n automatic machine in a manufacturing process is operating properly if the lengths of an important subcomponent are normally distributed with a mean of 116 cm and a standard deviation of 5.4 cm. A. Find the probability that one selected subcomponent is longer than 118 cm. Probability

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Answer:

0.3557 = 35.57% probability that one selected subcomponent is longer than 118 cm.

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.

Normally distributed with a mean of 116 cm and a standard deviation of 5.4 cm.

This means that [tex]\mu = 116, \sigma = 5.4[/tex]

Find the probability that one selected subcomponent is longer than 118 cm.

This is 1 subtracted by the pvalue of Z when X = 118. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{118 - 116}{5.4}[/tex]

[tex]Z = 0.37[/tex]

[tex]Z = 0.37[/tex] has a pvalue of 0.6443

1 - 0.6443 = 0.3557

0.3557 = 35.57% probability that one selected subcomponent is longer than 118 cm.