![](https://tex.z-dn.net/?f=)
Given:
population mean, μ =135
population standard deviation, σ = 15
sample size, n = 19
Assume a large population, say > 100,
we can reasonably assume a normal distribution, and a relatively small sample.
The use of the generally simpler formula is justified.
Estimate of sample mean
![\bar{x}=\mu=135](https://tex.z-dn.net/?f=%5Cbar%7Bx%7D%3D%5Cmu%3D135)
Estimate of sample standard deviation
![\s=\sqrt{\frac{\sigma^2}{n}}](https://tex.z-dn.net/?f=%5Cs%3D%5Csqrt%7B%5Cfrac%7B%5Csigma%5E2%7D%7Bn%7D%7D)
![=\sqrt{\frac{15^2}{19}}=3.44124](https://tex.z-dn.net/?f=%3D%5Csqrt%7B%5Cfrac%7B15%5E2%7D%7B19%7D%7D%3D3.44124)
to 5 decimal places.
Thus, using the normal probability table,
![P(125](https://tex.z-dn.net/?f=P%28125%3CX%3C130%29)
![=P(\frac{125-135}{3.44124}](https://tex.z-dn.net/?f=%3DP%28%5Cfrac%7B125-135%7D%7B3.44124%7D%3CZ%3C%5Cfrac%7B130-135%7D%7B3.44124%7D%29)
![=P(-2.90593](https://tex.z-dn.net/?f=%3DP%28-2.90593%3CZ%3C-1.45297%29)
![=P(Z](https://tex.z-dn.net/?f=%3DP%28Z%3C-2.90593%29%3D0.0018308)
![=P(Z](https://tex.z-dn.net/?f=%3DP%28Z%3C-1.45297%29%3D0.0731166)
Therefore
The probability that the mean weight is between 125 and 130 lbs
P(125<X<130)=0.0731166-0.0018308
=
0.0712858