Answer: 0.8945
Step-by-step explanation:
Let x represents the lifetime of smoke detectors that a company manufactures.
We assume that the lifetime of smoke detectors that a company manufactures is normally distributed.
Given : The average lifetime of smoke detectors that a company manufactures is 5 years, or 60 months, and the standard deviation is 8 months.
i.e. [tex]\mu=60\ \ \,\ \sigma= 8[/tex]
Sampl siez : n= 30
Then, the probability that a random sample of 30 smoke detectors will have a mean lifetime between 58 and 63 months will be :-
[tex]P(58<x<63)=P(\dfrac{58-60}{\dfrac{8}{\sqrt{30}}}<\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{63-60}{\dfrac{8}{\sqrt{30}}})\\\\=P(-1.37<z<2.05)\\\\=P(z<2.05)-P(z<-1.37)\\\\=P(z<2.05)-(1-P(z<1.37))\ \ [\because P(Z<-z)=1-P(Z<z)]\\\\=0.9798178-(1-0.9146565)\ \ [\text{By using p-value table for z}]\\\\=0.8944743\approx0.8945[/tex]
Hence, the probability that a random sample of 30 smoke detectors will have a mean lifetime between 58 and 63 months= 0.8945
