# Thread: Finding probability that sample mean differs from true mean?

1. ## Finding probability that sample mean differs from true mean?

I have the following problem:

"The breaking strength of a safety belt is known to be approximately normally distributed with a true mean (mu) of 8100 pounds and a true standard deviation (sigma) of 105 pounds. If a simple random sample of 40 belts is chosen, what is the probability that the sample mean (x-bar) will differ from the true mean (mu) by more than 30 pounds (in either direction)?"

I have an abnormally high z score, so I don't think I am doing it correctly.
n = 40 mu = 8100 sigma = 105
mu x-bar = 40
sigma x-bar = sigma/mu x-bar = 105/40 = 2.625

P(8070 < x-bar < 8130) = P((8070-40)/2.625 < z < (8130-40)/2.625) =
P(3059.05 < z < 3081.9)

Where have I gone wrong?

2. Originally Posted by oxyron
I have the following problem:

"The breaking strength of a safety belt is known to be approximately normally distributed with a true mean (mu) of 8100 pounds and a true standard deviation (sigma) of 105 pounds. If a simple random sample of 40 belts is chosen, what is the probability that the sample mean (x-bar) will differ from the true mean (mu) by more than 30 pounds (in either direction)?"

I have an abnormally high z score, so I don't think I am doing it correctly.
n = 40 mu = 8100 sigma = 105
mu x-bar = 40 Mr F says: Are you seriously trying to say that the mean of the sample mean is 40!!!

sigma x-bar = sigma/mu x-bar = 105/40 = 2.625

P(8070 < x-bar < 8130) = P((8070-40)/2.625 < z < (8130-40)/2.625) =
P(3059.05 < z < 3081.9)

Where have I gone wrong?
$\bar{X}$ ~ Normal $\left(\mu = 8100, \sigma = \frac{105}{\sqrt{40}}\right)$.

Calculate $\Pr(\bar{X} < 8070) + \Pr(\bar{X} > 8130) = 2\Pr(\bar{X} > 8130)$.