Question

Cell phone talk time between charges is advertised as 10 hours. Assume that talk time is approximately normally distributed with a mean of 10 hours and a standard deviation of 0.75 hours. a. Find the probability that talk time between charges for a randomly selected cell phone is above 11.35 hours.

2.b. Your phone gets only about 8.35 hours of talk time, what proportion of phones gets less talk time than yours?

2.c. How much talk time would you get between charges, if you had a phone that get just into the top 1%?

Answer 1

Solution :

Given ,

mean = = 10

standard deviation = = 0.75

P(x > 11.35) = 1 - P(x<11.35 )

= 1 - P[(x -) / < (11.35 -10) / 0.75]

= 1 - P(z <1.8 )

Using z table

= 1 - 0.9641

= 0.0359

probability= 0.0359

(b)

P(X<8.35 ) = P[(X- ) / < (8.35 -10) / 0.75]

= P(z < -2.2)

Using z table

= 0.0139

probability=0.0139

(c)

Using standard normal table,

P(Z > z) = 1%

= 1 - P(Z < z) = 0.01

= P(Z < z ) = 1 - 0. 01

= P(Z < z ) = 0.99

= P(Z < 2.33 ) = 0.99  

z = 2.33 (using standard normal (Z) table )

Using z-score formula  

x = z * +

x=2.33 *0.75+10

x= 11.7475

x=12