staticsCentral Limit Theorem with Mea

Question

An operation manager at an electronics company wants to test their amplifiers. The design engineer claims they have a mean output of 145 watts with a variance of 144.

What is the probability that the mean amplifier output would be greater than 146.9 watts in a sample of 97 amplifiers if the claim is true? Round your answer to four decimal places.

Answer 1

The probability of a value less than x is the cumulative distribution function φ(x) = (1 + erf((x – μ)/√2σ))/2, so the probability of a value greater than x is 1 – φ(x).

Answer 2

We have a population with mean μ = 145 and var = 144, thus σ = √144 = 12.

We want to find the probability that the mean of sample of 97 objects, Xmean, is higher than 146.9:

p(Xmean> 146.9) = ?

We will compute the Z score and use the Z tables (https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf)

Since the probability values provided in the Z tables above correspond to the probability of the sample mean falling to the left of the Z scores (lower), we will first write:

P(Xmean > 146.9) = 1- P(Xmean <=146.9)

The corresponding Z score = (Xmean- μ) / (σ/√n) = (146.9-145)/ (12/√97) = 1.9/12 * √97 = 1.559 ~ 1.56

From the Z table, we read p(z<1.56) = 0.94062

And thus P(Xmean >146.9) = 1- 0.9406 = 0.0594.

Does this value make sense? It may seem like a low probability for our sample mean being "just a bit higher" (146.9) than the population mean (145); however this makes sense in light of the sample size being large (97) for the claimed standard deviation (12).