Question

# Company XYZ know that replacement times for the portable MP3 players it produces are normally distributed...

Company XYZ know that replacement times for the portable MP3 players it produces are normally distributed with a mean of 2.5 years and a standard deviation of 0.7 years.

Find the probability that a randomly selected portable MP3 player will have a replacement time less than 0.4 years?
P(X < 0.4 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 1.5% of the portable MP3 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.

Solution :

Given that ,

mean = = 2.5

standard deviation = = 0.7

P(x < 0.4) = P[(x - ) / < (0.4 - 2.5) / 0.7]

= P(z < -3.000)

= 0.0013

P(x < 0.4) = 0.0013

Using standard normal table ,

P(Z < z) = 1.5%

P(Z < -2.170) = 0.015

z = -2.170

Using z-score formula,

x = z * +

x = -2.170 * 0.7 + 2.5 = 1.0

warranty = 1.0 years

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