SQQS1013 CHAPTER 5

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SQQS1013 Elementary Statistics

SPECI L

DISTRIBUTIONS

5.1 DISCRETE DISTRIBUTION

Consists of the values that a random variable can assume and the

corresponding probabilities of the values.

The probabilities are determined theoretically or by observation.

5.1.1 Binomial Probability Distribution

Binomial Probability Distribution concerns the outcomes of a binomialexperiment and the corresponding probabilities of these outcomes.

It is applied to find the probability that an outcome will occur x times in n

performances of experiment.

For example:

The probability of a defective laptop manufactured at a firm is 0.05, in

a random sample of ten.

The probability of 8 packages not arriving at its destination. To apply the binomial probability distribution, the random variable x must be

a discrete dichotomous random variable.

Each repetition of the experiment must result in one of two possible

outcomes.

Conditional of a Binomial Experiment

1. Each trial can have only two outcomes or outcomes that can be reduced to

two outcomes. These outcomes can be considered as either success or

failure.

2. There must be a fixed number of trials.

3. The outcomes of each trial must be independent of each other. In other

words, the outcome of one trial does not affect the outcome of another trial.

4. The probability of success is denoted by p and that of failure by q, and p +

q=1. The probabilities p and q remain constant for each trial.

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5.1.1.1 Calculating Binomial Probability

There are two methods to calculate the binomial probability:

a) Using Binomial Probability Distribution Formula

For a binomial experiment, the probability of exactly x successes in n trials

is given by the binomial formula:

P(x) =nCx p

xq

n-x

Where;

n = the total number of trialsp = probability of successq = 1-p = probability of failure

x = number of successes in n trialsn-x = number of failures in n trials

Compute the probabilities of X successes, using the binomial formula; n= 6, X= 3,p=0.03

Solution:

a) P(x=3) = 6C3 (0.03)3(1-0.03)6-3

=

6

C3 (0.03)

3

(0.97)

3

= 0.0005

A survey found that one out of five Malaysians says he or she has visited a doctor inany given month. If 10 people are selected at random, find the probability thatexactly 3 will have visited a doctor last month.

Solution:

Example 1

Example 2

FORMULA

The success does not mean that thecorresponding outcome is considered favorableor desirable and vice versa

The outcome to which the question refers is

called a success ; the outcome to which it doesnot refer is called a fa i lure .

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b) Using the Table of Binomial Probabilities

The probabilities for a binomial experiment can also be read from the table of

binomial probabilities.

For any number of trials n:

The binomial probability distribution is symmetric if p = 0.5 The binomial probability distribution is skewed to the right if p is less than

0.5 The binomial probability distribution is skewed to the left if p is greater than

0.5

Determining P (X 3) for n = 6 and p = 0.30

The table of cumulative binomial probabilities (less than)

p

n x 0.10 0.15 0.20 0.25 0.30 0.35 0.40

6 0 0.5314 0.3771 0.2621 0.1780 0.1176 0.0754 0.0467

1 0.8857 0.7765 0.6554 0.5339 0.4202 0.3191 0.2333

2 0.9842 0.9527 0.9011 0.8306 0.7443 0.6471 0.5443

3 0.9987 0.9941 0.9830 0.9624 0.9295 0.8826 0.8208

40.9999 0.9996 0.9984 0.9954

0.9891

0.9777 0.95905 1.0000 1.0000 0.9999 0.9998 0.9993 0.9982 0.9959

6 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

If you are using cumulative binomial probabilities table; P(X ≤ x); it is easier to calculatevalues from a binomial distribution.

0

( ) (1 ) x

n r n r

r

r

P X x C p p

p = 0.30

P (X 3) = 0.9295

n = 6

X 3

How to read the probability value from the table of binomial probabilities

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Compute the probability of X successes using the Binomial Table where n=2, p=0.30and X ≤1

Solution:

P( X ≤1) = 0.9100

25% of all VCR manufactured by a large electronics company are defective. A qualitycontrol inspector randomly selects three VCRs from the production line. What is theprobability that,

a) exactly one of these three VCRs is defective.

b) at least two of these VCRs are defective.

