Statistik topic8 special probability distribution

INTRODUCTION

In the last topic, we have learnt about probability function p(x) for discrete distribution, and f(x) for continuous distribution. We mentioned there that p(x) and f(x) is of any algebraic expressions which comply with the respective probability rules mentioned in last topic. However, in this topic we will discuss a special expression of p(x) for binomial distribution and a special expression of f(x) for normal distribution. Binomial distribution is an example of discrete distribution and normal distribution is a continuous distribution.

BINOMIAL DISTRIBUTION

The binomial distribution is one of the most commonly used discreteprobability distribution. It is used to obtain the exact probability of X successes in n repeated trials of a binomial experiment

8.1

TTooppiicc 88 SpecialProbabilityDistribution

LEARNING OUTCOMES

By the end of this topic, you should be able to:

1. use binomial distribution to solve binomial problem;

2. apply table of binomial distribution to calculate exact probability in binomial experiment;

3. formulate normal distribution to standard z-score; and

4. apply table of standard normal distribution to calculate probability.

TOPIC 8 SPECIAL PROBABILITY DISTRIBUTION 124

The trial in this experiment has only two outcomes which complementing each other. As an example, a trial of tossing of a Malaysian coin might land picture (G) or number (N). A student taking a final examination will either pass or fail.

Another feature of trial is that its results can be reduced to two outcomes which complement each other. For example, we are interested of getting even number in the trial of throwing a fair dice. The results can be reduced to getting event E (2 or 4 or 6) as one outcome, and event O (1, or, 3 or 5) as another outcome. These events E and O complement to each other.

8.1.1 Binomial Experiment

Binomial experiment is a probability experiment consisting of repetition of Bernoulli trial. Conducting a urine test on 50 students is an example of binomial experiment. In this experiment, the Bernoulli trial is the process of giving urine test on each student whose outcome is “positive” or “negative” result. The binomial experiment here is the repetition of the urine test 50 times.

The experiment of throwing fair dice 30 times where the interest of researcher is of getting even numbers on each throw is another example of binomial experiment. The Bernoulli trial here is “throwing the dice” whose outcomes can be reduced to getting event E i.e. {2,4,6} as one outcome and event O i.e. {1,3,5} as another outcome. This trial is repeated 30 times.

A trial which possess ONLY two outcomes one complementing each other is categorised as Bernoulli trial.

(i) Give two examples of Bernoulli trial which has ONLY two outcomes. For each trial, explain its outcomes and state how they are complementing each other.

(ii) Give one example of Bernoulli trial which has two reduced outcomes. For each trial, explain its outcomes and state how they are complementing each other.


Binomial Requirements

The binomial experiment should comply with the following requirements:

Sometimes, researcher tend to represent event success by using code “1” and event failure by code “0” The Pr(Success) = p, and Pr(Failure) = q = 1 – p. For example in one of the above examples, let event E, getting even number become the success, then p = Pr(E) = 0.5, and q = 0.5. Let digit ‘1’ represent success in a Bernoulli trial which has been repeated 10 times. For each trial, when event success occurs, a digit 1 is recorded, otherwise digit 0 is recorded. Then the following are example outcome of binomial experiment. It consists of strings of outcomes of repeated Bernoulli trial.

(a) 1010011101, this outcome of binomial experiment produces 6 successes, and 4 failures; with probability

Pr(1010011101) = Pr(6 successes and 4 failures)

= Pr(6 successes ) Pr(4 failures) = p6 q4

(b) 0111001010, this outcome of binomial experiment produces 5 successes,

and 5 failures, with probability

Pr(1010011101) = Pr(5 successes and 5 failures)

= Pr(5 successes ) Pr(5 failures) = p5 q5

However, for any x number of successes in n repetition Bernoulli trial, there are

xn

, or n choose x possible outcomes of binomial experiment each has a

combination of success and failures as given in (a) and (b) above.

It is not impossible to have all non-successful in a binomial experiment, which is represented by the string 0000000000 that consist of 10 failures. Likewise, it also

Each trial should have only two outcomes or outcomes that can be reduced to two outcomes. These outcomes which complement to each other can be considered as either success or failure. This trial is categorised as Bernoulli trial.

There must be a fixed number of repetitions, n of such trial.

The outcome of each trial must be independent of each other.


possible to get full string of 10 successes i.e. 1111111111. Thus, the number of success in a binomial experiment can range from zero to a maximum of nsuccesses. The outcomes of binomial experiment and their corresponding probabilities generate binomial distribution.

