Unformatted text preview: uld compute this by calculating
C(100,45)(0.5)45(0.5)55 + C(100,46)(0.5)46(0.5)54 + . . .,
but we can much more easily approximate it by looking at a normal distribution with mean
µ = 5 0 and standard deviation ß = 100·0.5·0.5 = 5. (Notice that three standard
deviations above and below the mean is the range 35 to 65, which is well within the range of
possible values for X, which is 0 to 100, so the approximation should be a good one.) Let Y
have this normal distribution. Then
P(45 ≤ X ≤ 55...
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 Spring '13
 Herbret
 Statistics, Normal Distribution, Standard Deviation, Probability theory, Binomial distribution

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