Solution:

Equally P(X = x), exactly that value, easy to use Poisson formula

At most All lower values and up to that x value, i.e. P(X ≤ x) , so take

directly from the table

Less than P(X < x) , that value x (and larger) is NOT included

At least P(X ≥ x), that value of x onwards

Greater than P(X > x), that value x (and smaller) is NOT included

From x1 to x2 P(x1 ≤ X ≤ x2), include both x1 and x2

Between x1 and x2 P(x1

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EXERCISE 5.1

1. A burglar alarm system has 6 fail-safe components. The probability of each failing is0.05. Find these probabilities:

a. Exactly 3 components will fail

b. Less than 2 components will fail

c. None will fail

2. A survey from Teenage Research Unlimited found that 30% of teenage consumersreceive their spending money from part-time jobs. If 5 teenagers are selected atrandom, find the probability that at least 3 of them will have part-time jobs.

3. R. H Bruskin Associates Market Research found that 40% of Americans do not think

that having a college education is important to succeed in the business world. If arandom sample of five American is selected, find these probabilities.

a. Exactly two people will agree with that statement.

b. At most three people will agree with that statement

c. At least two people will agree with that statement

d. Fewer than three people will agree with that statement.

4. It was found that 60% of American victims of health care fraud are senior citizens. If10 victims are randomly selected, find the probability that exactly 3 are senior citizens.

5. If 65% of the people in a community use the gym facilities in one year, find theseprobabilities for a sample of 10 people.

a. Exactly four people used the gym facilities.

b. At least six people not used the gym facilities.

6. In a poll of 12 to 18 year old females conducted by Harris Interactive for the GilletteCompany, 40% of the young female said that they expected the US to have a femalepresident within 10 years. Suppose a random sample of 15 females from this age

group selected. Find the probabilities that of young female in this sample who expecta female president within 10 years is;

a. at least 9

b. at most 5

c. 6 to 9

d. in between 4 and 8

e. less than 4

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7. In a Gallup Survey, 40% of the people interviewed were unaware that maintaining a

healthy weight could reduce the risk of stroke. If 15 people are selected at random, find

the probability that at least 9 are unaware that maintaining a proper weight could reduce

the risk of stroke.

5.1.1.2 Mean and Standard Deviation of the Binomial Probability Distribution

The mean, variance and standard deviation of a binomial distribution are:

Mean = = np

Variance = 2 = npq

Standard deviation = = npq

where;

n = the total number of trialsp = probability of successq = 1-p = probability of failure

Find the mean, variance and standard deviation for each of the values of n and p when

the conditions for the binomial distribution are met.

a. n=100, p=0.75

Solution:

a. = np = 100 (0.75) = 75

2 = npq = 100 (0.75) (0.25) = 18.75

= npq = 18.75 = 4.33

FORMULA

Example 5

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The Statistical Bulletin published by Metropolitan Life Insurance Co. reported that 2% of

all American births result in twins. If a random sample of 8000 births is taken, find the

mean, variance and standard deviation of the number of births that would result in

twins.

EXERCISE 5.2

1. It has been reported that 83% of federal government employees use e-mail. Ifa sample of 200 federal government employees is selected, find the mean,variance and standard deviation of the number who use e-mail.

2. A survey found that 25% of Malaysian watch movies at the cinema. Find themean, variance and standard deviation of the number of individuals who watchmovies at the cinema, if a random sample of 1000 Malaysian is selected.

5.1.2 Poisson Distribution

Named after the French mathematician Simeon D. Poisson. A discrete probability distribution that is useful when n is large and p is small

and when the independent variables occur over a period of time or given

area or volume.

Conditions to apply the Poisson Probability Distribution

X is a discrete random variable

The occurrences are random

The occurrences are independent

The following examples show the application of the Poisson probability

distribution.

The number of accidents that occur on a highway given during a one-week

period.

The number of customers entering a grocery store during a one-hour interval.

The number of television sets sold at a department store during a given week.

The number of typing errors per page.

A certain type of fabric made contains an average 0.5 defects per 500 yards.

