Question

A study indicates that​ 18- to​ 24- year olds spend a mean of 115 minutes watching video on their smartphones per month. Assume that the amount of time watching video on a smartphone per month is normally distributed and that the standard deviation is 10 minutes. Complete parts​ (a) through​ (d) below.

a. What is the probability that an​ 18- to​ 24-year-old spends less than 100 minutes watching video on his or her smartphone per​ month?

b.What is the probability that an​ 18- to​ 24-year-old spends between 100 and 160 minutes watching video on his or her smartphone per​ month?

c. What is the probability that an​ 18- to​ 24-year-old spends more than 160 minutes watching video on his or her smartphone per​ month?

d. one percent of all​ 18- to​ 24-year-olds will spend less than how many minutes watching video on his or her smartphone per​ month?

Answer 1

a) Let X be the time spent watching video on smart phone

then

z is a standard normal variate

To find , P( X<100)

From z table , P( z < -1.5) = 0.0668

Thus Probability that 18-24 yr old spend less than 100  minutes watching video on smartphone is 0.0668

b)  To find , P(100< X<160)

= P( -1.5 < z <0) +P( 0<z <4.5)

= 0.4332+ 0.5 ( from z table)

= 0.9332

Thus Probability that 18-24 yr old spend between 100 to 160 minutes watching video on smartphone is 0.9332  

c) To find , P( X>160)

P(z> 4.5) =0.000003 ( using excel formula) "=1-NORM.S.DIST(4.5,TRUE)"

P(z > 4.5) =0 (from z table)

Thus Probability that 18-24 yr old spend more than 160 minutes watching video on smartphone is 0.000003

d) To find c such that

P( X < c) =0.01

From z table

P(z < -2.33) =0.01

Thus

One percent of 18-24 yr old spend less than 91.7 minutes watching video on smart phone.