Question

A study on the latest fad diet claimed that the amounts of weight lost by all people on this diet had a mean of 20.720.7 pounds and a standard deviation of 4.14.1 pounds. Step 2 of 2 : If a sampling distribution is created using samples of the amounts of weight lost by 9090 people on this diet, "what would be the standard deviation of the sampling distribution of sample means"? Round to two decimal places, if necessary.

Answer 1

The standard deviation of the sampling distribution of sample means would be of 0.43 pounds.

Explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

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:

`\sigma = 4.1, n = 90`

So

`s = (\sigma)/(√(n)) = (4.1)/(√(90)) = 0.43`

The standard deviation of the sampling distribution of sample means would be of 0.43 pounds.