Introduction To Statistics And Data Analysis
Introduction To Statistics And Data Analysis
6th Edition
ISBN: 9781337793612
Author: PECK, Roxy.
Publisher: Cengage Learning,
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Textbook Question
Chapter 7, Problem 135CR

The longest “run” of S’s in the sequence SSFSSSSFFS has length 4, corresponding to the S’s on the fourth, fifth, sixth, and seventh positions. Consider a binomial experiment with n = 4, and let y be the length in the longest run of S’s.

  1. a. When p = 0.5, the 16 possible outcomes are equally likely. Determine the probability distribution of y in this case (first list all outcomes and the y value for each one). Then calculate μy.
  2. b. Repeat Part (a) for the case p = 0.6.
  3. c. Let z denote the longest run of either S’s or F’s. Determine the probability distribution of z when p = 0.5.

a.

Expert Solution
Check Mark
To determine

Find the probability distribution of y when p = 0.5.

Obtain μy.

Answer to Problem 135CR

The probability distribution of y is obtained as given below:

y01234
p(y)0.06250.43750.31250.12500.0625

The mean for the random variable y is 1.6875.

Explanation of Solution

Calculation:

It is given that, the longest “run” of S’s in the sequence SSFSSSSFFS has length 4. This length corresponds to S’s on the fourth, fifth, sixth, and seventh positions.

Define the random variable y as the length in the longest run of S’s. Here, the random variable y follows binomial distribution with n = 4.

When p = 0.5, there are 16 possible outcomes and these outcomes are equally likely to occur.

One of the possible outcomes is SSSS. Here, longest run of S is 4. Hence, the random variable y takes the value 4. The corresponding probability is, (0.5)(0.5)(0.5)(0.5)=0.0625. Similarly, the other outcomes and its corresponding y value and probability are listed in the below table.

OutcomeyProbability
SSSS4(0.5)(0.5)(0.5)(0.5)=0.0625
SSSF3(0.5)(0.5)(0.5)(0.5)=0.0625
SSFS2(0.5)(0.5)(0.5)(0.5)=0.0625
SSFF2(0.5)(0.5)(0.5)(0.5)=0.0625
SFSS2(0.5)(0.5)(0.5)(0.5)=0.0625
SFSF1(0.5)(0.5)(0.5)(0.5)=0.0625
SFFS1(0.5)(0.5)(0.5)(0.5)=0.0625
SFFF1(0.5)(0.5)(0.5)(0.5)=0.0625
FSSS3(0.5)(0.5)(0.5)(0.5)=0.0625
FSSF2(0.5)(0.5)(0.5)(0.5)=0.0625
FSFS1(0.5)(0.5)(0.5)(0.5)=0.0625
FSFF1(0.5)(0.5)(0.5)(0.5)=0.0625
FFSS2(0.5)(0.5)(0.5)(0.5)=0.0625
FFSF1(0.5)(0.5)(0.5)(0.5)=0.0625
FFFS1(0.5)(0.5)(0.5)(0.5)=0.0625
FFFF0(0.5)(0.5)(0.5)(0.5)=0.0625

The probability distribution of y is obtained as given below:

y01234
p(y)0.06250.0625×7  =0.43750.0625×5  =0.31250.0625×2  =0.12500.0625×1  =0.0625

Mean:

The mean for the random variable y is obtained as given below:

yp(y)yp(y)
00.06250
10.43750.4375
20.31250.625
30.1250.375
40.06250.25
y=1,...,4yp(y)=1.6875

From the above table, the mean is,

μy=y=1,...,4yp(y)=1.6875

Thus, the mean for the random variable y is 1.6875.

b.

Expert Solution
Check Mark
To determine

Find the probability distribution of y when p = 0.6.

Obtain μy.

Answer to Problem 135CR

The probability distribution of y is obtained as given below:

y01234
p(y)0.02560.3264 0.3456 0.17280.1296

The mean for the random variable y is 2.0544.

Explanation of Solution

Calculation:

Define the random variable y as the length in the longest run of S’s.

When p = 0.6, there are 16 possible outcomes and these outcomes are equally likely to occur.

One of the possible outcomes is SSSS. Here, S occurs for 4 times. Hence, the random variable y takes the value 4. The corresponding probability is, (0.6)(0.6)(0.6)(0.6)=0.1296. Similarly, the other outcomes and its corresponding y value and probability are listed in the below table.

