In: Statistics and Probability
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 |
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