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Suppose the random variable W represents the weight, and assume that W is Normally distributed with the given mean and standard deviation.

Then Prob(140 < W < 220) = Prob[(140 - 145)/31 < Z < (220 - 145)/31]

where Z has a standard Normal distribution.

that is, Prob(-0.1613 < Z < 2.4194) = 1 - [Prob(-0.1613 < Z) + Prob(Z < 2.4194)

= 1 - (0.4359 + 0.0078) = 0.5563

So percentage of women = 100*0.5563 = 56% approx.

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Q: How do you calculate the percentage of women that have weights between 140 lb and 220 lb with a mean of 145 lb and a standard deviation of 31 lb?
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