Question

Suppose that Motorola uses the normal distribution to determine the probability of defects and the number of defects in a particular production process. Assume that the production process manufactures items with a mean weight of 10 ounces. Calculate the probability of a defect and the suspected number of defects for a 1,000-unit production run in the following situations.
(a) The process standard deviation is 0.15, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects. If required, round your answer to four decimal places.
(b) Through process design improvements, the process standard deviation can be reduced to 0.05. Assume that the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects. If required, round your answer to four decimal places.
(c) What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean?

Answer 1

Answer:

$\mathbf{P(X < 9.85 \ or \ X> 10.15) \approx 0.3171}$     to four decimal places.

$\mathbf{P(X < 9.85 \ or \ X> 10.15) =0.0027}$  to four decimal places.

Step-by-step explanation:

a)

Assuming X to be the random variable which replace the amount of defectives and follows standard normal distribution whose mean  (μ) is 10 ounces and standard deviation (σ) is 0.15

The values of the random variable differ from mean  by ±  1 \such that the values are either greater than (10+ 0.15) or less than  (10-0.15)

= 10.15 or  9.85.

The  probability that the amount of defectives which are either greater than 10.15 or less than 9.85 can be calculated as follows:

$P(X < 9.85 \ or \ X> 10.15) = 1-P ( \dfrac{9.85-10}{0.15}< \dfrac{X-10}{0.15}< \dfrac{10.15-10}{0.15})$

$P(X < 9.85 \ or \ X> 10.15) = 1- \phi (1) - \phi (-1)$

Using the Excel Formula ( = NORMDIST (1) ) to calculate for the value of z =1 and -1 ;we have: 0.841345 and 0.158655 respectively

$P(X < 9.85 \ or \ X> 10.15) = 1- (0.841345-0.158655)$

$P(X < 9.85 \ or \ X> 10.15) =0.31731$

$\mathbf{P(X < 9.85 \ or \ X> 10.15) \approx 0.3171}$     to four decimal places.

b) Through process design improvements, the process standard deviation can be reduced to 0.05.

The  probability that the amount of defectives which are either greater than 10.15 or less than 9.85 can be calculated as follows:

$P(X < 9.85 \ or \ X> 10.15) = 1-P ( \dfrac{9.85-10}{0.05}< \dfrac{X-10}{0.05}< \dfrac{10.15-10}{0.05})$

$P(X < 9.85 \ or \ X> 10.15) = 1- \phi (3) - \phi (-3)$

Using the Excel Formula ( = NORMDIST (3) ) to calculate for the value of z =3 and -3 ;we have: 0.99865 and 0.00135 respectively

$P(X < 9.85 \ or \ X> 10.15) = 1- (0.99865-0.00135)$

$\mathbf{P(X < 9.85 \ or \ X> 10.15) =0.0027}$  to four decimal places.

(c)  What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean?

The main advantage of reducing the process variation is that the chance of getting the defecting item will be reduced as we can see from the reduction which takes place from a to b from above.