BASIC PRACTICE OF STATS-LL W/SAPLINGPLU
BASIC PRACTICE OF STATS-LL W/SAPLINGPLU
8th Edition
ISBN: 9781319216245
Author: Moore
Publisher: MAC HIGHER
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Chapter 15, Problem 15.32E

(a)

To determine

To find: The approximate distribution of x¯ according to the central limit theorem.

(a)

Expert Solution
Check Mark

Answer to Problem 15.32E

The approximate distribution of x¯ is normal with mean μ=8.8beatsperfiveseconds and the standard deviation is σn=0.2041beatsperfiveseconds .

Explanation of Solution

Given info:

The data shows that the mean heart rate (μ) is 8.8 beats per five seconds and the standard deviation (σ) is 1 beat per five seconds. The distribution of heart rates is not normal.

Calculation:

Central limit theorem:

It states that as the sample size n becomes large, the sampling distribution of x¯ follows normal distribution with mean μ and standard deviation σn .

Mean of the Sampling distribution of x¯ :

The mean of the sampling distribution of x¯ is μ . That is, 8.8.

Thus, the mean of the average heart rate x¯ is 8.8 beats per five seconds.

Standard deviation of the Sampling distribution of x¯ :

The standard deviation can be calculated by using the formula σn .

Substitute 1 for σ and 24 for n

σn=124=14.8990=0.2041

Thus, the standard deviation of x¯ is 0.2041 beats per five seconds.

Hence, by the central limit theorem,the approximate distribution of x¯ is normal with mean 8.8beatsperfiveseconds and standard deviation is 0.2041beatsperfiveseconds .

(b)

To determine

To find: The approximate probability that x¯ less than 8.

(b)

Expert Solution
Check Mark

Answer to Problem 15.32E

The approximate probability of x¯ less than 8 is 0.

Explanation of Solution

Calculation:

STATE:

The mean heart rate (μ) is 8.8 beats per five seconds and the standard deviation (σ) is 1 beat per five seconds. What is the approximate probability that x¯ is less than 8?

PLAN:

The distribution of heart rates is not normal. Then, the central limit theorem would be used to find the distribution of mean heart rate.

SOLVE:

Let X be the number of heart beats per five seconds and x¯ be the mean number of beats per five seconds.

From the part a, x¯ is normal with mean μ=8.8beatsperfiveseconds and the standard deviation is σn=0.2041beatsperfiveseconds .

Then,

P(x¯<8)=P(x¯μσn<88.8124)=P(Z<0.80.2041)=P(Z<3.92)0[FromTableA:StandardNormalCumulativeProportions]

Thus, the approximate probability of x¯ less than 8 is 0.

CONCLUSION:

The approximate probability of x¯ less than 8 is 0.

(c)

To determine

To find: The approximate probability that the heart rate of a runner is less than 100 beats per minute.

(c)

Expert Solution
Check Mark

Answer to Problem 15.32E

The approximate probability that the heart rate of a runner is less than 100 beats per minute is 0.0107.

Explanation of Solution

Calculation:

The number of five seconds in one minute is 605=12 .

The mean heart rate for five seconds is 10012=8.33beatsperfiveseconds .

Then,

P(x¯<8)=P(x¯μσn<8.338.8124)=P(Z<0.470.2041)=P(Z<2.30)=0.0107[FromTableA:StandardNormalCumulativeProportions]

Thus, the approximate probability that the heart rate of a runner is less than 100 beats per minute is 0.0107.

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