If you randomly purchase 11 items, what is the probability that their mean life will be longer than 3 years?

Question

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 3.2 years, and standard deviation of 0.7 years.

If you randomly purchase 11 items, what is the probability that their mean life will be longer than 3 years?

Answer 1

Hi Kaitlyn,

You need two main equations for this. The first is for standard error. Because you have a sample here, we need to compute sample standard error. To do this, the formula is:

SE=s/sqrt(n) where:

SE=standard error

s=standard deviation

n=sample size

For this problem

s=0.7

n=11

Thus:

SE=0.7/sqrt(11)

SE=0.21

Now, the second equation required is the classic z=(x-mu)/sigma, but with a twist for sample, not population, data. We need to substitute the sample standard error calculated above for sigma. We can still use the population mean we were given. Thus:

z=(x-mu)/SE where

x=3 years

mu=3.2 years

SE=0.21

z=(3-3.2)/0.21

z= -0.95

From z-table, we can get the probability that z is less than -0.95. Go to column on left for -0.95, row on top for 0.05.

P(Z<-0.95)=P(X<3)=0.1711

But we're not done yet. Recall that the question asked for probability of mean life GREATER than three years. Therefore, we just subtract:

P(X>3)=1-P(X<3)

P(X>3)=1-0.1711

P(X>3)=0.8289

I hope this helps. I encourage you to try some problems like this on your own. Good luck.