Answer:a. 0.13% b. 2.28%
Step-by-step explanation:
Let x denotes the useful life of batteries.
Given: [tex]\mu=56[/tex] months, [tex]\sigma=6[/tex]
Since X is normally distributed.
The probability that batteries with a useful life of less than 38 months
[tex]P(X<38)=P(\frac{X-\mu}{\sigma}<\frac{38-56}{6})\\\\=P(Z<-3) \ \ \ [Z=\frac{X-\mu}{\sigma}]\\\\=1-P(Z<3) \\\\=1- 0.9987\\\\=0.0013[/tex]
The percent of batteries with a useful life of less than 38 months =0.13%
The probability that batteries will last longer than 68 months
[tex]P(X>68)=P(\frac{X-\mu}{\sigma}>\frac{68-56}{6})\\\\=P(Z>2) \ \ \ [Z=\frac{X-\mu}{\sigma}]\\\\=1-P(Z<2)\\\\=1-0.9772=0.0228[/tex]
The percent of batteries that will last longer than 68 months =2.28%
