Question

The distribution of the amount of money in savings accounts for Florida State students has an average of 1,200 dollars and a standard deviation of 900 dollars. Suppose that we take a random sample of 10 Florida State students and ask them how much they have in their savings account. The sampling distribution of the sample mean amount of money in a savings account is

Answer 1

Answer:

By the Central Limit Theorem, the sampling distribution of the sample mean amount of money in a savings account is approximately normal with mean of 1,200 dollars and standard deviation of 284.6 dollars.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes 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 $s = \frac{\sigma}{\sqrt{n}}$.

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

Average of 1,200 dollars and a standard deviation of 900 dollars.

This means that $\mu = 1200, \sigma = 900$

Sample of 10.

This means that $n = 10, s = \frac{900}{\sqrt{10}} = 284.6$

The sampling distribution of the sample mean amount of money in a savings account is

By the Central Limit Theorem, approximately normal with mean of 1,200 dollars and standard deviation of 284.6 dollars.

Answer 2

The sampling distribution of the sample mean amount of money in a savings account is mean of 1,200 and standard deviation of 284.6 dollars.

Calculation of mean & standard deviation:

Since the distribution of the amount of money in savings accounts for Florida State students has an average of 1,200 dollars and a standard deviation of 900 dollars. And, a random sample of 10.

So here the mean should be $1,200

And, the standard deviation should be

$= 900 \div \sqrt10$

= $284.6

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