Question

7. The average life expectancy of dishwashers produced by a company is 8 years with a standard deviation of 6 months. Assume that the lives of dishwashers are normally distributed. a. What is the probability that a randomly selected dishwasher will have a life expectancy of at least 7 years? b. Dishwashers that fail operating in less than 4 1/2 years will be replaced free of charge. What percent of dishwashers are expected to be replaced free of charge? c. What are the minimum and the maximum life expectancy of the middle 95% of the dishwashers' lives? Give your answer in months. d. If 155 of this year's dishwasher production fail operating in less than 4 years and 4 months, how many dishwashers were produced this year?

Answer 1

a) P(X > 7)

= P(X - $\mu$ )/$\sigma$> (7 - $\mu$ )/$\sigma$)

= P(Z > (7 - 8)/0.5)

= P(Z > -2)

= 1 - P(Z < -2)

= 1 - 0.0228

= 0.9772

b) P(X < 4.5)

= P(X - $\mu$ )/$\sigma$ < (4.5 - $\mu$ )/$\sigma$)

= P(Z < (4.5 - 8)/0.5)

= P(Z < -7)

= 0

= 0.00%

c) P(X < x) = 0.025

or, P((X - $\mu$ )/$\sigma$ < (x - $\mu$ )/$\sigma$) = 0.025

or, P(Z < (x - 96)/6) = 0.025

or, (x - 96)/6 = -1.96

or, x = -1.96 * 6 + 96

or, x = 84.24

P(X > x) = 0.025

or, P((X - $\mu$ )/$\sigma$ > (x - $\mu$ )/$\sigma$) = 0.025

or, P(Z > (x - 96)/6) = 0.025

or, P(Z < (x - 96)/6) = 0.975

or, (x - 96)/6 = 1.96

or, x = 1.96 * 6 + 96

or, x = 107.76

d) P(X < 4.33)

= P(X - $\mu$ )/$\sigma$ < (4.33 - $\mu$ )/$\sigma$)

= P(Z < (4.33 - 8)/0.5)

= P(Z < -7.34)

= 0

Expected number of dishwashers were produced this year = 155 * 0 = 0