Question

Company XYZ know that replacement times for the DVD players it produces are normally distributed with a mean of 4.7 years and a standard deviation of 2.3 years.

Find the probability that a randomly selected DVD player will have a replacement time less than -0.8 years?
P(X < -0.8 years) =

Enter your answer accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

If the company wants to provide a warranty so that only 1.8% of the DVD players will be replaced before the warranty expires, what is the time length of the warranty?
warranty =  years

Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

Solution :

Given that ,

mean = = 4.7

standard deviation = = 2.3

P(x < - 0.8)

= P[(x - ) / < (- 0.8 - 4.7) / 2.3]

= P(z < - 2.391)

Using z table,

= 0.0084

Using standard normal table,

P(Z > z) = 1.8%

= 1 - P(Z < z) = 0. 018

= P(Z < z) = 1 - 0.018

= P(Z < z ) = 0.982

= P(Z < 2.097 ) = 0.982  

z = 2.097

Using z-score formula,

x = z * +

x = 2.097 * 2.3 + 4.7

x = 9.5