Question

We take a sample of size n=16 for a quality of life (QoL) measurement from a population with a normal distribution with m=8 and s=6.

a) What is the expected mean and the standard error for a sample size of n=16

b) What is the probability that the sample mean will be greater than 10?

Answer 1

a) The mean is 8 and the standard error is 1.5.

b) 9.18% probability that the sample mean will be greater than 10.

Explanation:

To solve this question, it is important to know the normal probability distribution and the Central Limit Theorem.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore 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 pvalue, 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 random variable X, with mean
`\mu` and standard deviation
`\sigma`, a large sample size can be approximated to a normal distribution with mean
`\mu` and standard deviation
`(\sigma)/(√(n))`

In this problem, we have that:

I call the population mean
`\mu` and the population standard deviation
`\sigma`

So

`\mu = 8, \sigma = 6`

a) What is the expected mean and the standard error for a sample size of n=16

By the Central Limit Theorem

`\mu = 8`

`s = (6)/(√(16)) = 1.5`

b) What is the probability that the sample mean will be greater than 10?

This probability is 1 subtracted by the pvalue of Z when X = 10.

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

Due to the Central Limit Theorem

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

`Z = (10 - 8)/(1.5)`

`Z = 1.33`

`Z = 1.33` has a pvalue of 0.9082.

So there is a 1-0.9082 = 0.0918 = 9.18% probability that the sample mean will be greater than 10.