distributions examples

8/3/2019 Distributions Examples

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1. NORMAL DISTRIBUTION (Continuous)

1. Find the percentage of the normally distributeddata that lies within 2 standarddeviations of the mean.

Solution: Read the percentages from the chart at the top of this page from -2 to +2

standard deviations.

4.4% + 9.2% + 15.0% + 19.1% + 19.1% + 15.0% + 9.2% + 4.4% = 95.4%

2. At the New Age Information Corporation, the ages of all new employees hiredduring the last 5 years are normally distributed. Within this curve, 95.4% of the

ages, centered about the mean, are between 24.6 and 37.4 years. Find the mean age

and the standard deviation of the data.

Solution: As was seen in Example 1, 95.4% implies a span of 2 standard deviations

from the mean. The mean age is symmetrically located between -2 standard

deviations (24.6) and +2 standard deviations (37.4).

The mean age is years of age.

From 31 to 37.4 (a distance of 6.4 years) is 2 standard deviations. Therefore, 1

standard deviation is (6.4)/2 = 3.2 years.


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3.The amount of time that Carlos plays video games in any givenweek is normally distributed. If Carlos plays video games an

average of 15 hours per week, with a standard deviation of 3

hours, what is the probability of Carlos playing video gamesbetween 15 and 18 hours a week?

Solution: The average (mean) is 15 hours. If the standard deviation is 3, the interval

between 15 and 18 hours is one standard deviation above the mean, which gives a

probability of 34.1% or 0.341, as seen in the chart at the top of this page.

4.The lifetime of a battery is normally distributedwith a mean life of

40 hours and a standard deviation of 1.2 hours. Find the probabilitythat a randomly selected battery lasts longer than 42 hours.

The most accurate answer to a problem such as this cannot be obtained by using the

chart at the top of this page. One standard deviation above the mean would be located at41.2 hours, 2 standard deviations would be at 42.4, and one and one-half standard

deviations would be at 41.8 standard deviations. None of these locations corresponds

exactly to the needed 42 hours. We need more power than we have in the chart to find

the most accurate answer. Calculator to the rescue!!

Solution: Graph the normal curve. We see from the location of 42 on the graph

that the answer is going to be quite small.


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Example 1

An average light bulb manufactured by the Acme Corporation lasts 300 days with a standard

deviation of 50 days. Assuming that bulb life is normally distributed, what is the probability that

an Acme light bulb will last at most 365 days?

Solution: Given a mean score of 300 days and a standard deviation of 50 days, we want to find

the cumulative probability that bulb life is less than or equal to 365 days. Thus, we know thefollowing:

The value of the normal random variable is 365 days. The mean is equal to 300 days. The standard deviation is equal to 50 days.


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We enter these values into the Normal Distribution Calculator and compute the cumulative

probability. The answer is: P( X < 365) = 0.90. Hence, there is a 90% chance that a light bulbwill burn out within 365 days.

Example 2

Suppose scores on an IQ test are normally distributed. If the test has a mean of 100 and a

standard deviation of 10, what is the probability that a person who takes the test will scorebetween 90 and 110?

Solution: Here, we want to know the probability that the test score falls between 90 and 110. The"trick" to solving this problem is to realize the following:

P( 90


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Imagesource

A die is tossed 3 times. What is the probability of

(a) No fives turning up?

(b) 1 five?

(c) 3 fives?

This is a binomial distribution because there are only 2 possible outcomes (we get a 5 or we

don't).

ANSWER

Now, n = 3 for each part. LetX = number of fives appearing.

(a) Here,x = 0.

(b) Here,x = 1.

(c) Here,x = 3.
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EXAMPLE 2

Hospital records show that of patients suffering from a certain disease, 75% die of it. What is the

probability that of 6 randomly selected patients, 4 will recover?

ANSWER

This is a binomial distribution because there are only 2 outcomes (the patient dies, or does not).

LetX = number who recover.

Here, n = 6 andx = 4. Letp = 0.25 (success - i.e. they live), q = 0.75 (failure, i.e. they die).

