Question

Company XYZ know that replacement times for the DVD players it produces are normally distributed with a mean of 7.8 years and a standard deviation of 1.4 years. Find the probability that a randomly selected DVD player will have a replacement time less than 3.5 years? P(X < 3.5 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 4.1% 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 = = 7.8

standard deviation = = 1.4

P(x < 3.5 )

= P[(x - ) / < ( 3.5 - 7.8 ) / 1.4 ]

= P(z < -3.071 )

Using z table,

= 0.0011

Probability = 0.0011

b.

The z - distribution of the 4.1% is,

P(Z < z) = 4.1%

= P(Z < z ) = 0.041

= P(Z < -1.739 ) = 0.041

z = -1.739

Using z-score formula,

x = z * +

x = -1.739 * 1.4 + 7.8

x = 5.3654

x = 5.4

Answer : x = 5.4