A synthetic fiber used in manufacturing carpet has tensile strength that is norm

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. Find the probability that a random sample of n = 6 fiber specimens will have sample mean tensile strength that exceeds 75.75 psi. A random sample has been taken from a normal distribution and the following confidence 2 intervals constructed using the same data: (38.53, 48.87) and (36.59, 50.81) What is the value of the sample? One of these intervals is 99% CI and the other is a 95% CI. Which one is the 95% CI and why?

Explanation / Answer

Solution

Back-up Theory

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then

Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)

P(X or t) = P[{(X - µ)/ } or {(t - µ)/ }] = P[Z or {(t - µ)/ }] .………(2)

X bar ~ N(µ, 2/n),…………………………………………………………….…….(3),

where X bar is average of a sample of size n from population of X.

So, P(X bar or t) = P[Z or {(n)(t - µ)/ }] …………………………………(4)

Q1

Let X = tensile strength (in psi). Then, we are given X ~ N(75.5, 3.52).

So, by (3) above, X bar ~ N(75.5, 3.52/6).

Now, probability that a sample of 6 has mean tensile strength exceeding 75.75 psi =

P(X bar 75.75) = P[Z {(6)(75.75 – 75.5)/3.5}] [by (4) above]

= P(Z 0.1750) = 0.4305 [using Excel Function] ANSWER

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.