AHL 4.15—Central limit theorem

Since $n=64>30$, the CLT applies, so $\bar{X}\sim N\left(500,\frac{64}{64}\right)=N(500,1)$.

$\text{SE}=\frac{\sigma }{\sqrt{n}}=\frac{8}{\sqrt{64}}=\frac{8}{8}=1$.

$Z=\frac{502-500}{1}=2$.

$P(\bar{X}>502)=P(Z>2)=1-0.9772=0.0228$.

$0.4013$ corresponds to using $Z=0.25$, likely from forgetting to divide by $\sqrt{n}$ and using $\sigma =8$ directly: $Z=2/8=0.25$.

$0.1587$ corresponds to $Z=1$, which would result from an arithmetic slip dividing $2/2$ instead of using the correct SE.

$0.3085$ corresponds to $Z=0.5$, another mis-computed standard error.

Only $0.0228$ correctly uses the standard error of $1$.

The CLT states $\bar{X}\sim N\left(\mu ,\frac{{\sigma }^2}{n}\right)$.

The standard deviation of $\bar{X}$ (the standard error) is $\frac{\sigma }{\sqrt{n}}$.

Standardising uses $Z=\frac{\bar{X}-\mu }{\sigma /\sqrt{n}}$.

This matches option B exactly.

This is the formula for standardising a single observation $X$, not the sample mean.

This multiplies by $\sqrt{n}$ instead of dividing, the opposite mistake.

This inverts the standard error incorrectly, using $n/\sigma$ instead of $\sigma /\sqrt{n}$.

Only B divides by the correct standard error $\sigma /\sqrt{n}$.