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 3.7 years and a standard deviation of 0.8 years. Find the probability that a randomly selected portable MP3 player will have a replacement time less than 1.5 years? P(X < 1.5 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 1.8% of the portable MP3 players will be replaced before the warranty expires, what is the time length of the warranty? warranty =

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.

We are given the distribution here as:

a) The probability here is computed as:
P(X < 1.5)

Converting it to a standard normal variable, we get here:

Getting it from the standard normal tables, we get here:

Therefore 0.0030 is the required probability here.

b) From standard normal tables, we have:

P( Z < -2.096) = 0.018

Therefore the length of warranty here is computed here as:
= Mean -2.096*Std Dev
= 3.7 - 2.096*0.8 = 2.0232

Therefore 2.0 years is the required warranty period such that only 1.8% of the portable players will be replaced.

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