Question

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.

Answer #1

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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