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The mean useful life of car batteries is 56 months. They have a standard deviation of 6. Assume the useful life of batteries is normally
distributed
a. Calculate the percent of batteries with a useful life of less than 38 months. (Round your answer to the nearest hundredth percent.)
b. Calculate the percent of batteries that will last longer than 68 months. (Round your answer to the nearest hundredth percent.)
a. Percent of batteries
%
b. Percent of batteries
%

Sagot :

JeanaShupp

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%

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