Answer to Question #293915 in Statistics and Probability for lyva

Question

A population has a mean of 75 and a standard deviation of 17.4. A sample of 35 items is randomly selected from this population. What is the probability that the mean of this samples lies between 72 and 79? 

Answer 1

Let $X=$ the mean of the samples: $X\sim N(\mu, \sigma^2/n)$

Given $\mu=75, \sigma=17.4, n=35$

$$P(72<X<79)=P(X<79)-P(X\le 72)$$

$$=P(X<\dfrac{79-75}{17.4/\sqrt{35}})-P(X\le \dfrac{72-75}{17.4/\sqrt{35}})$$

$$\approx P(X<1.360018)-P(X\le-1.020014)$$

$$\approx0.91309-0.15386=0.7592$$