Question

A large number of automobile batteries have
an average life length of 24 months. 34% of
them have 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 < Z < 0.44)= 0.17 and 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

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

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

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

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

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

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

$$\sigma\approx4.84891978$$

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

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

$$\approx0.15123348$$

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