Question

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.

Answer #1

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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