Question

on average,e U.S. adults spend 136 minutes per day using social media. Use a normal distribution with a standard deviation of 14.2 minutes to answer: what is the probability that an adult uses social media between 120 and 140 minutes a day, what is the probability that an adult uses social media more than 160 minutes per day, and how many minutes must an adult spend on social media in a day to be in the top 5% of all social media users?

Answer 1

Solution :

Given that,

mean = = 136

standard deviation = =14.2

a ) P (120 < x < 140 )

P ( 120 - 136 / 14.2 ) < ( x -  / ) < ( 140 - 136 / 14.2 )

P ( - 16 / 14.2 < z < 4 / 14.2 )

P (-1.23 < z < 0.28 )

P ( z < 0.28 ) - P ( z < -1.23)

Using z table

= 0.6109 - 0.1299

= 0.4810

Probability = 0.4810

b ) P (x > 160 )

= 1 - P (x < 160 )

= 1 - P ( x -  / ) < ( 160- 136 / 14.2 )

= 1 - P ( z < 24 / 14.2 )

= 1 - P ( z < 1.69 )

Using z table

= 1 - 0.9545

= 0.0455

Probability = 0.0455

d ) P( Z > z) = 5%

P(Z > z) = 0.05

1 - P( Z < z) = 0.05

P(Z < z) = 1 - 0.05

P(Z < z) = 0.95

z = 1.64

Using z-score formula,

x = z * +

x = 1.64* 14.2 + 136

= 159.288

x = 159