Question

Company XYZ know that replacement times for the portable MP3 players it produces are normally distributed with a mean of 3 years and a standard deviation of 0.5 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 4.4% 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 = = 3 years

standard deviation = = 0.5 years

a) P(x < 1.5) = P[(x - ) / < (1.5 - 3) / 0.5]

= P(z < -3.000)

Using z table,

= 0.0013

b) Using standard normal table,

P(Z < z) = 4.4%

= P(Z < z) = 0.044

= P(Z < -1.706) = 0.044

z = -1.706

Using z-score formula,

x = z * +

x = -1.706 * 0.5 + 3

x = 2.1 years

warranty = 2.1 years

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