MATH SOLVE

7 months ago

Q:
# Thompson and thompson is a steel bolts manufacturing company. their current steel bolts have a mean diameter of 128128 millimeters, and a standard deviation of 88 millimeters. if a random sample of 3131 steel bolts is selected, what is the probability that the sample mean would differ from the population mean by more than 0.90.9 millimeters? round your answer to four decimal places.

Accepted Solution

A:

First, let's make clear the numbers:

m = 128 mm

σ = 8 mm

n = 31

X - m = 0.9 mm

In order to find the probability requested, you need to compute the z-scores:

z = (X - m) / σ

Therefore:

z₁ = -0.9 / 8

= -0.1125

z₂ = 0.9 / 8

= 0.1125

Now, consider that:

p(z < z₁; z > z₂) = p(z < z₁) + p(z > z₂)

= p(z < z₁) + [1 - p(z < z₂)]

Look at a normal standard table to find the probabilities connected to the calculated z-scores (rounded to two decimal places):

p(z < -0.11) = 0.45620

p(z < +0.11) = 0.54380

Now, you can compute the total probability:

p(z < z₁; z > z₂) = 0.45620 + ( 1 - 0.54380)

= 0.9124

Therefore, the probability that the sample mean would differ from the population mean by more than 0.9 millimeters is p = 0.9124 which means 91.24%.

m = 128 mm

σ = 8 mm

n = 31

X - m = 0.9 mm

In order to find the probability requested, you need to compute the z-scores:

z = (X - m) / σ

Therefore:

z₁ = -0.9 / 8

= -0.1125

z₂ = 0.9 / 8

= 0.1125

Now, consider that:

p(z < z₁; z > z₂) = p(z < z₁) + p(z > z₂)

= p(z < z₁) + [1 - p(z < z₂)]

Look at a normal standard table to find the probabilities connected to the calculated z-scores (rounded to two decimal places):

p(z < -0.11) = 0.45620

p(z < +0.11) = 0.54380

Now, you can compute the total probability:

p(z < z₁; z > z₂) = 0.45620 + ( 1 - 0.54380)

= 0.9124

Therefore, the probability that the sample mean would differ from the population mean by more than 0.9 millimeters is p = 0.9124 which means 91.24%.