Respuesta :
Answer:
Probability that the mean of a sample of 90 computers would be less than 117.13 months is 0.0033.
Step-by-step 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 [tex]\bar X[/tex] = mean of a sample of 90 computers
The z score probability distribution for is given by;
Z = [tex]\frac{ \bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean life of a computer = 120 months
[tex]\sigma[/tex] = 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([tex]\bar X[/tex] < 117.13 months)
P([tex]\bar X[/tex] < 117.13) = P( [tex]\frac{ \bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{ 117.13-120}{\frac{10}{\sqrt{90} } }[/tex] ) = P(Z < -2.72) = 1 - P(Z [tex]\leq[/tex] 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.
Answer: P(x < 117.13) = 0.0033
Step-by-step 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
