Question

Thompson and thompson is a steel bolts manufacturing company. their current steel bolts have a mean diameter of 134 millimeters, and a variance of 36. if a random sample of 44 steel bolts is selected, what is the probability that the sample mean would be greater than 135.4 millimeters? round your answer to four decimal places.

Answer 1

We are required to evaluate the following probability that the mean would be greater than 135.4 mm.
the z-score is given by:
z=(x-μ)/σ
where:
μ- mean
σ-standard deviation
thus
z=(135.4-134)/36
z=0.03889

hence:
P(X≥135.4)=1-P(Z=0.03889)=1-0.4840=0.516~51.6%

Answer: 0.516~51.6%

Answer 2

The correct answer is:

0.0606.

Explanation:

The z-score is given by the formula

$z=\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}$,

where X is the value we're looking at, μ is the mean, σ is the standard deviation, and n is the sample size.

We have the variance instead of the standard deviation. The variance is the square of the standard deviation, so we find σ by taking the square root: 
√36=6.

We now have:
$z=\frac{135.4-134}{\frac{6}{\sqrt{44}}}=\frac{1.4}{\frac{6}{\sqrt{44}}}=0.9394$

This gives us the area to the left of this score, which is the probability that the mean is less than the score. We want the probability that the mean is greater than the score; this means we subtract from 1:

1-0.9394 = 0.0606.