Question

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 3.9 years, and standard deviation of 1 years.

If 10 items are picked at random, 4% of the time their mean life will be less than how many years?

Give your answer to one decimal place.

Answer 1

X :lifespan of items produced by the manufacturer.

X ~ N (3.9 , 1)

if a sample of size(n) 10 is selected, then the sample mean will follow normal distribution with,

mean = 3.9

C91co= - – pe

let , 4% of the time their mean life will be less than m years.

according to the problem,

P(X <m) = 0.04

$\Rightarrow P(\frac{\bar{X}-3.9}{0.3162}<\frac{m-3.9}{0.3162})=0.04$

m - 3.9 P(Z < 0.3162) = 0.04

т – 3.9, озво) = 0.04

$\Rightarrow \frac{m-3.9}{0.3162}=\Phi^{-1}(0.04)=-1.75069$[ in any blank cell of excel type =NORMSINV(0.04) , press enter]

>m=(-1.75069 * 0.3162) + 3.9 = 3.34643 3.3

If 10 items are picked at random, 4% of the time their mean life will be less than 3.3 years.

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