8.1 - 8.2: Introduction, Sampling Distribution of the
Sample Mean
Case 1: Normal Population with $\sigma$ known
1. A simple random sample of size $n = 8$ is taken from a population with mean 40 and
variance 12. The population distribution is assumed to be Normal. Let $\bar{X}$ be the
mean of the sample. Answer the following questions.
(a) What is the distribution of $\bar{X}$?
(b) Find $P(\bar{X} > 42)$.
(c) Suppose you need to choose a sample of size $n$ such that $P(\bar{X} > 42) < 0.025$.
What is the minimum $n$ required to achieve this?
Solution:
(a) $\bar{X} \sim N(40, \frac{12}{8}) = N(40, 1.5)$.
(b) $P(\bar{X} > 42) = P(Z > \frac{42 - 40}{\sqrt{1.5}}) = P(Z > 1.63) = 0.052$.
(c) $P(\bar{X} > 42) = P(Z > \frac{42 - 40}{\sqrt{12/\sqrt{n}}}) = 0.025$, implies, using $z_{0.025} = 1.96$, that
$\frac{42 - 40}{\sqrt{12/n}} = 1.96$. Solving for $n$ from this equation we get $n = 11.52$. Hence we
need at least a sample size of 12 to achieve the requirement.
