Answer:
to four decimal places.
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:
Using the Excel Formula ( = NORMDIST (1) ) to calculate for the value of z =1 and -1 ;we have: 0.841345 and 0.158655 respectively
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:
Using the Excel Formula ( = NORMDIST (3) ) to calculate for the value of z =3 and -3 ;we have: 0.99865 and 0.00135 respectively
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.