A synthetic fiber used in manufacturing carpet has tensile strength that is normally distributed with mean 75.5 psi and standard deviation 3.5 psi. For a random sample of n = 12 fiber specimens, find the value of k, such that P (75.35

Question

Tessa Gordon

Mathematics

8 April, 03:24

A synthetic fiber used in manufacturing carpet has tensile strength that is normally distributed with mean 75.5 psi and standard deviation 3.5 psi. For a random sample of n = 12 fiber specimens, find the value of k, such that P (75.35

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Answers (1)

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Graham Rich · Answer 1

k = 76.67

Step-by-step explanation:

P (75.35< x
Let us calculate the z-score for 75.35

z-score = (x-mean) / SD

From the question, mean = 75.5 and SD = 3.5

z-score = (75.35-75.5) / 3.5

z-score = - 0.0429

Let us calculate the z-score for k

z-score = (k-75.5) / 3.5

Thus, we have the probability range as;

P (-0.0429
P (z-0.0429) = 0.2

Let's say P (z<{ (k-75.5) / 3.5}) is F to combat ambiguity

from z score table P (z>-0.0429) = 0.48289

Hence,

F - 0.48291 = 0.2

F = 0.2 + 0.48291

F = 0.68291

From z-table, the z-score of is 0.476

Hence:

0.476 = (k - 75.5) / 3.5

k-75.5 = 3.5 (0.476)

k-75.5 = - 1.666

k = 75+1.666

k = 76.67