a weekly wage is a random variable X, $σ^2$ = 64, $µ$ = 75, p ≥ 0.9
use CLT to find the minimum sample size such that with probability at least 0.9, the average wage of this group is between 70 and 80.
so far I have:
P(70 ≤ X ≤ 80) ≥ 0.9 and I worked with this until I got to
P(-5 ≤ (X-µ) ≤ 5) ≥ 0.9
is this correct so far? if so where do I go from here?
edit
following on, I get
P(X ≤ |40√n|) ≥ 0.9
but I am unsure on how to solve this
Let $X$ be the sample mean. Using CLT, we have that $X$ is approximately a Gaussian random variable with mean equal to $\mu = 75$ and variance equal to $\sigma^2/n = 64/n$. We need to find minimum $n$ such that $\Pr(70 \leq X \leq 80) \geq 0.9$. Thus, we have,
\begin{align*} \Pr(70 \leq X \leq 80) & \geq 0.9 \\ \implies \Pr(-5 \leq X - 75 \leq 5) & \geq 0.9 \\ \implies \Pr\left(-\frac{5\sqrt{n}}{8} \leq \frac{X - 75}{8/\sqrt{n}} \leq \frac{5\sqrt{n}}{8}\right) & \geq 0.9 \\ \end{align*}
Since $X \sim \mathcal{N}(75, 8/\sqrt{n})$, $(X - 75)/(8/\sqrt{n}) \sim \mathcal{N}(0,1)$, a standard normal random variable, $Z$. Hence, we need to find $n$ such that \begin{align*} \Pr(-0.625 \sqrt{n} \leq Z \leq 0.625 \sqrt{n}) & \geq 0.9 \\ \implies 1 - 2 \Phi(-0.625 \sqrt{n}) & \geq 0.9 \\ \implies \Phi(-0.625 \sqrt{n}) & \leq 0.05 \\ \implies n & \geq \left(\frac{\Phi^{-1}(0.05)}{0.625} \right)^2. \end{align*}
Using standard tables, we obtain $n \geq 6.9262$.