[tex][/tex]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
[tex]\bar{x}=\mu=135[/tex]
Estimate of sample standard deviation
[tex]\s=\sqrt{\frac{\sigma^2}{n}}[/tex]
[tex]=\sqrt{\frac{15^2}{19}}=3.44124[/tex] to 5 decimal places.
Thus, using the normal probability table,
[tex]P(125<X<130)[/tex]
[tex]=P(\frac{125-135}{3.44124}<Z<\frac{130-135}{3.44124})[/tex]
[tex]=P(-2.90593<Z<-1.45297)[/tex]
[tex]=P(Z<-2.90593)=0.0018308[/tex]
[tex]=P(Z<-1.45297)=0.0731166[/tex]
Therefore
The probability that the mean weight is between 125 and 130 lbs
P(125<X<130)=0.0731166-0.0018308
=0.0712858
