A recent study of 1,215 randomly selected middle school students revealed that the average number of minutes they spent completing homework during the school week was 180 minutes with a standard deviation of 45 minutes.
1. Which of the following is the standard error, in minutes, of the sampling distribution of the mean number of minutes spent on homework per week for all middle school students?
O √(45)(55) / 1,215
O 45 / 1,215
O √45 / 1,215
O 45 / √1,215
O 1.96 (45 / √1,215)

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation, or standard error [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

[tex]\sigma = 45, n = 1215[/tex]

Then

[tex]s = \frac{\sigma}{\sqrt{n}} = \frac{45}{\sqrt{1215}}[/tex]

So the correct answer is:

O 45 / √1,215

dwivedisiddhant03 dwivedisiddhant03 · Answer 2 · 2022-01-11T06:38:04+01:00

To identify the the standard error, in minutes, we have to apply Central Limit Theorem.

The correct option is (c) [tex]\dfrac{45}{\sqrt{1215} }[/tex].

Central limit theorem

As per the central limit theorem, if population with mean and standard deviation take large random sample then the distribution of sample mean will normally distributed.

Given:

The standard deviation is [tex]\sigma =45[/tex]

The random variable is [tex]n=1215[/tex].

How to apply central limit theorem?

Write the formula to find the standard error.

[tex]s=\dfrac{\sigma}{\sqrt{n} }[/tex]

Substitute the value.

[tex]s=\dfrac{45}{\sqrt{1215} }[/tex]

Thus, the correct option is (c) [tex]\dfrac{45}{\sqrt{1215} }[/tex].

Learn more about Central Limit Theorem here:

https://brainly.com/question/898534