Answer :
Answer:
C. 317.4
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]\sigma = 0.15[/tex]
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 [tex]\mu = 9.85 + 0.15 = 10.15 - 0.15 = 10[/tex]
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
[tex]P = P(X \geq 10.15) + P(X \leq 9.85)[/tex]
[tex]P(X \geq 10.15)[/tex]
1 subtracted by the pvalue of Z when X = 10.15
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{10.15 - 10}{0.15}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413
1 - 0.8413 = 0.1587
So [tex]P(X \geq 10.15) = 0.1587[/tex]
[tex]P(X \leq 9.85)[/tex]
pvalue of Z when X = 9.85
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{9.85 - 10}{0.15}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587
Probability of a unit being defective
[tex]P = P(X \geq 10.15) + P(X \leq 9.85) = 0.1587 + 0.1587 = 0.3174[/tex]
Expected number for 1000
[tex]E(X) = np = 1000*0.3174 = 317.4[/tex]
So the correct answer is:
C. 317.4