Q.$X$ is normally distributed with mean 200 and standard deviation 5, i.e. $X \sim N(200, 25)$. Find the value of $x$ such that $P(X > x) = 0.10$.

This is a reverse problem: we are given a probability (10% in the upper tail) and must find the corresponding raw value $x$, using $x = \mu + z\sigma$. …

Step 1 — Convert the given probability into a cumulative (left-side) area. $P(X>x)=0.10 \Rightarrow P(X<x) = 1-0.10 = 0.90$, so we need the $z$ for which $\Phi(z) = 0.90$.

Step 2 — Look up this z in the table (Section 4). $\Phi(1.28) = 0.8997 \approx 0.90$, so $z = 1.28$.

Step 3 — Reverse the standardization formula.

$x = \mu + z\sigma = 200 + (1.28)(5) = 200 + 6.4 = 206.4$ …

Using z=1.28 directly as if it were the answer (forgetting to convert back to the original x-scale via x = μ + zσ), or looki …