Question

Assume the average price for a movie is ​$10.16. Assume the population standard deviation is ​$0.49 and that a sample of 32 theaters was randomly selected. Complete parts a through d below.

a. Calculate the standard error of the mean.

b. What is the probability that the sample mean will be less than 10.31​?

c. What is the probability that the sample mean will be less than $10.11​?

d. What is the probability that the sample mean will be more than ​$10.26​?

(Round everything to four decimal places as needed.)

Answer 1

Solution :

Given that ,

mean = = $10.16

standard deviation = = $0.49

a.

n = 32

= $10.16

= / n = 0.49 / 32 = 0.0866

b.

P( < $10.31) = P(( - ) / < (10.31 - 10.16) / 0.0866)

= P(z < 1.73)

= 0.9582

Probability = 0.9582

c.

P( < $10.11) = P(( - ) / < (10.11 - 10.16) / 0.0866)

= P(z < -0.58)

= 0.2810

Probability = 0.2810

d.

P( > $10.26) = 1 - P( < 10.26)

= 1 - P[( - ) / < (10.26 - 10.16) / 0.0866]

= 1 - P(z < 1.15)

= 1 - 0.8749

= 0.1251

Probability = 0.1251