Question

Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 13 ounces.

a) The process standard deviation is 0.1, and the process control is set at plus or minus 1.5 standard deviations. Units with weights less than 12.85 or greater than 13.15 ounces will be classified as defects. What is the probability of a defect (to 4 decimals)?

b)In a production run of 1000 parts, how many defects would be found (to 0 decimals)?

c) Through process design improvements, the process standard deviation can be reduced to 0.05. Assume the process control remains the same, with weights less than 12.85 or greater than 13.15 ounces being classified as defects. What is the probability of a defect (rounded to 4 decimals; getting the exact answer, although not necessary, will require Excel)?

d)In a production run of 1000 parts, how many defects would be found (to 0 decimals)?

Answer 1

Concepts and reason

Normal distribution:

Normal distribution is a continuous distribution of data that has the bell-shaped curve. The normally distributed random variable X has meanand standard deviation.

Also, the standard normal distribution represents a normal curve with mean 0 and standard deviation 1. Thus, the parameters involved in a normal distribution are mean and standard deviation.

Standardized z-score:

The standardized z-score represents the number of standard deviations the data point is away from the mean.

• If the z-score takes positive value when it is above the mean.

• If the z-score takes negative value when it is below the mean

Fundamentals

Let , then the standard z-score is found using the formula given below:

Where X denotes the individual raw score, denotes the population mean, and denotes the population standard deviation.

Procedure for finding the z-value is listed below:

1.From the table of standard normal distribution, locate the probability value.

2.Move left until the first column is reached.

3.Move upward until the top row is reached.

4.Locate the probability value, by the intersection of the row and column values gives the area to the left of z.

• Formula for finding the value of is, .

• Formula for finding the value of is,

.

(a)

The probability of a defect is obtained as shown below:

From the given information, a production process produces items with a mean weight of 13 ounces and process standard deviation is 0.1. That is, .

The required probability is,

From the “Standard normal table”, the area to the left of is 0.0668 and the area to the left of is 0.9331.

(b)

The numbers of defects in a production run is obtained below:

The required defects is,

(c)

The probability of a defect is obtained as shown below:

From the information given, the standard deviation is 0.05.

The required probability is,

From the “Standard normal table”, the area to the left of is 0.0013 and the area to the left of is 0.9986.

(d)

The numbers of defects in a production run is obtained below:

The required defects is,

Ans: Part a

Thus, the probability of a defect is 0.1337.