In: Statistics and Probability
A study indicates that 18- to 24- year olds spend a mean of 135 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 20 minutes.
a.
What is the probability that an 18- to 24-year-old spends less than 95 minutes watching video on his or her smartphone per month?
b.
What is the probability that an 18- to 24-year-old spends between 95 and 175 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 175 minutes watching video on his or her smartphone per month?
d.
OneOne
percent of all 18- to 24-year-olds will spend less than how many minutes watching video on his or her smartphone per month?
Solution :
Given that ,
mean =
= 135
standard deviation =
= 20
a) P(x < 95) = P[(x -
) /
< (95 - 135) / 20 ]
= P(z < -2.00)
Using z table,
= 0.0228
b) P(95 < x < 175) = P[(95 - 135)/ 20 ) < (x -
) /
<
(175 - 135) / 20) ]
= P(-2.00 < z < 2.00)
= P(z < 2.00) - P(z < -2.00)
Using z table,
= 0.9772 - 0.0228
= 0.9544
c) P(x > 175) = 1 - p( x< 175)
=1- p P[(x -
) /
< (175 - 135) / 20]
=1- P(z < 2.00)
Using z table,
= 1 - 0.9772
= 0.0228
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 * 20 + 135
x = 88.48
x = 88 minutes