= Solution
{c}
Condition on the total of 400 complaints. Assume each complaint independently has the same <probability> $p$ of being received on Monday; in particular, repeated complaints from one incident must not create material dependence. The Monday count $S$ then has a <binomial distribution> with parameters $(400,p)$. Test $H_0:p=0.2$ against $H_1:p>0.2$, rejecting for large $S$ as suggested by the <monotone likelihood ratio>.
Under the <null hypothesis>, $\mathbb E_0S=80$ and $\operatorname{Var}_0(S)=400(0.2)(0.8)=64$. Both expected successes and expected failures are large, so the <central limit theorem> gives a suitable <normal approximation>:
$$
Z=\frac{S-80}{8}\ \mathrel{\dot\sim}\ N(0,1),\qquad
\boxed{z_{\rm obs}=\frac{100-80}{8}=2.5}.
$$
Using the paper's supplied rounded table, the upper-tail <p-value> is approximately
$$
\boxed{p_{\rm approx}=1-\Phi(2.5)\approx0.005}.
$$
A more accurate evaluation of the <standard normal distribution> <cumulative distribution function> gives about 0.0062. For example, choose <significance level> 0.05 before examining the data; the supplied upper 5% <standard normal quantile> is approximately 1.6, and $2.5>1.6$. \b[Reject $H_0$ at 5%, and also at 1%, using this approximation.] The observed <binomial proportion> is $100/400=0.25$, providing evidence that the Monday <probability> exceeds 0.2 under the sampling assumptions.
If applying a <continuity correction>, approximate $\mathbb P_0(S\geq100)$ by $1-\Phi((99.5-80)/8)=1-\Phi(2.4375)\approx0.0074$; this gives the same 5% and 1% conclusions. The uncorrected value 0.005 is approximate, so a decision exactly at the 0.5% boundary is not warranted by the supplied rounded table.
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