## An oil exploration company purchases drill bits that have a life span that is approximately normally distributed with a mean equal to 74 hou

Question

An oil exploration company purchases drill bits that have a life span that is approximately normally distributed with a mean equal to 74 hours and a standard deviation equal to 12 hours. (Round your answers to four decimal places.) (a) What proportion of the company’s drill bits will fail before 65 hours of use? (b) What proportion will last at least 65 hours? (c) What proportion will have to be replaced after more than 83 hours of use?

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1 month 2021-09-13T21:11:40+00:00 1 Answer 0

a) 0.2266 = 22.66% of the company’s drill bits will fail before 65 hours of use.

b) 0.7734 = 77.34% will last at least 65 hours

c) 0.2266 = 22.55% will have to be replaced after more than 83 hours of use

Step-by-step explanation:

Problems of normally distributed samples are solved using 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.

In this problem, we have that: (a) What proportion of the company’s drill bits will fail before 65 hours of use?

This is the pvalue of Z when X = 65.    has a pvalue of 0.2266

0.2266 = 22.66% of the company’s drill bits will fail before 65 hours of use.

(b) What proportion will last at least 65 hours?

This is 1 subtracted by the pvalue of Z when X = 65.    has a pvalue of 0.2266

1 – 0.2266 = 0.7734

0.7734 = 77.34% will last at least 65 hours

(c) What proportion will have to be replaced after more than 83 hours of use?

This is 1 subtracted by the pvalue of Z when X = 65.    has a pvalue of 0.7734

1 – 0.7734 = 0.2266

0.2266 = 22.55% will have to be replaced after more than 83 hours of use