8.1.2 Binomial Probability Function

Let Y be a discrete random variable representing the total number of success in a binomial experiment with n repetition of Bernoulli trial. Then the probability of ysuccesses in n repetitions of Bernoulli trial is given by:

Formula 8.1

Where

Formula 8.1(a)

Then we say that Y follow binomial distribution written as Y b(y; n, p).

The probability of getting y successes in n repetition of Bernoulli trial can be obtained via Formula 8.1 or using table of binomial cumulative probability distribution.

y! (n-y)!n! n! = n (n–1) (n–2) … 2 1

yn

and,

(1–p)n-y,

others0

;...,,1,0 ny =pyn

P (y)y

ACTIVITY 8.1

1. Give two examples of binomial experiment; one for trial having two outcomes and one example for trial with reduced outcomes.

2. For each binomial trial, give samples of outcomes and its possible probability in term of p and q.


Example 8.1

Consider an experiment of throwing unbiased dice 10 times. Find the probability of obtaining even numbers six times.

Solution

The interested event in this experiment is: E = even numbers = {2,4,6}, The trial is: throwing an unbiased dice Trial outcomes: {E, O}, a Bernoulli trial with Pr(Success) = p = Pr(E) = 0.5. Here O stands for odd numbers. This trial is repeated 10 times to generate binomial experiment. Total umber of repetition = n = 10; Let Y be the number of successes in the experiment = Y b(y; 10, 0.5). Pr(getting 6 successes) is given by Using Formula 8.1 as

Pr(Y = 6) = p(6) = 46 )5.0()5.0(6

10 6 410! (0.5) (0.5)6! 4!

= 0.20508 0.2

About 20% of ten throws will result in even numbers.

The Mean and Variance of Binomial Distribution The mean of binomial distribution is given by:

pn

Formula 8.2

The variance is given by:

)1(2 pnp

Formula 8.3

8.1.3 Probability Table of Binomial Distribution

We are referring to the Table of Binomial Distribution published by OUM. In this table Y is referred as binomial discrete random variable. The number of success is represented by r, and Pr(success) = . For each n, the column of r is ranging


from 0 to n. The content of the table is accumulating probabilities of getting 0 until r successes written in probability language as Pr(Y r). It is given by the following formula:

Formula 8.4

For example, Let Y is from binomial distribution with n = 10, p = = 0.2. Find the probability of the following events:(a) Obtain 3 or less successes. (b) Obtain more than 3 successes. (c) Obtain exactly 4 successes.

Refer to page 5 of the table published by OUM, look at the block on column n = 10 (See Table 8.1) below. (a) Pr(Y 3) = 0.8791. Look at row r = 3, and column = 0.2. (b) By complement, Pr( Y > 3) = 1 - Pr(Y 3) = 1 – 0.8791 = 0.1209 (c) Pr(r = 4) = Pr(Y 4) - Pr(Y 3) = 0.9672 – 0.8791 = 0.0881(The answer

obtained by Formula 8.1 is 0.08808).

Table 8.1: Binomial Distribution

n r 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.5010 0

123456789

0.59870.91390.98850.99900.99991.1.1.1.1.

0.34870.73610.92980.98720.99840.99991.1.1.1.

0.19690.54430.82020.95000.99010.99860.99991.1.1.

0.10740.37580.67780.87910.96720.99360.99910.99991.1.

0.05630.24400.52560.77590.92190.98030.99650.99961.1.

0.02820.14930.38280.64960.84970.95270.98940.99840.99991.

0.01350.08600.26160.51380.75150.90510.97400.99520.99951.

0.00600.04640.16730.38230.63310.83380.94520.98770.99830.9999

0.00250.02330.09960.26600.50440.73840.89800.97260.99550.9997

0.00100.01070.05470.17190.37700.62300.82810.94530.98930.9990

11 0 12

0.56880.89810.9848

0.31380.69740.9104

0.16730.49220.7788

0.08590.32210.6174

0.04220.19710.4552

0.01980.11300.3127

0.00880.06060.2001

0.00360.03020.1189

0.00140.01390.0652

0.00050.00590.0327

P (r) = Pr (Y r) = p (y)r

y=0


Probabilities RULES using Table of Discrete Distribution

In general, we have the following properties in getting probabilities using table:

Rule 5

Pr (Y < r) = Pr (Y r –1) = P (r –1)

Rule 6

P(r) = Pr (Y = r) = Pr (Y r) = P (r –1) – Pr (Y r–1) = P(r) – P (r –1)

Rule 7

Pr (Y > r) = 1 – Pr (Y r) = 1 – P(r)

Rule 8

Pr (Y r) = 1 –Pr (Y < r) = 1 –Pr (Y r –1) = 1 –P(r –1)

In this table Greek letter (red theta) has been used for the probability of success which ranges from 0 to 0.5.