The probability that 4 will recover:

1. A test consists of 10 multiple choice questions with fivechoices for each question. As an experiment, you GUESS on

each and every answer without even reading the questions.

What is the probability of getting exactly 6 questions correct on

this test?

Solution:n = 10

r= 6

nr= 4

p = 0.20 = probability of guessing the correct answer on a question

q = 1 - p = 0.80 = probability of not guessing the correct answer on a

question


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2. Whenrolling a

die 100

times,

what is the

probabilit

y of

rolling a

"4"

exactly 25

times?

Solution:n = 100

r= 25nr= 75

p = 1/6 = probability of rolling a "4"

q = 1 - p = 5/6 = probability of not rolling a "4"

3. At a

certainintersection,

the light for

eastbound

traffic is red

for 15

seconds,

yellow for 5

seconds, and

green for 30

seconds. Find the

probability

that out of the

next eight

eastbound

cars that


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arrive

randomly at

the light,

exactly three

will bestopped by a

red light.

Solution:n = 8

r= 3

nr= 5

p = 15/50 = probability of a red light

q = 1 - p = 35/50 = probability of not a red light

3. POISSON DISTRIBUTION (Discrete)Example:

Consider, in an office 2 customers arrived today. Calculate the possibilities for exactly 3 customers to

be arrived on tomorrow.

Step1: Find e-.

where, =2 and e=2.718

e- = (2.718)-2 = 0.135.

Step2: Find x.

where, =2 and x=3.

x = 23 = 8.

Step3: Find f(x).


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f(x) = e-x / x!

f(3) = (0.135)(8) / 3! = 0.18.

Hence there are 18% possibilities for 3 customers to be arrived on tomorrow.

EXAMPLE 1

A life insurance salesman sells on the average 3 life insurance policies per week. Use Poisson'slaw to calculate the probability that in a given week he will sell

(a) some policies

(b) 2 or more policies but less than 5 policies.

(c) Assuming that there are 5 working days per week, what is the probability that in a given day

he will sell one policy?

Here, = 3

(a) "Some policies" means "1 or more policies". We can work this out by finding 1 minus the

"zero policies" probability:

P(X> 0) = 1 P(x0)

Now so

So


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(b)

(c) Average number of policies sold per day:

So on a given day,

EXAMPLE 2

Twenty sheets of aluminum alloy were examined for surface flaws. The frequency of the numberof sheets with a given number of flaws per sheet was as follows:

Number of flaws Frequency

0 4

1 3

2 5

3 2

4 4

5 1

6 1

What is the probability of finding a sheet chosen at random which contains 3 or more surfaceflaws?

The total number of flaws is given by:

(0 4) + (1 3) + (2 5) + (3 2) + (4 4) + (5 1) + (6 1) = 46


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So the average number of flaws for the 20 sheets is given by:

The required probability is:

EXAMPLE 3

If electricity power failures occur according to a Poisson distribution with an average of 3

failures every twenty weeks, calculate the probability that there will not be more than one failureduring a particular week.

The average number of failures per week is:

"Not more than one failure" means we need to include the probabilities for "0 failures" plus "1

failure".

EXAMPLE 4

Vehicles pass through a junction on a busy road at an average rate of 300 per hour.

1. Find the probability that none passes in a given minute.2. What is the expected number passing in two minutes?3. Find the probability that this expected number actually pass through in a given two-

minute period.


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4. The average number of cars per minute is:5. (a)6.

7. (b)E(X) = 5 2 = 108. (c) Now, with = 10, we have:

EXAMPLE 5

A company makes electric motors. The probability an electric motor is defective is 0.01. What is

the probability that a sample of 300 electric motors will contain exactly 5 defective motors?

he average number of defectives in 300 motors is = 0.01 300 = 3

The probability of getting 5 defectives is:

NOTE: This problem looks similar to abinomial distributionproblem, that we met in the last

section.

If we do it using binomial, with n = 300,x = 5,p = 0.01 and q = 0.99, we get:

P(X= 5) = C(300,5)(0.01)5(0.99)295 = 0.10099

We see that the result is very similar. We can use binomial distribution to approximate Poissondistribution (and vice-versa) under certain circumstances.
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