Question

The quality control manager at a computer manufacturing company believes that the mean life of a computer is 120 months, with a standard deviation of 10 months. If he is correct, what is the probability that the mean of a sample of 90 computers would be less than 117.13 months? Round your answer to four decimal places.

Answer 1

Answer: P(x < 117.13) = 0.0033

Explanation:

Let us assume that the mean life of the computers are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ/√n

Where

x = lifetime of the computers in months.

µ = mean lifetime

σ = standard deviation

n represents the number of samples

From the information given,

µ = 120 months

σ = 10 months

n = 90

The probability that the mean life of a sample of 90 computers would be less than 117.13 months is expressed as

P(x < 117.13)

For x = 117.13

z = (117.3 - 120)/10/√90

z = - 2.87/1.0541 = - 2.72

Looking at the normal distribution table, the probability corresponding to the z score is 0.0033

Answer 2

Probability that the mean of a sample of 90 computers would be less than 117.13 months is 0.0033.

Explanation:

We are given that the quality control manager at a computer manufacturing company believes that the mean life of a computer is 120 months, with a standard deviation of 10 months.

Also, a random sample of 90 computers is selected.

Firstly, Let
`\bar X` = mean of a sample of 90 computers

The z score probability distribution for is given by;

Z =
`( \bar X - \mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\mu` = mean life of a computer = 120 months

`\sigma` = standard deviation = 10 months

n = sample of computers = 90

Probability that the mean of a sample of 90 computers would be less than 117.13 months is given by = P(
`\bar X` < 117.13 months)

P(
`\bar X` < 117.13) = P(
`( \bar X - \mu)/((\sigma)/(√(n) ) )` <
`( 117.13-120)/((10)/(√(90) ) )` ) = P(Z < -2.72) = 1 - P(Z
`\leq` 2.72)

= 1 - 0.99674 = 0.0033

Therefore, probability that the mean of a sample of 90 computers would be less than 117.13 months is 0.0033.