Example 8.2

Let Y is from binomial distribution with n = 6, p = = 0.2. Find the probability of the following events:

(a) Obtain 2 or less successes

(b) Obtain less than 2 successes

(c) Obtain exactly 2 successes.


Solution

Refer to page 5 of the binomial table published by OUM, with p = = 0.35, n = 6.

(a) Pr (Y 2) = 0.6471, Refer at block n = 6, row r = 2; column = 0.35.

(b) By Rule 5, Pr( r< 2) = Pr(r 1), since r discrete;

Pr (r< 2) = Pr(r 1) = 0.3191. Refer at block n = 6, row r = 1,column = 0.35.

(c) By Rule 6; Pr( r = 2) = Pr ( r 2) – Pr ( r 1) = 0.6471 – 0.3191 = 0.3280.

Using Table by Complementary Method when > 0.5

The table in the book is prepared for probability of success 0.5 or less. In practice, we may have probability of success greater than 0.5. For > 0.5, a complementary method has to be used as follows:

When r n, and > 0.5, then the followings are true:

(a) Event {Y = r successes} is equivalent to Event {X = (n – r) failures}

Where Y b(y; n, ), and X b(x; n, 1- ). Thus,

Pr(Y = r) Pr(X = n - r), Y success, X failures

Formula 8.5

For example, let Y b(y; 6, = 0.8), and find Pr(Y = 4).

Define X b(x; 6, 1- = 0.2), and X is number of failures. Then from table page 5, and Rule 6, we have:

Pr(Y = 4) Pr(X = 6 - 4) = Pr (X = 2) = Pr(r 2) – Pr(r 1) = 0.9011 – 0.6554 = 0.2457

(b) Event {Y r successes} is equivalent to Event {X (n – r ) failures}

where Y b(y; 6, ), and X b(x; 6, 1- ). Thus,

Pr(Y r ) Pr{X (n – r )}= 1- Pr( X < n – r)

= 1- Pr( X n – r -1) for discrete X.

Formula 8.6


For example, let Y b(y; 6, = 0.8), and find Pr(Y 2 ).

Define X b(x; 6, 1- = 0.2).

Pr(Y 2) Pr{X (6 – 2)} = 1- Pr( X < 6 – 2) = 1- Pr( X 6 – 2 -1)

= 1 - Pr(X 3) = 1 - 0.9830 = 0.0170

(By the exact Formula 8.1, the answer is 0.01696).

Example 8.3

Let Y be a binomial distribution with n = 6, p = = 0.6. Find the probability of obtaining 3 or less successes.Solution

Define X b(x; 6, 1- = 0.4).

By using Formula 8.6, we have

Pr(Y 3) Pr{X (6 – 3)} = 1- Pr(X < 6 – 3) = 1- Pr(X 6 – 3 - 1)

= 1- 0.5443 = 0.4557.


ACTIVITY 8.2

1. Compute the probability of Y successes using exact probability Formula 8.1. (a) n = 5, y = 3; p = 0.1 (b) n = 8, y = 5; p = 0.4 (c) n = 7, y = 4; p = 0.7

2. Compute the probability of Y successes in Question 1 by using table of binomial distribution published by OUM. Make a comparison of their values.

3. It is found that 40% of the first year students are using learnerstudy system in one semester. Find the probability in a sample of10 students, exactly 5 of which use learner study system.

4. Given that Y~ b(y; 4, 0.4) by using table of binomial distribution, find the probabilities of the following events: (a) (Y< 2), (c) (Y > 2) (b) (Y = 2), (d) (Y 2)

5. Given that Y~ b(y; 4, 0.65) by using table of binomial distribution, find the probabilities of the following events: (a) (Y< 2), (b) (Y = 2) (c) (Y 2)

6. A student answers 8 questions of MCQ type. Each question has 5 answers with only 1 correct answer. Compute the probability the student obtaining 4 correct questions.

