Question

The lifetime of a certain type of battery is normally distributed with mean value 13 hours and standard deviation 1 hour. There are nine batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages? (Round your answer to two decimal places.) 

Answer 1

Let X be the random variable that denotes the lifetime of a certain type of battery.

X $\sim$ Normal ($\mu$ = 13, $\sigma$ ² = 1)

There are nine batteries in a package. Therefore, the sample size is n = 9.

According to the central limit theorem, if a random variable X is normally distributed with mean $\mu$ and variance $\sigma$ ², then the distribution of the random variable $\sum$ X is approximately normally distributed with mean n$\mu$ and variance n$\sigma$², where n is the sample size.

Using the central limit theorem, $\sum$ X $\sim$ Normal (n$\mu$ = 9*13 = 117, n$\sigma$² = 9*1 = 9)

Let the required lifetime value be c.

P($\sum$X > c) = 0.05

$\therefore$ P(Z > (c - 117) / $\sqrt{9}$ ) = 0.05

$\therefore$ 1 - P(Z < (c - 117) / $\sqrt{9}$ ) = 0.05

$\therefore$ P(Z < (c - 117) / $\sqrt{9}$ ) = 0.95

From the standard normal table, P(Z < 1.645) = 0.95

$\therefore$ (c - 117) / $\sqrt{9}$ ) = 1.645

$\therefore$ c - 117 = 3 * 1.645

$\therefore$ c = 117 + 4.935

$\therefore$ c = 121.935

Therefore, the required lifetime value is 121.94 hours.