Example 6

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5.1.2.1 Calculating Poisson Probability

a) Using Poisson Probability Distribution Formula

According to the Poisson probability distribution, the probability of x occurrences in

an interval is:

P(X)=!

x

e

x

where;

: mean number of occurrences in that interval

e : approximately 2.7183Find e x on your calculator

Find probability P(X; ), using the Poisson formula for P(5;4)

Solution:

P(X=5) =

4 5(4 )

5!

e

= 0.1563

If there are 200 typographical errors randomly distributed in a 500-page manuscript, findthe probability that a given page contains exactly three errors.

Solution:

FORMULA

Example 7

Example 8

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b) Using the Table of Poisson Probabilities

The probabilities for a Poisson distribution can also be read from the table

of Poisson probabilities.

Determining P (X 3) for = 1.5

The table of cumulative Poisson probabilities (less than)

x 1.1 1.2 1.3 1.4 1.5 1.6 1.7

0 0.3329 0.3012 0.2725 0.2466 0.2231 0.2019 0.1827

1 0.6990 0.6626 0.6268 0.5918 0.5578 0.5249 0.49322 0.9004 0.8795 0.8571 0.8335 0.8088 0.7834 0.7572

3 0.9743 0.9662 0.9569 0.9463 0.9344 0.9212 0.90684 0.9946 0.9923 0.9893 0.9857 0.9814 0.9763 0.9704

5 0.9990 0.9985 0.9978 0.9968 0.9955 0.9940 0.9920

6 0.9999 0.9997 0.9996 0.9994 0.9991 0.9987 0.9981

7 1.0000 1.0000 0.9999 0.9999 0.9998 0.9997 0.9996

If you are using a cumulative Poisson probabilities table; P(X ≤ x); it is easier to

calculate.

Find the probability P(X; ); using Poisson table; P(10;7)

Solution:

P(X = 10) = P(X 10) – P(X 9)

= 0.9015 – 0.8305

= 0.0710

How to read the probability value from the table of Poisson probabilities

= 1.50

( )!

k x

k

e P X x

k

X 3 P (X 3) = 0.9344

Example 9

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A sales firm receives an average of three calls per hour on its toll-free number. For any

given hour, find the probability that it will receive the following:

a) at most 3 callsb) at least 3 callsc) five or more callsd) between 1 to 4 calls in 2 hours

Solution:

= 3 calls per hour

a) P(X 3) = 0.6472

b) P(X 3) = 1 – P(X 2)

= 1 – 0.4232

= 0.5768

c) P(X 5) = 1 – P(X 4)

= 1 – 0.8153

= 0.1847

= 6 calls in 2 hours

d) P(1< X 4) = P(X 4) – P(X 1)

= 0.2851 – 0.0174

= 0.2677

Example 1

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Chapter 5: Special Distribution 11

EXERCISE 5.3

1. On average a household receives 2 telemarketing phone calls per week. Using thePoisson distribution formula, find the probability that a randomly selected householdreceives:

a. exactly six telemarketing phone calls during a given week.

b. less than three telemarketing phone calls in one month.

2. A washing machine at LaundryMat breaks down on average three times per month.Using the Poisson probability distribution formula, find the probability that thismachine will have

a) exactly two breakdowns per month

b) at most one breakdown per month

c) no breakdown in 2 months

3. In an airport between 19:00 and 20:00 hours, the number of flights that land followsa Poisson distribution with a mean 0.9 per five minutes interval. Find the probabilitythat the number of flights that land is:

a) one or less between 19:00 and 19:05 hours

b) more than three between 19:15 and 19:30 hours

4. An average of 4.8 customers visit Malaysia Savings and Loan every half hour. Findthe probability that during a given hour, the number of customer is

a) exactly twob) at most 2

c) none

d) more than 5 in one hour

6. Sports Score Jay receives, on average, eight calls per hour requesting the latestsports score. For any randomly selected hour, find the probability that the companywill receive

a) at least eight calls

b) three or more calls

c) at most seven calls.

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5.1.2.2 Mean and Standard Deviation of the Poisson ProbabilityDistribution

Mean,

=

Variance, 2 =

Standard Deviation,

=

An auto salesperson sells an average of 0.9 cars per day. Find the mean, variance

and standard deviation of cars sold per day by this salesperson.

