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 less than 67.5?

Answer 1

Let X = the population random variable

$\mu$ = E(X) = 70 and $\sigma$² = Var(X) = 6² =36

Let $\bar{x}$ be the mean of the random sample of size n taken from this population.

By Central Limit Theorem we know that $\bar{x}$ ~ N(μ, σ²/n) asymptotically i.e. for large n.

Therefore, Z = $\frac{(\bar{x}-\mu)}{\sigma/\sqrt{n}}$ ~ N(0, 1) asymptotically.

Here, n = 36

Now, the probability that the sample mean is less than 67.5

= P($\bar{x}$ < 67.5)

= P($\frac{(\bar{x}-70)}{6/\sqrt{36}}$ < $\frac{67.5-70}{6/\sqrt{36}}$)

= P(Z < - 2.5)

= $\Phi$(- 2.5)

= 0.0062 [obtained from standard normal distribution table]

Answer: The probability that the sample mean is less than 67.5 is 0.0062.