Question

a) Average talk time between charges of a cell phone is advertised as 4.6 hours. Assume that talk time is normally distributed with a standard deviation of 0.6 hours.

Find the probability that talk time between charges for a randomly selected cell phone is either more than 5.7 hours or below 2.8 hours. (If you use the z table, round the "z" value to 2 decimal places. Round your final answer to 4 decimal places. Do NOT express as a percentage.)

b) Average talk time between charges of a cell phone is advertised as 4.6 hours. Assume that talk time is normally distributed with a standard deviation of 0.6 hours.

Twenty-nine percent of the time, talk time between charges is below a particular value. What is this value? (Round "z" value to 2 decimal places and final answer to 2 decimal places.)

c) Average talk time between charges of a cell phone is advertised as 4.6 hours. Assume that talk time is normally distributed with a standard deviation of 0.6 hours.

Find the probability that talk time between charges for a randomly selected cell phone is below 3.7 hours. (If you use the z table, round  "z" value to 2 decimal places. Round your final answer to 4 decimal places. Do NOT express as a percentage.)

Answer 1

(a)

= 4.6

= 0.6

To find P(2.8<X<5.7)

Case 1: For X from 2.8 to mid value:
Z = (2.8 - 4.6)/0.6 = - 3.00

Table of Area Under Standard Normal Curve gives area = 0.4987

Case : For X from mid value to 5.7:
Z = (5.7 - 4.6)/0.6 = 1.83

Table of Area Under Standard Normal Curve gives area = 0.4664

So,

P(2.8 < X < 5.7) = 0.4987 + 0.4664 = 0.9651

(b)

= 4.6

= 0.6

29% corresponds to area = 0.50 - 0.29 = 0.21 from mid value to Z on LHS.

Table of Area Under Standard Normal Curve gives Z = - 0.55
Z = - 0.55 = (X - 4.6)/0.6

So,

X = 4.6 - (0.55 X 0.6) = 4.27

So,

Answer is:

4.27

(c)

= 4.6

= 0.6

To find P(X<3.7):
Z = (3.7 - 4.6)/0.6 = - 1.50

Table of Area Under Standard Normal Curve gives area = 0.4332

So,

P(X<3.7) = 0.5 - 0.4332 = 0.0668