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The life span of computer monitors manufactured by XYZ company is normally distributed with mean 8 years and standard deviation 1.5 years. What percentage of XYZ monitors will remain operational after 9 years?(Need to solve the problem process, this is a solution)



Answer :

The percentage that a randomly selected hot water heater has a life span of 9 years is about 65.62%.

According to the question;

Mean (μ) = 8 years

Standard deviation (σ) = 1.5 years

The z score can be used to find the probability of a random variable occurring over a normally distributed parameter if its mean and standard deviation is given. z = x- μ /  σ

The z score can then be used to find the probability of a randomly selected hot water heater lifespan being below that value: P ( Z < z)

First, find the z score for 9

z = 12 - 13 / 1.5 = - 0.6667

z = 15 - 13 / 1.5 = 1.333

Therefore,

P ( X < 12) = P ( Z < - 0.6667 ) ≈ 0.2525

P ( X < 15) = P ( Z < 1.333 ) ≈ 0.9087

To find the probability of a randomly selected hot water heater having a lifespan between 12 and 15 over this distribution, subtract the P for the lower value from the P for the higher value.

P ( - 0.6667 < X < 1.333 )  = P ( Z < 1.333 ) - P ( Z < - 0.6667 ) = 0.9087 - 0.2525 = 0.6562

P ( 12 < X < 15 ) ≈ 65.62%

Learn more about Z-scores here

brainly.com/question/14887272

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