Answer:
(a) The standard error of the mean is 0.091.
(b) The probability that the sample mean will be less than $7.75 is 0.0107.
(c) The probability that the sample mean will be less than $8.10 is 0.9369.
(d) The probability that the sample mean will be more than $8.20 is 0.0043.
Step-by-step explanation:
We are given that the average price for a movie in the United States in 2012 was $7.96.
Assume the population standard deviation is $0.50 and that a sample of 30 theaters was randomly selected.
Let = <u><em>sample mean price for a movie in the United States</em></u>
The z-score probability distribution for the sample mean is given by;
Z = ~ N(0,1)
where, = population mean price for a movie = $7.96
= population standard deviation = $0.50
n = sample of theaters = 30
(a) The standard error of the mean is given by;
Standard error = =
= 0.091
(b) The probability that the sample mean will be less than $7.75 is given by = P( < $7.75)
P( < $7.75) = P( < ) = P(Z < -2.30) = 1 - P(Z 2.30)
= 1 - 0.9893 = <u>0.0107</u>
The above probability is calculated by looking at the value of x = 2.30 in the z table which has an area of 0.9893.
(c) The probability that the sample mean will be less than $8.10 is given by = P( < $8.10)
P( < $8.10) = P( < ) = P(Z < 1.53) = <u>0.9369</u>
The above probability is calculated by looking at the value of x = 1.53 in the z table which has an area of 0.9369.
(d) The probability that the sample mean will be more than $8.20 is given by = P( > $8.20)
P( > $8.20) = P( > ) = P(Z > 2.63) = 1 - P(Z 2.63)
= 1 - 0.9957 = <u>0.0043</u>
The above probability is calculated by looking at the value of x = 2.63 in the z table which has an area of 0.9957.