SOLUTION: Americans watch an average of 3.5 hours of television per person per day. If the standard deviation for the number of hours of television watched per day is 3.9 and a random sample

Question 631182: Americans watch an average of 3.5 hours of television per person per day. If the standard deviation for the number of hours of television watched per day is 3.9 and a random sample of 266 Americans is selected, the mean of this sample belongs to a sampling distribution.
(a) What is the shape of this sampling distribution?
(1)skewed right normal
(2)symmetrical
(3)skewed left
(4)uniform

(b) What is the mean of this sampling distribution? (Give your answer correct to one decimal place.)

(c) What is the standard deviation of this sampling distribution? (Give your answer correct to two decimal places.)
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
(a) What is the shape of this sampling distribution?
(1)skewed right normal
(2)symmetrical
(3)skewed left
(4)uniform

Answer choice marked in red.

This is because the distribution will approach a normal distribution (ie become more like it) as you increase the sample size.
Remember that normal distributions are symmetrical.

The other choices that involved the word "skewed" are incorrect because skewed is the opposite of symmetrical. So that's out.

Also, it's not uniform because the more extreme cases (eg: those who watch barely any TV or those who watch too much TV) will not have the same proportion as those who watch the average amount of TV
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(b) What is the mean of this sampling distribution? (Give your answer correct to one decimal place.)

The mean of the sampling distribution is equal to the population mean. So the mean of this sampling distribution is 3.5 hours.
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(c) What is the standard deviation of this sampling distribution? (Give your answer correct to two decimal places.)

The standard deviation of the sampling distribution with population standard deviation and size is

$\LARGE \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

Plug in and and simplify...

$\LARGE \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

$\LARGE \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

$\LARGE \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

So the standard deviation of this sampling distribution is approximately 0.24 (rounded to two decimal places).

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