.(2nd DISTR 3:invNorm(0.32, 100, 16).)

(c) If P (Y2 < φ0.75) = 0.75, then φ0.75 ≈ (circle one)(i) 106.5 (ii) 133.5 (iii) 125.4 (iv) 142.3.(2nd DISTR 3:invNorm(0.75, 120, 20).)


f(x)

x100 16th percentile?

(a)

N(100,256)

f(x)

f(x)f(x)

x100

89th percentile?(b)

(d)

x120

N(120,400)

(c)

x120

N(120,400)

N(100,256)

16th percentile?

0.16

0.89

0.16

89th percentile?

0.89

Figure 4.8: Normal percentiles: IQ scores

(d) Consider the following table of percentiles and possible values of per-centiles. Use Figure 4.7.

Column I Column II

(a) If P (Y1 < φ0.16) = 0.16, then φ0.16 ≈ (a) 119.6(2nd DISTR 3:invNorm(0.16, 100, 16))

(b) If P (Y1 < φ0.89) = 0.89, then φ0.89 ≈ (b) 84.1(2nd DISTR 3:invNorm(0.89, 100, 16))

(c) If P (Y2 < φ0.16) = 0.16, then φ0.16 ≈ (c) 144.5(2nd DISTR 3:invNorm(0.16, 120, 20))

(d) If P (Y2 < φ0.89) = 0.89, then φ0.89 ≈ (d) 100.1(2nd DISTR 3:invNorm(0.89, 120, 20))

Using your calculator and the figure above, match the four items in columnI with the items in column II.

Column I (a) (b) (c) (d)Column II

(e) Since the normal distribution for the sixteen year olds is symmetric, cen-tered at 100, the 50th percentile must be(circle one) (i) below 100 (ii) equal to 100 (iii) above 100.

(f) The 75th percentile is that IQ score such that there is a 75% chance of theIQ scores being below this IQ score and so a 25% chance of the IQ scoresbeing above this IQ score.(i) True (ii) False


(g) The 75th percentile for the sixteen year olds must be(circle one) (i) below 100 (ii) equal to 100 (iii) above 100.

(h) The 75th percentile for the twenty year olds must be(circle one) (i) below 120 (ii) equal to 120 (iii) above 120.

4. Standard normal.Normal densities of IQ scores for 16 year olds, Y1, and 20 year olds, Y2, aregiven by

f(y1) =1

16√2π

e−(1/2)[(y−100)/16]2 ,

f(y2) =1

20√2π

e−(1/2)[(y−120)/20]2 .

Both densities may be transformed to a standard normal with µ = 0 and σ = 1using the following equation,

Z =Y − µ

σ.

60 80 100 120 140 160 180 X, nonstandard

-3 -2 -1 0 1 2 3 Z, standard

52 68 84 100 116 132 148 X, nonstandard

-3 -2 -1 0 1 2 3 Z, standard

16 year olds

mean 100, SD 16

20 year olds

mean 120, SD 20

110

Figure 4.9: Standard normal and (nonstandard) normal

(a) Since µ = 100 and σ = 16, a 16 year old who has an IQ of 132 isz = 132−100

16= (choose one) (i) 0 (ii) 1 (iii) 2 (iv) 3

standard deviations above the mean IQ, µ = 100.

(b) A 16 year old who has an IQ of 84 isz = 84−100

16= (choose one) (i) −2 (ii) −1 (iii) 0 (iv) 1

standard deviations below the mean IQ, µ = 100.


(c) Since µ = 120 and σ = 20, a 20 year old who has an IQ of 180 isz = 180−120

20= (choose one) (i) 0 (ii) 1 (iii) 2 (iv) 3

standard deviations above the mean IQ, µ = 120.

(d) A 20 year old who has an IQ of 100 isz = 100−120

20= (choose one) (i) −3 (ii) −2 (iii) −1 (iv) 0

standard deviations below the mean IQ, µ = 120.

(e) Although both the 20 year old and 16 year old scored the same, 110, onan IQ test, the 16 year old is clearly brighter relative to his/her age groupthan is the 20 year old relative his/her age group because

z1 =110− 100

16= 0.625 > z2 =

110− 120

20= −0.5.

(i) True (ii) False

(f) The probability a 20 year old has an IQ greater than 90 is

P (Y2 > 90) = P

(

Z2 >90− 120

20

)

= P (Z2 > −1.5) =

(i) 0.93 (ii) 0.95 (iii) 0.97 (iv) 0.99.(nonstandard (x): 2nd DISTR 2:normalcdf(90, 2nd EE 99, 120, 20)

or standard (z): 2nd DISTR 2:normalcdf( 90−120

20, 2nd EE 99, 0, 1).)

(g) The probability a 20 year old has an IQ between 125 and 135 is

P (125 < Y2 < 135) = P

(

125− 120

20< Z2 <

135− 120

20

)

= P (0.25 < Z2 < 0.75) =

(i) 0.13 (ii) 0.17 (iii) 0.27 (iv) 0.31.(nonstandard (x): 2nd DISTR 2:normalcdf(125, 135, 120, 20)

or standard (z): 2nd DISTR 2:normalcdf(

125−120

20, 135−120

20, 0, 1

)

.)

(h) If a normal random variable Y with mean µ and standard deviation σ

can be transformed to a standard one Z with mean µ = 0 and standarddeviation σ = 1 using

Z =Y − µ

σ,

then Z can be transformed to Y using

Y = µ+ σZ.

(i) True (ii) False

(i) A 16 year old who has an IQ which is three (3) standards above the meanIQ has an IQ of y1 = 100 + 3(16) = (choose one)(i) 116 (ii) 125 (iii) 132 (iv) 148.


(j) A 20 year old who has an IQ which is two (2) standards below the meanIQ has an IQ of y2 = 120− 2(20) = (choose one)(i) 60 (ii) 80 (iii) 100 (iv) 110.

(k) A 20 year old who has an IQ which is 1.5 standards below the mean IQhas an IQ of y2 = 120− 1.5(20) = (choose one)(i) 60 (ii) 80 (iii) 90 (iv) 95.

(l) The probability a 20 year old has an IQ greater than one (1) standarddeviation above the mean is

P (Z2 > 1) = P (Y2 > 120 + 1(20)) = P (Y2 > 140) =

(choose one) (i) 0.11 (ii) 0.13 (iii) 0.16 (iv) 0.18.(standard (z): 2nd DISTR 2:normalcdf(1, 2nd EE 99, 0, 1)

or nonstandard (x): 2nd DISTR 2:normalcdf(120 + 1(20), 2nd EE 99, 120, 20).)

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