7. It is known that only 60% of a defected computer can be repaired. A sample of 8 computers is selected randomly, find the probability of(a) At most 3 computers can be repaired. (b) 5 or less can be repaired (c) None can be repaired


NORMAL DISTRIBUTION

It is a continuous distribution having the following properties:

(a) Its probability curve is bell-shaped.

(b) Its mean, mode, and median are equal and located at the centre of the distribution.

(c) It is a unimodal distribution.

(d) The curve is symmetrical about the mean. Thus it has same shape on both sides of a vertical line drawn through the centre.

(e) The curve never touches the horizontal axis or mathematically it touches at infinity.

(f) The total area under the curve is equal to 1.0.

(g) The area under the normal curve that lies within one standard deviation of the mean is 0.68 (or 68%); within two standard deviations of the mean is 0.95 (or 95%); within three standard deviations of the mean is 0.997 (or 99.7%). See Figure 8.1

(h) The normal distribution has two parameters known as its mean and variance 2 .

(i) Continuous random variable X from normal distribution with mean andvariance 2 is denoted by X N( , 2 ).

Figure 8.1: Areas under the normal distribution curve

8.2


(j) Figure 8.2, shows some other properties. Let X N ( 1 , 21 ), and X

N( 2 , 22 ). If 2 greater than 1 , then normal curve 2 is located on the

right of the normal curve 1. If 21 less than 2

2 , then normal curve 2 is more spread than normal curve 1.

Figure 8.2: Some other properties, comparison of parameters


8.2.1 Standard Normal Distribution

Let X be a continuous random variable from a normal distribution N( , 2), then X can be transformed to Z score of standard normal distribution by the following Formula:

XZ

Formula 8.7

The random score Z is said to have standard normal distribution with mean 0 and a known variance 1. Standard normal curve preserves the same normal properties. Figure 8.3 depicts the above transformation.

Figure 8.3: Transformation of X to Z score

ACTIVITY 8.3

Using the same X-Y axes, sketch the following normal curves:

N(10, 2), N(10,4), N(10, 16), N(20, 2), N(20, 9)


Table of Standard Normal Curve N(0, 1)

With the above transformation, almost all probability problems can be solved easily by using standard normal table. In the table published by OUM, standard normal is given on page 23. It is developed based on cumulative distribution given in Formula 8.8.

z

duuzZzZz )()Pr()Pr()(

Formula 8.8

A portion of this table is shown in Table 8.2 where )(u is standard normal density function and )(z is the area on the left of z as shown in Figure 8.4. This formula is for positive z.

Figure 8.4: The area (z)

For example let random variable X come from normal distribution with mean 20 and variance 16. Find the Pr(X < 22). It is the area to the left of Z = (22-20)/4 = 0.5. So from the table on page 23 (see Table 8.1), look at row 0.50 and under column 0.00, the answer is 0.69146.


Table 8.2 z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.00 0.50000 0.50399 0.51197 0.51595 0.51994 0.52392 0.52790 0.53188 0.53188 0.53586 0.10 0.53983 0.54380 0.54776 0.55172 0.55567 0.55962 0.56356 0.56749 0.57142 0.57535 0.20 0.57926 0.58706 0.59095 0.59483 0.59871 0.60257 0.60642 0.61026 0.61026 0.61409 0.30 0.61791 0.62172 0.62552 0.62930 0.63307 0.63683 0.64058 0.64431 0.64803 0.65173 0.40 0.65542 0.65910 0.66276 0.66640 0.67003 0.67364 0.67724 0.68082 0.68439 0.68793

0.50 0.69146 0.69497 0.69847 0.70194 0.70540 0.70884 0.71226 0.71566 0.71904 0.72240 0.60 0.72575 0.72907 0.73237 0.73565 0.73891 0.74215 0.74537 0.74857 0.75175 0.75490 0.70 0.75804 0.76115 0.76424 0.76730 0.77035 0.77337 0.77637 0.77935 0.78230 0.78524 0.80 0.78814 0.79103 0.79389 0.79673 0.79955 0.80234 0.80511 0.80785 0.81057 0.81327 0.90 0.81594 0.81859 0.82121 0.82381 0.82639 0.82894 0.83147 0.83398 0.83646 0.83891

1.00 0.84134 0.84375 0.84614 0.84849 0.85083 0.85314 0.85543 0.85769 0.85993 0.86214 1.10 0.86433 0.86650 0.86864 0.87076 0.87286 0.87493 0.87698 0.87900 0.88100 0.88298 1.20 0.88493 0.88686 0.88877 0.89065 0.89251 0.89435 0.89617 0.89796 0.88730 0.90147 1.30 0.90320 0.90490 0.90658 0.90824 0.90988 0.91149 0.91308 0.91466 0.91621 0.91774 1.40 0.91924 0.92073 0.92220 0.92364 0.92507 0.92647 0.92785 0.92922 0.93056 0.93189

