Answer to Question #111453 in Statistics and Probability for Subhajit Paul

Question

A large number of automobile batteries have an average life length of 24 months. 34% of them have than a months, is average life between 22 and 26 months and 272 of them last longer than 29 months. If life of the batteries follows normal distribution, how many batteries were tested? What is the standard deviation?

Given that P(0<=3 <=0.44) = 0.17, P(0<=Z<=1.1) = 0.3643.

Answer 1

 

$X\sim N(\mu, \sigma^2)$

Then

$$Z={X-\mu\over \sigma}\sim N(0, 1)$$

Given

$$\mu=24, P(22<X<26)=0.34,P(X>29)=0.27$$

$$P(22<X<26)=P(X<26)-P(X\leq22)=$$

$$=P(Z<{26-24\over \sigma})-P(Z\leq{22-24\over \sigma})=$$

$$=1-2P(Z\leq{22-24\over \sigma})=0.34$$

$$P(Z\leq{-2\over \sigma})=0.33$$

$${-2\over \sigma}\approx-0.412463$$

$$\sigma\approx4.84891978$$

$$P(X>29)=1-P(Z\leq{29-24\over \sigma})\approx$$

$$\approx1-P(Z\leq{5\over 4.84891978})\approx1-P(Z\leq1.0311575)\approx$$

$$\approx1-0.84876652\approx0.15123348={272\over n}$$

$$n\approx{272 \over 0.15123348}\approx1799$$