The times to pop a regular bag of microwave popcorn without burning it are normally distributed with a mean time of 140 seconds and a standard deviation of 20 seconds. The times to pop a mini bag of microwave popcorn without burning it are normally distributed with a mean time of 90 seconds and a standard deviation of 15 seconds. Two independent random samples, 25 of each, are taken, and the mean popping times are calculated.

Let [tex]\(\bar{x}_R\)[/tex] be the mean popping time of a randomly selected regular-sized bag, and [tex]\(\bar{x}_m\)[/tex] be the mean popping time of a mini-sized bag.

Which best describes the standard deviation of the sampling distribution of [tex]\(\bar{x}_R - \bar{x}_m\)[/tex]?

A. 1 second
B. 5 seconds
C. 25 seconds
D. 35 seconds



Answer :

To determine the standard deviation of the sampling distribution of the difference in means [tex]\(\bar{x}_R - \bar{x}_M\)[/tex], we can use the properties of standard deviations and sampling distributions.

Given:
- [tex]\(\bar{x}_R\)[/tex] is the mean popping time for a regular bag of microwave popcorn.
- [tex]\(\bar{x}_M\)[/tex] is the mean popping time for a mini bag of microwave popcorn.
- Mean popping time for regular bags, [tex]\(\mu_R = 140\)[/tex] seconds, with standard deviation [tex]\(\sigma_R = 20\)[/tex] seconds.
- Mean popping time for mini bags, [tex]\(\mu_M = 90\)[/tex] seconds, with standard deviation [tex]\(\sigma_M = 15\)[/tex] seconds.
- Sample size for regular bags, [tex]\(n_R = 25\)[/tex].
- Sample size for mini bags, [tex]\(n_M = 25\)[/tex].

The standard deviation of the sampling distribution of the difference in sample means ([tex]\(\bar{x}_R - \bar{x}_M\)[/tex]) is given by the formula:

[tex]\[ \sigma_{\bar{x}_R - \bar{x}_M} = \sqrt{\frac{\sigma_R^2}{n_R} + \frac{\sigma_M^2}{n_M}} \][/tex]

Let's plug in the given values:

1. Calculate the variance for the regular bags of popcorn:
[tex]\[ \frac{\sigma_R^2}{n_R} = \frac{20^2}{25} = \frac{400}{25} = 16 \][/tex]

2. Calculate the variance for the mini bags of popcorn:
[tex]\[ \frac{\sigma_M^2}{n_M} = \frac{15^2}{25} = \frac{225}{25} = 9 \][/tex]

3. Sum these variances to get the variance of the difference in sample means:
[tex]\[ 16 + 9 = 25 \][/tex]

4. Take the square root of this sum to find the standard deviation of the difference in sample means:
[tex]\[ \sigma_{\bar{x}_R - \bar{x}_M} = \sqrt{25} = 5 \][/tex]

Therefore, the standard deviation of the sampling distribution of [tex]\(\bar{x}_R - \bar{x}_M\)[/tex] is [tex]\(5\)[/tex] seconds. The correct answer is:

5 seconds