Question

According to a research​ institution, the average hotel price in a certain year was

​$107.64

.

Assume the population standard deviation is

​$20.00

and that a random sample of

46

hotels was selected. Complete parts a through d below.

a. Calculate the standard error of the mean.

sigma Subscript x overbar

equals​$nothing

​(Round to two decimal places as​ needed.)

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

​$109

​?

Upper P left parenthesis x overbar less than $ 109 right parenthesis

equalsnothing

​(Round to four decimal places as​ needed.)

c. What is the probability that the sample mean will be more than

​$113

​?

Upper P left parenthesis x overbar greater than $ 113 right parenthesis

equalsnothing

​(Round to four decimal places as​ needed.)

d. What is the probability that the sample mean will be between

​$106

and

​$108

​?

Upper P left parenthesis $ 106 less than or equals x overbar less than or equals $ 108 right parenthesis

equalsnothing

​(Round to four decimal places as​ needed.)

Answer 1

Solution :

Given that ,

mean = = 107.64

standard deviation = = 20.00

n = 46

a) = = 107.64

= / n = 20.00 / 46 = 2.95

b) P( < 109) = P(( - ) / < (109 - 107.64) / 2.95)

= P(z < 0.46)

Using z table

= 0.6772

c) P( > 113) = 1 - P( < 113)

= 1 - P[( - ) / < (113 - 107.64) / 2.95]

= 1 - P(z < 1.82)

Using z table,    

= 1 - 0.9656

= 0.0344

d) P(106 < < 108)  

= P[(106 - 107.64) / 2.95 < ( - ) / < (108 - 107.64) / 2.95)]

= P( -0.56 < Z < 0.12)

= P(Z < 0.12) - P(Z < -0.56)

Using z table,  

= 0.5478 - 0.2877

= 0.2601