Question

% 7.4.12-T Question Help Assume the average price for a movie is $11.62. Assume the population standard deviation is $0.55 and that a sample of 36 theaters was randomly selected. Complete parts a through d below. a. Calculate the standard error of the mean. 0- = 5.0917 (Round to four decimal places as needed.) b. What is the probability that the sample mean will be less than $11.79? P(x<$11.79) = 9678 (Round to four decimal places as needed.) c. What is the probability that the sample mean will be less than $11.57? P(x<$11.57) = 0.2927 (Round to four decimal places as needed.) d. What is the probability that the sample mean will be more than $11.73? P(>$11.73) - N (Round to four decimal places as needed.)

Answer 1

Solution :

Given that ,

mean = $\mu$ = $11.62

standard deviation = $\sigma$ = $0.55

a.

$\sigma$$\bar x$ = $\sigma$ / $\sqrt$ n = 0.55 / $\sqrt$ 36 = 0.0917

b.

P($\bar x$ < $11.79) = P(($\bar x$ - $\mu$ $\bar x$ ) / $\sigma$ $\bar x$ < (11.79 - 11.62) / 0.0917)   

P(z < 1.85)

= 0.9678

P($\bar x$ < $11.79) = 0.9678

c.

P($\bar x$ < $11.57) = P(($\bar x$ - $\mu$ $\bar x$ ) / $\sigma$ $\bar x$ < (11.57 - 11.62) / 0.0917)   

P(z < -0.55)

= 0.2912

P($\bar x$ < $11.57) = 0.2912

d.

P($\bar x$ > $11.73) = 1 - P($\bar x$ < 11.73)

= 1 - P[($\bar x$ - $\mu$ $\bar x$ ) / $\sigma$ $\bar x$ < (11.73 - 11.62) / 0.0917]

= 1 - P(z < 1.20)

= 1 - 0.8849

= 0.1151

P($\bar x$ > $11.73) = 0.1151