1.50 0.93319 0.93448 0.93574 0.93699 0.93822 0.93943 0.94062 0.94179 0.94295 0.94408 1.60 0.94520 0.94630 0.94738 0.94845 0.94950 0.95053 0.95154 0.95352 0.95352 0.95449 1.70 0.95543 0.95637 0.95728 0.95818 0.95907 0.95994 0.96080 0.96246 0.96246 0.96327 1.80 0.96407 0.96485 0.96562 0.96637 0.96712 0.96784 0.96856 0.96995 0.96995 0.97062

1.90 0.97128 0.97193 0.97257 0.97320 0.97381 0.97441 0.97500 0.97558 0.97615 0.97670

Area on the Left of Negative z

The area on the left of negative z can be obtained by symmetry property and using complement method as follows.

Pr (Z < – z) = Pr (Z > z) = 1 – (z)

Formula 8.9

For example let random variable X is coming from normal distribution with mean 20, and variance 16. Find the Pr(X < 18). It is the area to the left of Z = (18-20)/4 = -0.5. So from Formula 8.9, it is equivalent to the area on the right of Z = 0.5, and by complement the area is 1 – 0.69146 = 0.30854.

Area under the Normal Curve Lying between Two Values of z

Pr( a < Z < b) = (b) - (a)

Formula 8.10


It is the shaded area given in Figure 8.5.

Figure 8.5: Area between two values of z

Example 8.4

Using standard normal table, find the Pr(a<Z<b) for the following values of a and b.

(i) a =1.5, b = 2.55

(ii) a = –2.0, b = –1.5

(iii) a = –1.5, b = 1.5

Solution

Figure 8.6 depicts the necessary shaded area and from table we have:

(1.50) = 0.93319, (2.00) = 0.97725, (2.55) = 0.99461

From Formula 8.10, we have:

(i) Pr(1.50< Z<2.55) = (2.55) – (1.50)

= 0.06142


(ii) Pr (–2.00 < Z < –1.50) = (–1.50) - (–2.00),

= {1– (1.50)} – {1- (2.00)}

= 0.06681 – 0.02275 = 0.04406

(iii) Pr (–1.50 < Z < 1.50) = (1.50) – (–1.50),

= (1.50) – (1- 1.50)

= 0.93319 – 0.06681 = 0.86638

Example 8.5

Let a continuous random variable X follows normal distribution N(4.0, 16). Find

(a) Pr(X < 2).

(b) Pr(X > 4.6).

Solution

Use Formula 8.7 to get the corresponding z score. Then use Formula 8.8, 8.9, 8.10.

(a)4

4216

4Pr)2Pr( XX = )5.0Pr(Z =1 - )5.0(

= 1 – 0.69146 = 0.380854

(b)4

46.416

4Pr)6.4Pr( XX = )15.0Pr(Z =1 - )15.0(

= 1 – 0.55962 = 0.44038


Example 8.6

The long-distance calls made by executives of a private university are normally distributed with a mean 10 minutes and a standard deviation of 2.0 minutes. Find the probability that a call:

(a) lasts between 8 and 13 minutes

(b) lasts more than 9.0 minutes

(c) lasts less than 7 minutes.

Solution

Let random variable X represents duration of long-distance call and X N(10, 4).

(a)2

10132

102108Pr)138Pr( XX 5.10.1Pr Z

= (1.5) – (-1.0)

= (1.5) – {1 - (1.0)}

= (1.5) + (1.0) – 1 = 0.93319 + 0.84134 – 1 = 0.77453

(b)2109Pr)0.9Pr( ZX 5.0Pr Z

= (0.5) = 0.69146

(c)2107Pr)0.7Pr( ZX 5.1Pr Z

= 1 – (1.5) = 1 – 0.93319 = 0.06681


8.2.2 Application to Real Problems

Binomial and normal distributions are widely used in solving various day to day problems. If sample size is large, normal distribution can be used to replace binomial distribution in solving binomial problems. However, some numerical corrections have to be done.

