Question

For a population with µ = 80 and σ = 20, the distribution of sample means based on n = 16 will have an expected value of ____ and a standard error of ____. Select one: a. 5; 80 b. 80; 5 c. 20; 20 d. 80; 1.25 2) Samples of size n = 9 are selected from a population with µ = 80 with σ = 18. What is the standard error for the distribution of sample means? Select one: a. 80 b. 18 c. 6 d. 2 3) For a normal population with a mean of µ = 80 and a standard deviation of σ = 10, what is the probability of obtaining a sample mean greater than M = 75 for a sample of n = 25 scores?

Select one:
A. p = 0.0062
B. p = 0.9938
C. p = 0.3085
D. p = 0.6915

Answer 1

1. b. 80; 5

2. c. 6

3. A. p = 0.0062

Explanation:

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".

Part 1

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

`X\sim N(\mu =80,\sigma =20)`

We select samples of size n=16. That represent the sample size.

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

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

So the expected value would be 80 and the standard error Se=20/sqrt(16)=5.

b. 80; 5

Part 2

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

`X\sim N(\mu =80,\sigma =18)`

We select samples of size n=9. That represent the sample size.

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

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

So the expected value would be 80 and the standard error Se=18/sqrt(9)=6.

c. 6

Part 3

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

`\bar X \sim N(\mu=80, (\sigma)/(√(n))=2)`

And we want to calculate the probability that the mean for this sample would be greater than 75. So we want to find:

`P(\bar X >75)`

We can use the z score formula given by:

`z=(\bar x -\mu_o)/((\sigma)/(√(n)))`

`P(\bar X >75)=1-P(\bar x<75) = 1-P(Z<(75-70)/(2))=1-P(Z<2.5)=1-0.994=0.0062`