Question

Consider a population with a mean of 70 and a standard deviation of 6. A random sample of size
36 is drawn. What is the probability that the sample mean is between 70:5 and 71:5 ?

Answer 1

If $n>30,$ the Central Limit Theorem can be used.

Let $X_1,X_2,...,X_n$ be a random sample from a distribution with mean $\mu$ and variance $\sigma^2.$ Then if $n$ is sufficiently large, has approximately a normal distribution with $\mu_{\bar{X}}=\mu$ and $\sigma_{\bar{X}}=\sigma^2/n.$

Give that $n=36, \mu=70, \sigma=6.$

$n=36>30=>X\sim(\mu;\sigma^2/n)$. Then

$$Z={\bar{X}-\mu \over \sigma/\sqrt{n}}\sim N(0;1)$$

$$P(70.5<\bar{X}<71.5)=P(Z<{71.5-70\over 6/\sqrt{36}})-P(Z<{70.5-70\over 6/\sqrt{36}})=$$

$$=P(Z<{1.5})-P(Z<0.5)\approx0.93319-0.69146\approx0.2417$$

The probability that the sample mean is between $70.5$ and $71.5$ is $0.2417.$