According to a marketing website, adults in a certain country average 54 minutes

Question

According to a marketing website, adults in a certain country average 54 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 11 minutes. Complete parts a through c below.

a. What is the probability that the amount of time spent today on mobile devices by an adult is less than 60

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 between 37 and 52 minutes?

____ (Round to four decimal places as needed.)

c. What amount of time spent today on mobile devices by an adult represents the 85th percentile?

An amount of time of ____minutes represents the 85th percentile.

(Round to two decimal places as needed.)

Explanation / Answer

Mean = 54 minutes

Standard deviation = 11 minutes

P(X < A) = P(Z < (A - mean)/standard deviation)

A) P(X < 60) = P(Z < (60-54)/11)

= P(Z < 0.55)

= 0.7088

B) P(37 < X < 52) = P(X < 52) - P(X < 37)

= P(Z < (52 - 54)/11) - P(Z < (37 - 54)/11)

= P(Z < -0.18) - P(Z < -1.55)

= 0.4286 - 0.0606

= 0.3680

C) Let K represent the 85th percentile

P(X < K) = 0.85

P(Z < (K - 54)/11) = 0.85

Take the value of z corresponding to 0.85 from standard normal distribution table

(K - 54)/11 = 1.04

K = 65.44 minutes

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