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Suppose that the weight of a person selected at random from some population is normally distributed with parameters $\mu$ (mean) and $\sigma$ (variance). Suppose also that $P(X \le 160) = 1/2$ and $P(X \le 140) = 1/4.$ Find $\mu$ and $\sigma,$ and find $P(X \ge 200).$ Of all the people in the population weighing at least 200 pounds, what percentage will weigh over 220 pounds?

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  • $\begingroup$ I formatted your question. I'm thinking you may have intended to say that $\sigma$ is the standard deviation, not the variance. If so, please edit. $\endgroup$ – BruceET Apr 12 '18 at 1:55
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HINTS

  1. If $X \sim \mathcal{N}(0, 1)$ what are the values of $a,b$ such that $\mathbb{P}[X \le a] = 1/2$ and $\mathbb{P}[X\le b] = 1/4$?
  2. How do you convert $Y \sim \mathcal{N}(\mu, \sigma)$ to $X \sim \mathcal{N}(0, 1)$?
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From $P(X \le 160) = 1/2,$ you can infer that $\mu = 160.$ Then from $$P(X \le 140) = \left(\frac{X-\mu}{\sigma}\le \frac{140 - 160}{\sigma} \right) = P\left(Z \le \frac{-20}{\sigma} \right) = 1/4,$$ You can find $-20/\sigma$ and thus $\sigma.$

Then, knowing both $\mu$ and $\sigma,$ you are ready to find $P(X \ge 200).$

Note: In principle, if you are given values of any two probabilities of the type $P(X \le a)$ and $P(X \le b),$ where $a \ne b$ are both known, then you can solve two equations in two unknowns to get $\mu$ and $\sigma.$

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