Q.If $X \sim N(50, 100)$ — i.e. a normal distribution with mean 50 and variance 100 — find the standard normal variate $z$ corresponding to $x = 65$, and hence find $P(X < 65)$.

Since the variance is 100, the standard deviation is $\sigma = \sqrt{100} = 10$. Standardizing $x=65$ against $\mu=50,\sigma=10$ converts the problem into a standard-table lookup.

Step 1 — Identify parameters. $\mu = 50$, variance $=100 \Rightarrow \sigma = \sqrt{100} = 10$.

Step 2 — Standardize.

Step 3 — Read the table (Section 4). $P(X<65) = P(Z<1.5) = \Phi(1.5) = 0.9332$.

Independent second method (dual-solve check): using the complement rule, $P(X>65) = 1-\Phi(1.5) = 1-0.9332 = 0.0668$, i.e. only 6.68% of the distribution lies above $x=65$. Since $z=1.5$ is well past $+1\sigma$ (68.27% within $\pm1\sigma$, leaving 15.865% beyond $+1\sigma$ alone) but has not yet reached $+2\sigma$ (95.45% within $\pm2\sigma$, leaving only 2.275% beyond $+2\sigma$), an upper-tail probability of 6.68% is exactly consistent with $z=1.5$ sitting between those two benchmarks — confirming the table read is correct.

Alternative check: verify z arithmetic by reversing it — x = μ + zσ = 50 + 1.5(10) = 65, which matches the given x, confirming no sign or arithmetic slip in the standardization step.

A common slip is standardizing using $\sigma^2$ (the variance, 100) instead of $\sigma$ (the SD, 10) in the denominator — always take the square root of a stated variance before dividing. Another slip is reading $\Phi(1.5)$ as the area in the RIGHT tail instead of the LEFT — the table always gives the cumulative area to the left of z.