Question

The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probability of each of the following occurring from this population. Appendix A Statistical Tables a. A random sample of size 36 yielding a sample mean of 78 or more b. A random sample of size 150 yielding a sample mean of between 71 and 77 c. A random sample of size 219 yielding a sample mean of less than 74.2Incorrect. The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probability of each of the following occurring from this population. Appendix A Statistical Tables a. A random sample of size 36 yielding a sample mean of 78 or more b. A random sample of size 150 yielding a sample mean of between 71 and 77 c. A random sample of size 219 yielding a sample mean of less than 74.2

Answer 1

a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean of a population is 74 and the standard deviation is 15.

This means that
`\mu = 74, \sigma = 15`

Question a:

Sample of 36 means that
`n = 36, s = (15)/(√(36)) = 2.5`

This probability is 1 subtracted by the pvalue of Z when X = 78. So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (78 - 74)/(2.5)`

`Z = 1.6`

`Z = 1.6` has a pvalue of 0.9452

1 - 0.9452 = 0.0548

0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

Question b:

Sample of 150 means that
`n = 150, s = (15)/(√(150)) = 1.2247`

This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So

X = 77

`Z = (X - \mu)/(s)`

`Z = (77 - 74)/(1.2274)`

`Z = 2.45`

`Z = 2.45` has a pvalue of 0.9929

X = 71

`Z = (X - \mu)/(s)`

`Z = (71 - 74)/(1.2274)`

`Z = -2.45`

`Z = -2.45` has a pvalue of 0.0071

0.9929 - 0.0071 = 0.9858

0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c. A random sample of size 219 yielding a sample mean of less than 74.2

Sample size of 219 means that
`n = 219, s = (15)/(√(219)) = 1.0136`

This probability is the pvalue of Z when X = 74.2. So

`Z = (X - \mu)/(s)`

`Z = (74.2 - 74)/(1.0136)`

`Z = 0.2`

`Z = 0.2` has a pvalue of 0.5793

0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2