Correction for Conversion Binomial Discrete Variable to Normal Continuous Variable

Let Y be discrete random variable from a binomial distribution b(y: n, p) with mean np and variance np(1-p). Then, the corresponding continuous X follows normal distribution with mean np and variance np(1-p). The following are necessary corrections when shifting discrete Y to continuous X. The normal distribution will give a good approximation to binomial, if n is large and np > 5; also np(1-p)> 5. The approximation is recommended because when n is large the numerical calculation using binomial function become tedious.

ACTIVITY 8.4

1. Let a continuous random variable X follows normal distribution N(4.0, 16). Find Pr(2 < X < 3)

2. Let a continuous random variable X follow normal distribution N(50.0, 4). Find(a) Pr( X < 45) (b) Pr(X>55) (c) Pr(45.0 < X < 55.0)

3. A continuous random variable X is normally distributed with mean 50 and standard deviation 5. What value of K is such that Pr(X < K) = 0.08.


Table 8.3

Discrete Variable Y; y is Observed Value Corrections in Continuous Variable X, y is the Same Observed Value for Discrete Y

(a) Y = y y - 0.5 < X < y + 0.5

(b) Y y X y + 0.5

(c) Y < y X < y – 0.5

(d) Y y X y – 0.5

(e) Y > y X > y + 0.5

Correspondingly, we have the equivalent probability statement in the following Table 8.4.

Table 8.4

Y b(y;n, p); y is Observed Value X N(np, np(1-p)); y is the Same Observation Value for Discrete Y

(a) Pr(Y = y) Pr(y - 0.5 < X < y + 0.5)

(b) Pr(Y y) Pr(X y + 0.5)

(c) Pr(Y < y) Pr(X < y – 0.5)

(d) Pr(Y y) Pr(X y – 0.5)

(e) Pr(Y > y) Pr(X > y + 0.5)

Example 8.7

Let Y from binomial distribution b(y; 100, 0.4). Find

(a) Pr(Y > 50).

(b) Pr(50< Y < 60)

Solution

Y is binomial with n = 100; mean = np = 40, variance = np(1-p) = 24. Define continuous variable X from normal N(40, 24). The observation value is y = 50.


(a) From conversion Table 8.4(e), we have

Pr(Y > 50) Pr(X > 50 + 0.5) = Pr(X > 50.5) = 1 – Pr(X 50.5)

= 1 – Pr(X 24

405.50 ) = 1 – (2.143)

= 1 - (2.14) 1 – 0.98382 = 0.01618.

(b) In this problem, Y is strictly greater than 50 and strictly less than 60, so use conversion Table 8.3 (c) & (e). Thus we have,

Pr( 50 < Y < 60) Pr( 50 + 0.5 < X < 60 – 0.5)

= 24

405.59 - 24

405.50 = (3.98) – (2.14)

= 0.99997 – 0.98382 = 0.01615

Correction Due to Rounding Error in Observation

Sometimes the observations are recorded after being round up or round down to the nearest integer, or figure. Thus correction is recommended as in the following example.

Example 8.8

Let the lifetime of electric bulb follows normal distribution with mean 1000 hours and standard deviation 100 hours. The observation of the lifetime is recorded to the nearest hours. An electric bulb is selected at random, find the probability that it will have lifetime between 850 hours and 1050 hours.

Solution

Let X be the lifetime of the electric bulb, then X N(1000, 1002 ).

100000,15.050,1

100000,15.849Pr)5.050,15.849Pr( ZX

)505.1()505.0(

= 0.69497 + 0.93448 – 1 = 0.62945


Two types of distributions have been introduced. A binomial distribution is of discrete type and normal distribution of continuous type. It has been clearly described the difference between these two types of distributions. The binomial distribution is concerning exact probability whose values can be obtained through Formula 8.1 or by using table of binomial distribution. For the normal distribution, probability value can best be obtained by using table of standard normal. As such, all normal problems can be transformed to standard normal via Formula 8.7 and then the standard normal table can be used to determine the probability. At the end of the topic, approximation to binomial by normal distribution has been introduced. However, some correction has to be done when converting discrete variable to continuous variable as given in Table 8.3 & 8.4.

ACTIVITY 8.5

1. The length of yard stick is said to follow normal distribution with mean 150mm, and standard deviation 10mm. The length is recorded to the nearest integer. A yard stick is taken at random:

(a) find the probability its length falls in the following intervals: (i) between 125mm and 155mm;

(ii) more than 180mm.

(b) If 500 such yard stick has been selected randomly, find the number of them who has lengths in the respective intervals given in (a).

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Statistik topic8 special probability distribution