91Ó°ÊÓ

The Gaussian curve is given by the formula:

$y=\frac{1}{\mathrm{σ}\sqrt{2\mathrm{π}}}{e}^{-{\left(x-\mathrm{μ}\right)}^{2}2{\mathrm{σ}}^2}$

Where,

$\mathrm{μ}$is approximated by$\stackrel{¯}{x}$

$\mathrm{σ}$is approximated by s

e is the base of the natural logarithm

$1/\mathrm{σ}\sqrt{2\mathrm{π}}$is normalization factor.

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

$z=\frac{x-\mathrm{μ}}{\mathrm{σ}}≈\frac{x-\stackrel{¯}{x}}{s}$

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

(a)

The estimated percentage of brakes that are 80% worn:

Between$x=-∞$ 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 $-∞$to -2.1000 is the same as the area from +2.1000 to $+∞$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 $z=∞$is 0.5000 , hence the area between$z=2.100$ and $z=∞$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% .