Question

A random sample is drawn from a population with mean μ = 66 and standard deviation σ = 5.5. use table 1.
a. is the sampling distribution of the sample mean with n = 16 and n = 36 normally distributed? yes, both the sample means will have a normal distribution. no, both the sample means will not have a normal distribution. no, only the sample mean with n = 16 will have a normal distribution. no, only the sample mean with n = 36 will have a normal distribution.
b. can you use the standard normal distribution to calculate the probability that the sample mean falls between 66 and 68 for both sample sizes? yes, for both the sample sizes, standard normal distribution could be used. no, for both the sample sizes, standard normal distribution could not be used. no, only for the sample size with n = 16, standard normal distribution could be used. no, only for the sample size with n = 36, standard normal distribution could be used.
c. calculate the probability that the sample mean falls between 66 and 68 for n = 36. (round intermediate calculations to 4 decimal places, "z" value to 2 decimal places, and final answer to 4 decimal places.)

Answer 1

Using the normal distribution, we have that:

a) No, only the sample mean with n = 36 will have a normal distribution.

b) No, only for the sample size with n = 36, standard normal distribution could be used.

c) There is a 0.4854 = 48.54% probability that the sample mean falls between 66 and 68 for n = 36.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

• Bu the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$, as long as the underlying distribution is normal or n > 30.

The mean and the standard deviation are given, respectively, by:

$\mu = 66, \sigma = 5.5$.

We have no information if the underlying distribution is normal or not, hence, for items a and b, we can only use n = 36.

Item c:

The standard error, considering n = 36, is found as follows:

$s = \frac{5.5}{\sqrt{36}} = 0.9167$

The probability is the p-value of Z when X = 68 subtracted by the p-value of Z when X = 66, hence:

X = 68:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{68 - 66}{0.9167}$

Z = 2.18

Z = 2.18 has a p-value of 0.9854.

X = 66:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{66 - 66}{0.9167}$

Z = 0

Z = 0 has a p-value of 0.5.

0.9854 - 0.5 = 0.4854.

There is a 0.4854 = 48.54% probability that the sample mean falls between 66 and 68 for n = 36.

More can be learned about the normal distribution at https://brainly.com/question/4079902

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