Solution:

= = 0.9 2 = = 0.9

= = 0.9 = 0.9487

An insurance salesperson sells an average of 1.4 policies per day.

a) Find the probability that this salesperson will sell no insurance policy on a

certain day.

b) Find the mean, variance and standard deviation of the probability this

salesperson will sell the policies per day.

Solution:

FORMULA

Example 11

Example 12

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5.2 CONTINUOUS DISTRIBUTION

In Chapter 4, we defined a continuous random variable as a random variable

whose values are not countable.

A continuous random variable can assume any value over an interval orintervals because the number of values contained in any interval is infinite.

The possible number of values that a continuous random variable can

assume is also infinite.

Example of continuous random variables:

The life of battery, heights of people, time taken to complete an examination,

amount of milk in a gallon, weights of babies, prices of houses.

The probability distribution of continuous random variable has two

characteristics:

1. The probability that X assumes a value in any interval lies in the range 0 to 1.

Figure 1: Area under a curve between two points.

2. The total probability of all the (mutually exclusive) intervals within which X can

assume a value is 1.0.

Figure 2: Total area under a probability distribution.

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5.2.1 Normal Distribution

The normal distribution is the most important and most commonly used

among all of probability distributions.

A large number of phenomena in the real world are normally distributedeither exactly or approximately.

The normal probability distribution or the normal curve is a bell-shaped

(symmetric) curve.

Its mean is denoted by while its standard deviation is denoted by .

Figure 3: Normal distribution with mean; and standard deviation; .

Normal probability distributions give bell-shaped curves which can be

illustrated:

a) The total area under the curve is 1.0. b) The curve is symmetric about the

mean.

c)

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c) The two tails of the curve extend indefinitely which means that the curve never

touch x- axis.

5.2.1.1 The Standard Normal Distribution

Is a normal distribution with = 0 and = 1.

The value under the curve indicates the proportion of area in each section.

(example figure 2; pg 13)

The units for the standard normal distribution curve are denoted by z and

are called the z values or z scores.

The z value or z score is actually the number of standard deviation that a

particular x value is away from the mean.

The area under a standard normal distribution curve is used to solve

practical application problems such as:

finding the percentage of adult woman whose height is between 5 feet 4

inches and 5 feet 7 inches.

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A. Finding areas under the standard normal distribution curve

The standard normal distribution table lists the areas under the

standard normal curve to the left of z -values from –3.49 to 3.49.

Although the z -values on the left side of the mean are negative, the area

under the curve is always positive.

Determine the area under the standard normal curve to the left of z = 1.95

How to read the probability value from the standard normal distribution table

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Note:z = 1.95 can be interpreted as area to the left of 1.95.

P(z < 1.95) = 0.9744 or P(z < 1.95) = P(z 1.95) = 0.9744The probability that a continuous random variable assumes a single value iszero. Therefore, P(z = 1.95) = 0.

Find the area under the standard normal curve:a) To the left of z = 1.56b) To the right of z = -1.32c) From z = 0.85 to z = 1.95

Solution:

a) To the left of z = 1.56

P(z < 1.56) = 0.9406

b) To the right of z = -1.32

P(z > -1.32) = 1 – P(z < -1.32)

= 1 - 0.0934

= 0.9066

c) From z = 0.85 to z = 1.95

P(0.85 z 1.95)= P(z 1.95) – P(z 0.85)

= 0.9744 – 0.8023

= 0.1721

Example 13

0 1.56

-1.32 0

0.85 1.95

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EXERCISE 5.4

Find the area under the standard normal curve:

a) To the left of z = -2.87

b) To the right of z = 2.45

c) Between z = -2.15 and z =1.67

B. Converting an x value to a z value

For a normal variable X , a particular value of X can be converted to its

corresponding z value by using the formula:

z = x

Where and are the mean and standard deviation of the normal distribution of

X , respectively.

FORMULA

Remember!

The z value for the mean of a normal

distribution is always zero.

