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Sagot :
Answer:
84.13% of the shipment is defective
Step-by-step explanation:
When the distribution is normal, we use 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.
Normally distributed with mean 39.3 and standard deviation 0.2
This means that [tex]\mu = 39.3, \sigma = 0.2[/tex]
What percentage of the shipment is defective
Less than 39.5 or greater than 40.5.
Less than 39.5:
This is the pvalue of Z when X = 39.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{39.5 - 39.3}{0.2}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413.
Greater than 40.5:
1 subtracted by the pvalue of Z when X = 40.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{40.5 - 39.3}{0.2}[/tex]
[tex]Z = 6[/tex]
[tex]Z = 6[/tex] has a pvalue of 1
1 - 1 = 0
Total:
0.8413 + 0 = 0.8413
84.13% of the shipment is defective
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