Question

2. According to a marketing website, adults in a certain country average 62 minutes per day on mobile devices this year. Assume that minutes per day on mobile devices follow the normal distribution and has a standard deviation of 8 minutes. Complete parts a through d below.

a. What is the probability that the amount of time spent today on mobile devices by an adult is less than 65 minutes?

(Round to four decimal places as needed.)

b. What is the probability that the amount of time spent today on mobile devices by an adult is more than 55 minutes?

(Round to four decimal places as needed.)

c. What is the probability that the amount of time spent today on mobile devices by an adult is between 45 and 58 minutes?

(Round to four decimal places as needed.)

d. What amount of time spent today on mobile devices by an adult represents the 60th percentile?

An amount of time of ....... minutes represents the 60th percentile.

(Round to two decimal places as needed.)

3. According to a recent survey, the average daily rate for a luxury hotel is $238.94. Assume the daily rate follows a normal probability distribution with a standard deviation of $21.09. Complete parts a through d below.

a. What is the probability that a randomly selected luxury hotel's daily rate will be less than $ 255?

(Round to four decimal places as needed.)

b. What is the probability that a randomly selected luxury hotel's daily rate will be more than $ 262?

(Round to four decimal places as needed.)

c. What is the probability that a randomly selected luxury hotel's daily rate will be between $ 246 and $ 266 ?

(Round to four decimal places as needed.)

d. The managers of a local luxury hotel would like to set the hotel's average daily rate at the 75 th percentile, which is the rate below which 75% of hotels' rates are set. What rate should they choose for their hotel?

The managers should choose a daily rate of $........

(Round to the nearest cent as needed.)

Answer 1

P ( X < 65 )

Standardizing the value

Z = ( 65 - 62 ) / 8

Z = 0.38

P ( X < 65 ) = P ( Z < 0.38 )

P ( X < 65 ) = 0.648

Part b)

P ( X > 55 ) = 1 - P ( X < 55 )

Standardizing the value

Z = ( X - \mu ) / \sigma

Z = ( 55 - 62 ) / 8

Z = -0.88

P ( Z > -0.88 )

P ( X > 55 ) = 1 - P ( Z < -0.88 )

P ( X > 55 ) = 1 - 0.1894

P ( X > 55 ) = 0.8106

Part c)

P ( 45 < X < 58 )

Standardizing the value

Z = ( 45 - 62 ) / 8

Z = -2.12

Z = ( 58 - 62 ) / 8

Z = -0.5

P ( -2.12 < Z < -0.5 )

P ( 45 < X < 58 ) = P ( Z < -0.5 ) - P ( Z < -2.12 )

P ( 45 < X < 58 ) = 0.3085 - 0.0168

P ( 45 < X < 58 ) = 0.2917

Part d) P ( Z < ? ) = 60% = 0.60

Looking for the probability of 0.60 i standard normal table to find the critical value Z

Z = 0.25

X = 64

64 minutes spent today on mobile devices by an adult represents the 60th percentile.