Question

According to a research​ institution, the average hotel price in a certain year was ​$95.82 . Assume the population standard deviation is ​$18.00 and that a random sample of 36 hotels was selected. Complete parts a through d below. a. Calculate the standard error of the mean. sigma Subscript x overbar equals​$nothing ​(Round to two decimal places as​ needed.) b. What is the probability that the sample mean will be less than ​$98 ​? Upper P left parenthesis x overbar less than $ 98 right parenthesis equalsnothing ​(Round to four decimal places as​ needed.) c. What is the probability that the sample mean will be more than ​$100 ​? Upper P left parenthesis x overbar greater than $ 100 right parenthesis equalsnothing ​(Round to four decimal places as​ needed.) d. What is the probability that the sample mean will be between ​$94 and ​$96 ​? Upper P left parenthesis $ 94 less than or equals x overbar less than or equals $ 96 right parenthesis equalsnothing ​(Round to four decimal places as​ needed.

Answer 1

Solution :

Given that ,

mean = = 95.82

standard deviation = = 18.00

n = 36

a) = = 95.82

= / n = 18.00 / 36  = 3.00

b) P( < 98) = P(( - ) / < (98 - 95.82) / 3.00 )

= P(z < 0.73)

Using z table

= 0.7673

c) P( > 100) = 1 - P( < 100)

= 1 - P[( - ) / < (100 - 95.82) / 3.00]

= 1 - P(z < 1.39)

Using z table,    

= 1 - 0.9177

= 0.0823

d) P(94 < < 96)  

= P[(94 - 95.82) / 3.00 < ( - ) / < (96 - 95.82) / 3.00)]

= P(-0.61 < Z < 0.06)

= P(Z < 0.06) - P(Z < -0.61)

Using z table,  

= 0.5239 - 0.2709

= 0.2530