Question

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 15 years, and standard deviation of 3.7 years. If you randomly purchase 22 items, what is the probability that their mean life will be longer than 13 years?

Answer 1

The probability that their mean life will be longer than 13 years = .9943 or 99.43%

Explanation:

Given -

Mean
`(\\u )` = 15 years

Standard deviation
`(\sigma )` = 3.7 years

Sample size ( n ) =22

Let
`\overline{X}` be the mean life of manufacturing items

the probability that their mean life will be longer than 13 years =

`P(\overline{X}> 13)` =
`P(\frac{\overline{X} - \\u }{(\sigma )/(√(n))}> (13 - 15 )/((3.7)/(√(22))))`
`(Z =\frac{\overline{X} - \\u }{(\sigma )/(√(n))})`

=
`P(Z> -2.53)`

=
`1 - P(Z < -2.53)` Using Z table

= 1 - .0057

= .9943