OutcomeyProbability
SSSS4(0.6)(0.6)(0.6)(0.6)=0.1296
SSSF3(0.6)(0.6)(0.6)(0.4)=0.0864
SSFS2(0.6)(0.6)(0.4)(0.6)=0.0864
SSFF2(0.6)(0.6)(0.4)(0.4)=0.0576
SFSS2(0.6)(0.4)(0.6)(0.6)=0.0864
SFSF1(0.6)(0.4)(0.6)(0.4)=0.0576
SFFS1(0.6)(0.4)(0.4)(0.6)=0.0576
SFFF1(0.6)(0.4)(0.4)(0.4)=0.0384
FSSS3(0.4)(0.6)(0.6)(0.6)=0.0864
FSSF2(0.4)(0.6)(0.6)(0.4)=0.0576
FSFS1(0.4)(0.6)(0.4)(0.6)=0.0576
FSFF1(0.4)(0.6)(0.4)(0.4)=0.0384
FFSS2(0.4)(0.4)(0.6)(0.6)=0.0576
FFSF1(0.4)(0.4)(0.6)(0.4)=0.0384
FFFS1(0.4)(0.4)(0.4)(0.6)=0.0384
FFFF0(0.4)(0.4)(0.4)(0.4)=0.0256

The probability distribution of y is obtained as given below:

yp(y)
00.0256
10.0567+0.0567+0.0384+0.0576+0.0384+0.0384+0.0384=0.3264
20.0864+0.0567+0.0864+0.0576+0.0576=0.3456
30.0864+0.0864=0.1728
40.1296

Mean:

The mean for the random variable y is obtained as given below:

yp(y)yp(y)
00.02560
10.32640.3264
20.34560.6912
30.17280.5184
40.12960.5184
y=1,...,4yp(y)=2.0544

From the above table, the mean is,

μy=y=1,...,4yp(y)=2.0544

Thus, the mean for the random variable y is 2.0544.

c.

Expert Solution
Check Mark
To determine

Find the probability distribution of z when p = 0.5.

Answer to Problem 135CR

The probability distribution of z is obtained as given below:

z1234
p(z)0.1250 0.50000.25000.1250

Explanation of Solution

Calculation:

Define the random variable z as the length in the longest run either S’s or F’s.

When p = 0.5, there are 16 possible outcomes and these outcomes are equally likely to occur.

One of the possible outcomes is SSSS. Here, S occurs for 4 times. Hence, the random variable y takes the value 4. The corresponding probability is, (0.5)(0.5)(0.5)(0.5)=0.0625. Similarly, the other outcomes and its corresponding y value and probability are listed in the below table.

OutcomezProbability
SSSS4(0.5)(0.5)(0.5)(0.5)=0.0625
SSSF3(0.5)(0.5)(0.5)(0.5)=0.0625
SSFS2(0.5)(0.5)(0.5)(0.5)=0.0625
SSFF2(0.5)(0.5)(0.5)(0.5)=0.0625
SFSS2(0.5)(0.5)(0.5)(0.5)=0.0625
SFSF1(0.5)(0.5)(0.5)(0.5)=0.0625
SFFS2(0.5)(0.5)(0.5)(0.5)=0.0625
SFFF3(0.5)(0.5)(0.5)(0.5)=0.0625
FSSS3(0.5)(0.5)(0.5)(0.5)=0.0625
FSSF2(0.5)(0.5)(0.5)(0.5)=0.0625
FSFS1(0.5)(0.5)(0.5)(0.5)=0.0625
FSFF2(0.5)(0.5)(0.5)(0.5)=0.0625
FFSS2(0.5)(0.5)(0.5)(0.5)=0.0625
FFSF2(0.5)(0.5)(0.5)(0.5)=0.0625
FFFS3(0.5)(0.5)(0.5)(0.5)=0.0625
FFFF4(0.5)(0.5)(0.5)(0.5)=0.0625

The probability distribution of z is obtained as given below:

z1234
p(z)0.0625×2  =0.12500.0625×8  =0.50.0625×4  =0.2500.0625×2  =0.125

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Chapter 7 Solutions

Introduction To Statistics And Data Analysis

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