Use Table 4 - 1 for this exercise. Suppose that the mileage at which 10000sets of automobile brakes had been worn through was recorded. The average was 62700, and the standard deviation was 10400miles.

(a) What fraction of brakes is expected to be 80% worn in less than 45800miles?

(b) What fraction is expected to be80%worn at a mileage between 60000and 70000miles?

The Gaussian curve is given by the formula:

$y=\frac{1}{\sigma \sqrt{2\mathrm{\pi }}}{e}^{-{\left(x-\mu \right)}^{2}2{\sigma }^2}$

Where,

$\mu$is approximated by$\overline{x}$

$\sigma$is approximated by s

e is the base of the natural logarithm

$1/\sigma \sqrt{2\mathrm{\pi }}$is normalization factor.

The variations from the mean value are stated in multiples of the standard deviation, as follows:

$z=\frac{x-\mu }{\sigma }\approx \frac{x-\overline{x}}{s}$

The area under the whole curve from $z=-\infty$to $z=+\infty$must be unity.

(a)

The estimated percentage of brakes that are 80% worn:

Between$x=-\infty$ and$x=-40860$ miles, find the proportion of the area of the Gaussian curve.

When$$\begin{gathered} x=40860 \\ z=\frac{\left(40860-62700\right)}{10400} \\ =-2.1000 \end{gathered}$$ ,

The area from $-\infty$to -2.1000 is the same as the area from +2.1000 to $+\infty$because the Gaussian curve is symmetric. From table 4 - 1 , we know that the area between z = 0 and z = 2.1 is$0.4821$ .

The area between $z=0$and $z=\infty$is 0.5000 , hence the area between$z=2.100$ and $z=\infty$is$0.500-0.4281=0.0179$ .

The fraction of brakes expected to be 80% worn in less than 40860 miles is 0.0179 or 1.79% .

(b)

Fraction of brakes that is expected to be 80% worn:

At 57500 miles,

$$\begin{gathered} z=\frac{\left(57500-62700\right)}{10400} \\ =-0.5000 \end{gathered}$$

$$\begin{gathered} z=\frac{\left(71020-62700\right)}{10400} \\ =+0.8000 \end{gathered}$$

The area from z = -0.5000 to z = 0 is the same as the area from z = 0 to z = +0.5000 since the Gaussian curve is symmetric.

The area between z = 0 and z = +0.5000 is 0.1915 , according to table 4 - 1 .

Since the distance between z = 0 and z = +0.800 is 0.2881 , the distance between z = -0.5000 and z = +0.8000 is 0.1915 + 0.2881 = 0.4796 .

Hence, the fraction of brakes expected to be 80% worn between 57500 and 71020 miles is 0.4796 or 47.96% .