= Solution
The <normal distribution> standardized statistic for the <binomial distribution> count is
$$
Z=\frac{5150-5000}{\sqrt{10000/4}}=\boxed{3}.
$$
Because the alternative is two-sided, the approximate <p-value> is $2\{1-\Phi(3)\}\approx0.0027$. A continuity correction replaces $3$ by $(149.5)/50=2.99$ and gives approximately $0.0028$, with the same conclusion. Thus \b[the result is statistically significant at both the 5% and 1% levels]. The <p-value> is a null tail probability, not a <posterior probability> that the <null hypothesis> is true.
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