1. a) 0.3174 = 31.74% probability of a defect
1. b) The expected number of defects for a 1,000-unit production run is of 317.
2. a) 0.0026 = 0.26% probability of a defect
2. b) The expected number of defects for a 1,000-unit production run is of 3.
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean and standard deviation , the zscore of a measure X is given by:
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.
Question 1:
We have that:
a. Calculate the probability of a defect.
Less than 9.88 or greater than 10.12. These probabilities are equal, so we find one and multiply by 2.
Probability of less than 9.88:
This is the pvalue of Z when X = 9.88. So
has a pvalue of 0.1587
2*0.1587 = 0.3174
0.3174 = 31.74% probability of a defect
b. Calculate the expected number of defects for a 1,000-unit production run.
The expected number of defects is 31.74% of 1000. So
0.3174*1000 = 317.4
Rounding to the nearest integer
The expected number of defects for a 1,000-unit production run is of 317.
Question 2:
The mean remains the same, but the standard deviation is now
a. Calculate the probability of a defect.
Less than 9.88 or greater than 10.12. These probabilities are equal, so we find one and multiply by 2.
Probability of less than 9.88:
This is the pvalue of Z when X = 9.88. So
has a pvalue of 0.0013
2*0.0013 = 0.0026
0.0026 = 0.26% probability of a defect
b. Calculate the expected number of defects for a 1,000-unit production run.
The expected number of defects is 31.74% of 1000. So
0.0026*1000 = 2.6
Rounding to the nearest integer
The expected number of defects for a 1,000-unit production run is of 3.
