Question

A variable of a population has a mean of $\mu =100$and a standard deviation of$\sigma =28.$

a. Identify the sampling distribution of the sample mean for samples of size $49.$

b.In answering part (a), what assumption did you make about the distribution of the variable?

c. Can you answer part (a) if the sample size is $16$instead of $49?$

Why or why not?

Short answer

Part a) Sample mean follows Normal with a mean $100$and standard deviation $4(varianceis16).$

Part b) We did not make any assumptions.

Part c) No, for a sample of size $16$, we cannot find the distribution of the sample mean.

Step-by-step solution

Part a) Step 1: Given information

The population mean$\mu =100.$

The population standard deviation, $\sigma =28.$

Simplification

The sample size is $49.$

By the application of the central limit theorem, we know, for a large sample, the sample mean follows Normal distribution having mean, ${\mu }_{\overline{x}}=\mu ands\tan darddeviation,{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}},n=samplesize.$

Therefore, for the given sample of size $49,$

$$\begin{gathered} {\mu }_{\overline{x}}=\mu =100 \\ {\sigma }_{\overline{x}}=\frac{28}{\sqrt{49}} \\ =\frac{28}{7} \\ =4. \end{gathered}$$

Therefore, sample mean, $\overline{X}~N\left({\mu }_{\overline{x}},{\sigma }_{\overline{x}}\right)\equiv N(100,4^2).$

Part b) Step 1: Explanation

We did not make any assumptions. Since the sample size is greater than $30,$ easily consider the sample as a large sample and hence by using the central limit theorem, the distribution of the sample mean is Normal.

Part c) Step 1: Explanation

No, for a sample of size $16$, we cannot find the distribution of the sample mean.

The sample size $16(<30)$ is so small to consider it a larger sample and since the population distribution is not normal, we cannot consider the distribution of sample mean is Normal distribution unless the sample is a large sample.