A variable of two populations has a mean of 20 and a standard deviation of 48 for one of the populations and a mean of 20 and a standard deviation of 15 for the other population. Complete parts (a)
through (c).
a. For independent samples of size 16 and 9, respectively, find the mean and standard deviation of $x_1-x_2$. (Assume that the sampling is done with replacement or that the population is
large enough.)
The mean of $x_1-x_2$ is $\boxed{}$.
(Type an integer or a decimal. Do not round.)
The standard deviation of $x_1-x_2$ is $\boxed{}$.
(Round to four decimal places as needed.)
A
b. Must the variable under consideration be normally distributed on each of the two populations for you to answer part (a)? Choose the correct answer below.
A. No, the formulas for the mean and standard deviation of $x_1-x_2$ hold regardless of the distributions of the variable on the two populations, as long as the sampling is done with replacement
or that the population is large enough.
OB. Yes, the variable must be approximately normally distributed on each of the two populations for the formulas for the mean and standard deviation of $x_1-x_2$ to hold.
OC. No, the variable must be approximately normally distributed on at least one of the two populations for the formulas for the mean and standard deviation of $x_1 - x_2$ to hold, as long as the
sampling is done with replacement.
OD. No, the variable does not need to be normally distributed for the formulas for the mean and standard deviation of $x_1-x_2$ to hold as long as the sample sizes are large enough, as long as the
sampling is done with replacement.
c. Can you conclude that the variable $x_1 - x_2$ is normally distributed? Explain your answer.
Choose the correct answer below.
A. Yes, $x_1-x_2$ is always normally distributed because of the central limit theorem.
B. No, $x_1-x_2$ is normally distributed only if x is normally distributed on each of the two populations or if the sample sizes are large enough.
OC. Yes, $x_1-x_2$ is always normally distributed because it is calculated using parameters.
OD. No, since $x_1 - x_2$ must be greater than or equal to 0, the distribution is right skewed, so cannot be normally distributed.
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