Question

According to a research​ institution, the average hotel price in a certain year was ​$99.07. Assume the population standard deviation is ​$20.00 and that a random sample of 31 hotels was selected. Complete part d below.

d. What is the probability that the sample mean will be between ​$97 and ​$99​? _______​(Round to four decimal places as​ needed.)

Answer 1

Solution :

Given that,

mean = = $99.07

standard deviation = = $20.00

n = 31

= $99.07

= / n = 20.00 / 31 = 3.5921

d. P($97 < < $99) = P((97 - 99.07) / 3.5921) <( - ) / < (99 - 99.07) / 3.5921))

= P(-0.5763 < Z < -0.0195)

= P(Z < -0.0195) - P(Z < -0.5763) Using z table,

= 0.4922 - 0.2822

= 0.2100

Probability = 0.2100