Respuesta :
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 [tex]\bar X[/tex] = sample mean price for a movie in the United States
The z-score probability distribution for the sample mean is given by;
Z = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean price for a movie = $7.96
[tex]\sigma[/tex] = population standard deviation = $0.50
n = sample of theaters = 30
(a) The standard error of the mean is given by;
Standard error = [tex]\frac{\sigma}{\sqrt{n} }[/tex] = [tex]\frac{0.50}{\sqrt{30} }[/tex]
= 0.091
(b) The probability that the sample mean will be less than $7.75 is given by = P([tex]\bar X[/tex] < $7.75)
P([tex]\bar X[/tex] < $7.75) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{7.75-7.96}{\frac{0.50}\sqrt{30} } }[/tex] ) = P(Z < -2.30) = 1 - P(Z [tex]\leq[/tex] 2.30)
= 1 - 0.9893 = 0.0107
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([tex]\bar X[/tex] < $8.10)
P([tex]\bar X[/tex] < $8.10) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{8.10-7.96}{\frac{0.50}\sqrt{30} } }[/tex] ) = P(Z < 1.53) = 0.9369
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([tex]\bar X[/tex] > $8.20)
P([tex]\bar X[/tex] > $8.20) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{8.20-7.96}{\frac{0.50}\sqrt{30} } }[/tex] ) = P(Z > 2.63) = 1 - P(Z [tex]\leq[/tex] 2.63)
= 1 - 0.9957 = 0.0043
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.
