Question

In: Statistics and Probability

Use the following information to answer questions 1-3. A population of 1,000 students spends an average...

Use the following information to answer questions 1-3.

A population of 1,000 students spends an average of $10.50 a day on dinner. The standard deviation of the expenditure is $3. A simple random sample of 64 students is taken.

Determine the expected value of the sample mean.

a.

$3.00

b.

$10.50

c.

$5.25

Determine the standard deviation of the sample mean.

a.

0.363

b.

3.000

c.

0.375

Determine the probability that these 64 students will spend a combined total of more than $716.80.

a.

0.973

b.

0.027

c.

0.062

Solutions

Expert Solution

Let X be a random variable representing the expenditure. Then the population mean is E(X)=   =10.5

and the population standard deviation is sample size is n=64

sample mean =    where are samples.

the expected value of the sample mean=

the standard deviation of the sample mean=

because

So,

Standard deviation =

the probability that these 64 students will spend a combined total of more than $716.80

=

here as the sample size is greater than 30 so we use normal approximation to find the probability.

Z is a standard normal random variable.

the answer does not match any given option. So I think you can take 0.027 as the answer which is close to 0.031


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