Question

Consider a parent population with mean 75 and a standard deviation 7. The population doesn’t appear to have extreme skewness or outliers.

a What are the mean and standard deviation of the distribution of sample means for n = 40?
b What’s the shape of the distribution? Explain your answer in terms of the Central Limit Theorem.
c What proportion of the sample means of size 40 would you expect to be 77 or less? If you use your calculator, show what you entered.
d Draw a sketch of the probability you found in part C.

Answer 1

a)
`\bar X \sim N(\mu=375, \sigma={\bar X}=(7)/(√(40))=1.107)`

b) Since the sample size is large enough n>30 and the original distribution for the random variable X doesn’t appear to have extreme skewness or outliers, the distribution for the sample mean would be bell shaped and symmetrical.

c)
`P(\bar X \leq 77)=P(Z<(77-75)/(1.107))=P(Z<1.807)=0.965`

d) See figure attached

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable of interest. We know from the problem that the distribution for the random variable X is given by:

`E(X) = 75`

`sd(X) = 7`

We take a sample of n=40 . That represent the sample size

Part a

From the central limit theorem we know that the distribution for the sample mean
`\bar X` is also normal and is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

`\bar X \sim N(\mu=375, \sigma={\bar X}=(7)/(√(40))=1.107)`

Part b

Since the sample size is large enough n>30 and the original distribution for the random variable X doesn’t appear to have extreme skewness or outliers, the distribution for the sample mean would be bell shaped and symmetrical.

Part c

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

`z=(\bar X- \mu)/((\sigma)/(√(n)))`

And we want to find this probability:

`P(\bar X \leq 77)=P(Z<(77-75)/(1.107))=P(Z<1.807)=0.965`

We can us the following excel code: "=NORM.DIST(1.807,0,1,TRUE)"

Part d

See the figure attached.