Answer to Question #333599 in Statistics and Probability for tea

Question

A random sample of size 36 was taken from the population with a mean of 3.4 and a variance of 4. Approximately the probability that the sample mean is greater than 3.0 but less than 3.6

Answer 1

Let $X=$ the sample mean, $X\sim N(\mu, \sigma^2/n).$

$$P(3.0<X<3.6)=P(X<3.6)-P(X\le 3.0)$$

$$=P(Z<\dfrac{3.6-3.4}{\sqrt{4}/\sqrt{36}})-P(Z\le\dfrac{3.0-3.4}{\sqrt{4}/\sqrt{36}})$$

$$=P(Z<0.6)-P(Z\le-1.2)$$

$$\approx0.7257469-0.1150697$$

$$\approx 0.610677$$