Problem 9-10 The elongation of a steel bar under a particular tensile load may be assumed to be normally distributed, with a mean of .06 in. and standard deviation of .008 in. A sample of n=100 bars is subjected to the test. Find the probability that the sample mean elongation is between .0585 in. and .0605 in.

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funkeyusuff52 funkeyusuff52 · Answer 1 · 2019-11-10T18:45:56+01:00

Answer:

The answer is 0.7036.

Step-by-step explanation:

Check the attached file for the computations.

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raphealnwobi raphealnwobi · Answer 2 · 2022-04-12T12:34:59+02:00

The probability that the mean life of a random sample mean elongation is between .0585 in. and .0605 in. is 70.57%

Z score is used to determine by how many standard deviations the raw score is above or below the mean.

It is given by:

z = (raw score - mean) / (standard deviation÷√sample)

Mean = 0.06, standard deviation = 0.008, sample = 100.

For x = 0.0585:

z = (0.0585 - 0.06)/ (0.008 ÷√100) = -1.88

For x = 7.2:

z = (0.0605 - 0.06)/ (0.008 ÷√100) = 0.623

P(-1.88 < z < 0.63) = P(z < 0.63) - P(z < -1.88) = 0.7357 - 0.03 = 0.7057

The probability that the mean life of a random sample mean elongation is between .0585 in. and .0605 in. is 70.57%

Find out more on z score at: https://brainly.com/question/25638875