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Sagot :
Answer:
0.8029 = 80.29% probability that the thickness is between 3.0 and 7.0 mm.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Mean of 4.6 millimeters (mm) and a standard deviation of 1.5 mm.
This means that [tex]\mu = 4.6, \sigma = 1.5[/tex]
Find the probability that the thickness is between 3.0 and 7.0 mm.
This is the pvalue of Z when X = 7 subtracted by the pvalue of Z when X = 3. So
X = 7
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{7 - 4.6}{1.5}[/tex]
[tex]Z = 1.6[/tex]
[tex]Z = 1.6[/tex] has a pvalue of 0.9452
X = 3
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3 - 4.6}{1.5}[/tex]
[tex]Z = -1.07[/tex]
[tex]Z = -1.07[/tex] has a pvalue of 0.1423
0.9452 - 0.1423 = 0.8029
0.8029 = 80.29% probability that the thickness is between 3.0 and 7.0 mm.
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