Question

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

Answer 1

k = 76.67

Explanation:

P(75.35< x<k) = 0.2

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<z<{(k-75.5)/3.5} = 0.2

P(z<{(k-75.5)/3.5}) - 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