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Let X be a continuous random variable that has a normal distribution with a mean of 50

and a standard deviation of 10. Convert the following X values to z values.

a) 55 b) 35

Solution:

For the given normal distribution, = 50 and =10

a) The z value for X = 55 is computed as follows:

b) z = 35 – 50 =10

Let X be a continuous random variable that is normally distributed with a mean of 65 and

a standard deviation of 15. Find the probability that X assumes a value:

a) less than 43

b) greater than 74

c) between 56 and 71

Solution:

Example 14

Example 15

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EXERCISE 5.5

1. Let X denote the time taken to run a road race. Suppose X is approximately normallydistributed with a mean of 190 minutes and a standard deviation of 21 minutes. Ifone runner is selected at random, what is the probability that this runner willcomplete this road race:

a) in less than 150 minutes?

b) in 205 to 245 minutes?

2. The mean number of hours a student spends on the computer is 3.1 hours per day. Assuming a standard deviation of 0.5 hour, find the percentage of students whospend less than 3.5 hours on the computer. Assume the variable is normallydistributed.

3. The scores of 6000 candidates in a certain examination are found to be

approximately normal distributed with a mean of 55 and a standard deviation of 10:

a) If a score of 75 or more is required for passing with a distinction, estimate

the number of grades with distinction.

b) Calculate the probability that a candidate selected at random has a score

between 45 and 65.

5.3 INTRODUCTION TO t - DISTRIBUTION

The t distribution is very similar to the standardized normal distribution.

Both distributions are bell-shaped and symmetrical.

However, the t distribution has larger area in the tails and smaller area in

the centre than does the standardized normal distribution.

This is because is unknown and s is used to estimate it.

Because the value of is uncertain, the values of t that are observed will be

more variable than for z .

Standard normalt distribution for 5 degrees of freedom

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Chapter 5: Special Distribution

Tutorial21

As the number of degrees of freedom increases, the t distribution gradually

approaches the standardized normal distribution until the two are virtually

identical.

This happens because s becomes a better estimate of as the sample size

gets larger.

With a sample size of about 120 or more, s estimates precisely enough

that there is little difference between the t and Z distributions.

For this reason, most statisticians use z instead of t when the sample size is

greater than 120.

EXERCISE 5.6

1. 10 % of the bulbs produced by a factory are defective. A sample of 5 bulbs isselected randomly and tested for defect. Find the probability that

a) two bulbs are defective

b) at least one bulb is defective

2. In a university, 20 percent of the students fail statistics test. If 20 students from theuniversity are interviewed, what is the probability of getting:

a) less than 3 students who fail the test

b) more than 3 students who fail the test

c) exactly 4 students who fail the test

3. A financial institution in Kuala Lumpur has a job vacancy for a risk analyst and eachapplicant must seat for a written test. Based on the management experience, only

40% of the applicants pass the test and qualify for the interview session. There are20 applicants who have applied for the jobs. Find the probability that there are morethan 50% of the applicants will pass the test.

4. An Elementary Statistics class has 75 students. If there is a 12% absentee rate perlesson, find the mean, variance and standard deviation of the number of studentswho will be absent from each class.

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10. Encik Ahmad works as a lawyer at his own law firm which is situated at BandarKenangan. He drives to his work place every day. The estimated time taken to hiswork place is normally distributed with a mean of 24 minutes and a standarddeviation of 3.8 minutes.

a) Calculate the probability of estimated time taken of at least half and hour.

b) Find the probability of estimated time taken from 20 minutes until 25

minutes

c) Find the percentage that the estimated time taken of more than 25

minutes.

d) Find the probability of estimated time taken of less than 10 minutes.

11. Assume X is the time for a runner to finish his 2 km run. Given X is normallydistributed with a mean of 15 minutes and a standard deviation of 3 minutes. If onerunner is selected at random, find the probability that the runner can finish his 2kmrun in;

a) less than 13 minutes

b) not more than 16 minutes

c) within 14 minutes and 17 minutes.

12. Given the systolic blood pressure for the obesity group has a mean 132 mmHgand a standard deviation 8 mmHg. Assuming the variable normally distributed,

find the probability an obese person that has been selected at random have asystolic blood pressure:

a) More than 130 mmHg.

b) Less than 140 mmHg.

c) Between 131 mmHg and 136 mmHg.

13. The number of passenger for domestic flight from Alor Setar to Kuala Lumpur isnormally distributed with mean 80 and standard deviation 12. If one domesticflight is selected at random, find the probability the flight carries:

a) less than 90 passengers

b) at least 75 passengers

c) between 79 to 95 passengers

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