Question

Mean shifts on a production line. Six Sigma is a comprehensive approach to quality goal setting that involves statistics. An article in Aircraft Engineering and Aerospace Technology (Vol. 76, No. 6, 2004) demonstrated the use of the normal distribution in Six Sigma goal setting at Motorola Corporation. Motorola discovered that the average defect rate for parts produced on an assembly line varies from run to run and is approximately normally distributed with a mean equal to 3 defects per million. Assume that the goal at Motorola is for the average defect rate to vary no more than 1.5 standard deviations above or below the mean of 3. How likely is it that the goal will be met?

Short answer

There is an 87% chance that the goal will be met

Step-by-step solution

Given information

Assume that Motorola discovered the average defect rate for parts produced on an assembly line varies from run to run and it is approximately normally distributed with a mean of 3 and standard deviation sigma.

Calculating for probability

Here, x follows a normal distribution with $\mu =3and\sigma =\sigma$

$x=3+1.5\sigma$

The z-score is,

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{3+1.5\sigma -3}{\sigma } \\ =\frac{1.5\sigma }{\sigma } \\ =1.5 \end{gathered}$$

Again,$x=3-1.5\sigma$

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{3-1.5\sigma -3}{\sigma } \\ =\frac{-1.5\sigma }{\sigma } \\ =-1.5 \end{gathered}$$

The probability is,

$$\begin{gathered} P(\mu -1.5\sigma <x<\mu +1.5\sigma )=P(3-1.5\sigma <x<3+1.5\sigma ) \\ =P(x<3+1.5\sigma )-P(x<31.5\sigma ) \\ =P(z<1.5)-(1-P(z<1.5\left)\right) \\ =0.9332-1+0.9332 \\ =0.8664 \\ \approx 0.87 \\ P(\mu -1.5\sigma <x<\mu +1.5\sigma )=0.87a \end{gathered}$$

Therefore, there is an 87% chance that the goal will be met