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Let Xsub1, Xsub2, ..., Xsub48 be a random sample of size 48 from the distribution with pdf f(x) = 1/(x^2), 1 < x < infinity. Approximate the probability that at most 10 of these random variables have values greater than 4.

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If $X$ is one of the $X_i$, then $\Pr(X\gt 4)=\int_4^\infty \frac{dx}{x^2}=\frac{1}{4}$.

Now we have $48$ independent experiments, each with probability of success $\frac{1}{4}$. We want the probability of $10$ or fewer successes. An exact expression for this is $$\sum_{k=0}^{10}\binom{48}{k}(1/4)^k(3/4)^{48-k}.$$ Now we can compute. For example, Wolfram Alpha will do it, as will many other programs. Or we can cross our fingers and use the normal approximation to the binomial, preferably with continuity correction.

The relevant normal has mean $(48)(1/4)=12$ and variance $(48)(1/4)(3/4)=9$, so standard deviation $3$. Then find the probability that a normal with mean $12$ and standard deviation $3$ is $\le 10.5$.

Remark: The fact that the standard deviation is so "nice," and $\frac{10.5-12}{3}=-0.5$ may be evidence that it is intended that you use the normal approximation with continuity correction!

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