(Help)X is a normally distributed random variable with mean 35 and standard deviation 16. What is the probability that X is between 67 and 83? Use the 0.68-0.95-0.997 rule and write your answer as a decimal.
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Notice that
67 = 35 + 32 = 35 + 2•16
and
83 = 35 + 48 = 35 + 3•16
so 67 is +2 standard deviations from the mean, and 83 is +3 standard deviations from the mean.
The 68-95-99.7 rule says that, for any normal distribution,
• approximately 95% of the distribution lies within 2 s.d. of the mean
• approx. 99.7% lies within 3 s.d. of the mean
This is to say,
• P(35 - 2•16 < X < 35 + 2•16) = P(3 < X < 67) ≈ 0.95
• P(35 - 3•16 < X < 35 + 3•16) = P(-13 < X < 83) ≈ 0.997
Subtracting these gives
P(-13 < X < 83) - P(3 < X < 67) ≈ 0.047
This subtraction effectively removes the within-2-s.d. part of the distribution from the within-3-s.d. part. In other words, we're now talking about the part of the distribution between -3 and -2 s.d. from the mean *and* between +2 and +3 s.d. from the mean. So, this equation is the same as
P(-13 < X < 3) + P(67 < X < 83) ≈ 0.047
Normal distributions are symmetric about their means, so the two probabilities here are equal, and in particular
2 P(67 < X < 83) ≈ 0.047
so that
P(67 < X < 83) ≈ 0.0235