Question

A Nielsen study indicates that 18-to-34 year olds spend a mean of 93 minutes whatching video on their smartphones per week. Assume that the amount of time watching video on a smartphone per week is normally distributed and that the standard deviation is 15 minutes.

1- What is the probability that an 18- to 34 year old spends less than 77 minutes watching video on his or her smartphone per week?

2- What is the probability that an 18- to 34- year old spends between 77 minutes and 109 minutes watching video on his or her smartphone per week?

3- What is the probability that an 18- to 34 year old spends more than 109 minutes watching video on his or her smartphone per week?

4- One percent of all 18- to 34- year olds will spend less than how many minutes watching video on his or her smartphone per week?

(Type it if possible)

Answer 1

Solution :

Given that ,

mean = = 93

standard deviation = =15

(a)P(x <77 ) = P[(x - ) / < (77 - 93) /15 ]

= P(z <-1.07 )

Using z table,

=0.1423

(b)P(77 < x < 109) = P[(77 - 93)/15 ) < (x - ) /  < (109 - 93) 15/ ) ]

= P( -1.07< z < 1.07)

= P(z <1.07 ) - P(z <-1.07 )

Using z table,

=0.8577 -  0.1423

=0.7154

(c)P(x >109 ) = 1 - p( x<109 )

=1- p P[(x - ) / < (109 - 93) /15 ]

=1- P(z < 1.07)

Using z table,

= 1 -0.8577

=0.1423

(d)Using standard normal table,

P(Z < z) = 1%

= P(Z < z) = 0.01

= P(Z <-2.326 ) = 0.01

z = -2.326

Using z-score formula  

x = z * +

x = -2.326 *15+93

x= 58.11