Answer:
The probability that their mean life will be longer than 13 years = .9943 or 99.43%
Step-by-step explanation:
Given -
Mean [tex](\nu )[/tex] = 15 years
Standard deviation [tex](\sigma )[/tex] = 3.7 years
Sample size ( n ) =22
Let [tex]\overline{X}[/tex] be the mean life of manufacturing items
the probability that their mean life will be longer than 13 years =
[tex]P(\overline{X}> 13)[/tex] = [tex]P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}> \frac{13 - 15 }{\frac{3.7}{\sqrt{22}}})[/tex] [tex](Z =\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}})[/tex]
= [tex]P(Z> -2.53)[/tex]
= [tex]1 - P(Z < -2.53)[/tex] Using Z table
= 1 - .0057
= .9943
