Answer to Question #216525 in Statistics and Probability for aasa

Question

As part of its quality assurance program, the Autolite Battery Company

conducts tests on battery life. For a particular D-cell alkaline battery, the mean life is 19 hours.

Answer 1

Whole problem:

As part of its quality assurance program, the Autolite Battery Company conducts tests on battery life. For a particular D-cell alkaline battery, the mean life is 19 hours. The useful life of the battery follows a normal distribution with a standard deviation of 1.2 hours.

Answer the following questions:

1. About 68 percent of the batteries failed between what two values?

2. About 95 percent of the batteries failed between what two values?

3. Virtually all of the batteries failed between what two values?

X = life time of battery

$X ~N(\mu=19, \sigma=1.2)$

1.

$P(x_1<X<x_2) =0.68, P(-z<Z<z) = 0.68 \\ P(Z<z) -P(Z<-z) = 0.68 \\ P(Z<z) -(1 -P(Z<z)) =0.68 \\ 2P(Z<z) = 1.68 \\ P(Z<z) = 0.84$

The value of z that leaves an area of 0.84 to the right is z = -0.995 and the left is z = 0.995.

$-0.995 = \frac{x_1 -19}{1.2} \\ x_1 = 19 -0.995 \times 1.2 \\ x_1 = 17.806 \\ 0.995 = \frac{x_2 -19}{1.2} \\ x_2 = 19 + 0.995 \times 1.2 \\ x_2 = 20.194 \\ P(17.8<X<20.19) = 0.68$

2.

$P(x_1<X<x_2) = 0.95, P(-z<Z<z) = 0.95 \\ P(Z<z) -P(Z<-z) = 0.95 \\ P(Z<z) -(1 -P(Z<z)) =0.95 \\ 2P(Z<z) = 1.95 \\ P(Z<z) = 0.975$

The value of z that leaves an area of 0.975 to the right is z = -1.96 and the left is z = 1.96.

$-1.96 = \frac{x_1 -19}{1.2} \\ x_1 = 19 -1.96 \times 1.2 \\ x_1 = 16.648 \\ 1.96 = \frac{x_2 -19}{1.2} \\ x_2 = 19 + 1.96 \times 1.2 \\ x_2 = 21.352 \\ P(16.65<X<21.35) = 0.95$

3.

$P(x_1<X<x_2) =0.99, P(-z<Z<z) = 0.99 \\ P(Z<z) -P(Z<-z) = 0.99 \\ P(Z<z) -(1 -P(Z<z)) =0.99 \\ 2P(Z<z) = 1.99 \\ P(Z<z) = 0.995$

The value of z that leaves an area of 0.995 to the right is z = -2.58 and the left is z = 2.58.

$-2.58 = \frac{x_1 -19}{1.2} \\ x_1 = 19 -2.58 \times 1.2 \\ x_1 = 12.904 \\ 2.58 = \frac{x_2 -19}{1.2} \\ x_2 = 19 + 2.58 \times 1.2 \\ x_2 = 22.096 \\ P(12.9<X<22.1) = 0.99$