Question

According to a marketing website, adults in a certain country average 61 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 d below.

a. What is the probability that the amount of time spent today on mobile devices by an adult is less than 70

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 50

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 40and 56

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.

Answer 1

Solution :

Given that ,

mean = = 61

standard deviation = = 11

(a)

P(x < 70) = P((x - ) / < (70 - 61) / 11) = P(z < 0.81)

Using standard normal table,

P(x < 70) = 0.7910

Probability = 0.7910

(b)

P(x > 50) = 1 - P(x < 50)

= 1 - P((x - ) /  < (50 - 61) / 11)

= 1 - P(z < -1)

= 1 - 0.1587

= 0.8413

P(x > 50) = 0.8413

Probability = 0.8413

(c)

P(40 < x < 56) = P((40 - 61 / 11) < (x - ) / < (56 - 61) / 11) )

= P(-1.90 < z < -0.45)

= P(z < -0.45) - P(z < -1.90)

= 0.3264 - 0.0287 = 0.2977

Probability = 0.2977

(d)

P(Z < z) = 60%

P(Z < 0.25) = 0.60

z = 0.25

Using z-score formula,

x = z * +

x = 0.25 * 11 + 61 = 63.75

60th percentile = 63.75