Question

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. Calculate the expected number of defects for a 1000-unit production.

A. 524.8
B. 742.1
C. 317.4
D. 500

Answer 1

To calculate the expected number of defects for a 1000-unit production, we need to determine the proportion of units that will be classified as defects. The process control is set at plus or minus one standard deviation, with units outside the acceptable range being considered defects. Using the normal distribution, we can calculate the proportion of units falling outside the acceptable range and then multiply it by the total number of units. The expected number of defects is 317.4.

Step-by-step explanation:

To calculate the expected number of defects for a 1000-unit production, we need to determine the proportion of units that will be classified as defects. In this case, units with weights less than 9.85 or greater than 10.15 ounces are considered defects.

The process control is set at plus or minus one standard deviation, which means the acceptable range for weights is 9.85 plus or minus 0.15 ounces.

Let's calculate the proportion of units that fall outside this range:

Likelihood of units being classified as defects = P(X < 9.85 or X > 10.15)

Using the normal distribution, we can calculate the z-scores for 9.85 and 10.15:

Z-score for 9.85 = (9.85 - 10) / 0.15 = -1.00

Z-score for 10.15 = (10.15 - 10) / 0.15 = 1.00

The area under the standard normal curve to the left of -1.00 is approximately 0.1587, and the area under the standard normal curve to the right of 1.00 is also approximately 0.1587.

Therefore, the likelihood of a unit being classified as a defect is 0.1587 + 0.1587 = 0.3174.

The expected number of defects for a 1000-unit production can be calculated by multiplying the proportion of defects by the total number of units:

Expected number of defects = 0.3174 * 1000 = 317.4

Therefore, the expected number of defects for a 1000-unit production is 317.4.

Answer 2

C. 317.4

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\sigma = 0.15`

Tthe 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.

So
`\mu = 9.85 + 0.15 = 10.15 - 0.15 = 10`

Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects. Calculate the expected number of defects for a 1000-unit production.

The first step is finding the probability that a unit is defective.

This is

`P = P(X \geq 10.15) + P(X \leq 9.85)`

`P(X \geq 10.15)`

1 subtracted by the pvalue of Z when X = 10.15

`Z = (X - \mu)/(\sigma)`

`Z = (10.15 - 10)/(0.15)`

`Z = 1`

`Z = 1` has a pvalue of 0.8413

1 - 0.8413 = 0.1587

So
`P(X \geq 10.15) = 0.1587`

`P(X \leq 9.85)`

pvalue of Z when X = 9.85

`Z = (X - \mu)/(\sigma)`

`Z = (9.85 - 10)/(0.15)`

`Z = -1`

`Z = -1` has a pvalue of 0.1587

Probability of a unit being defective

`P = P(X \geq 10.15) + P(X \leq 9.85) = 0.1587 + 0.1587 = 0.3174`

Expected number for 1000

`E(X) = np = 1000*0.3174 = 317.4`

So the correct answer is:

C. 317.4