Distinguish CLT estimates from exact probabilities and strict thresholds

$E(X)=10(0.4)=4$ and $\operatorname{Var}(X)=10(0.4)(0.6)=2.4$. For 50 independent observations, the source uses $\bar X\mathrel{\dot\sim}N(4,2.4/50)=N(4,0.048)$ by the CLT. Hence $P(\bar X<4.5)\approx\Phi(0.5/\sqrt{0.048})=0.98876056\ldots\approx0.989$, to 3 s.f. This retains the printed unadjusted normal approximation.

The total $T=\sum_{j=1}^{50}X_j$ is exactly $B(500,0.4)$ for independent observations. Since $T$ is an integer, $\bar X<4.5$ means $T\le224$. Thus the exact probability is $\sum_{k=0}^{224}\binom{500}{k}(0.4)^k(0.6)^{500-k}=0.9869935684\ldots$. The source’s 0.989 is an approximation, not this exact binomial value.

$\bar Y\sim N(4,3/5)$ exactly. Therefore $P(\bar Y<5)=\Phi(1/\sqrt{3/5})\approx0.902$, to 3 s.f.

The printed answer is a large sample, using the course convention $n\ge30$. With independent identically distributed observations and finite variance, the CLT justifies an approximate normal sample mean as $n$ becomes large. Thirty is a practical convention, not a universal accuracy guarantee for every population shape.

The uniform model is a new population with mean 6 and variance 12. For large $n$, $\bar Y\mathrel{\dot\sim}N(6,12/n)$. The inequality becomes $P(Z>-\sqrt{n/12})>0.99$, or $\sqrt{n/12}>\Phi^{-1}(0.99)=2.32634787\ldots$. Thus $n>12[\Phi^{-1}(0.99)]^2\approx64.9427$ and the least integer under this approximation is 65, which also meets the source’s large-sample convention.

The source uses normalcdf(5,10^99,6,√(12/X)), with lower 5, upper $10^{99}$, mean 6 and standard deviation $\sqrt{12/X}$. Its printed comparison table is:

The rounded 0.9900 at $n=65$ alone cannot establish a strict inequality. The unrounded normal value is about 0.9900273, which is greater than 0.99. At 64 it is below 0.99. The surrounding screen values are retained below.

Source screen values: