Question

According to a research​ institution, the average hotel price in a certain year was ​$97.39. Assume the population standard deviation is ​$18.00 and that a random sample of 34 hotels was selected. Complete parts a through d below.

a. Calculate the standard error of the mean. ​(Round to two decimal places as​ needed.)

b. What is the probability that the sample mean will be less than ​$98​? ​(Round to four decimal places as​ needed.)

c. What is the probability that the sample mean will be more than ​$101​? ​(Round to four decimal places as​ needed.)

d. What is the probability that the sample mean will be between ​$96 and ​$97​? ​(Round to four decimal places as​ needed.)

Answer 1

= $97.39

= $18.00

Sample size, n = 34

a) Standard error, =

=

= 3.087

b) For sampling distribution of mean, P( < A) = P(Z < ( - )/)

P(sample mean will be less than ​$98​) = P( < 98)

= P(Z < (98 - 97.39)/3.087)

= P(Z < 0.20)

= 0.5793

c) P(sample mean will be more than ​$101​) = P( > 101)

= 1 - P( < 101)

= 1 - P(Z < (101 - 97.39)/3.087)

= 1 - P(Z < 1.17)

= 1 - 0.8790

= 0.1210

d) P(sample mean will be between ​$96 and ​$97) = P(96 < < 97)

= P( < 97) - P( < 96)

= P(Z < (97 - 97.39)/3.087) - P(Z < (96 - 97.39)/3.087)

= P(Z < -0.13) - P(Z < -0.45)

= 0.4483 - 0.3264

= 0.1219