Question 29.22: A random variable, h, has a normal distribution with mean 7 ......
A random variable, h, has a normal distribution with mean 7 and standard deviation 2. Calculate the probability that
(a) h > 9 (b) h < 6 (c) 5 < h < 8
Step-by-Step
(a) Applying the transformation gives
9\to{\frac{9-7}{2}}=1So h > 9 has the same probability as x > 1, where x is a random variable with a standard normal distribution:
P(h > 9) = P(x > 1) = 1 − P(x < 1) = 1 − 0.8413 = 0.1587
(b) Applying the transformation gives
6\rightarrow{\frac{6-7}{2}}=-0.5So h < 6 has the same probability as x < −0.5:
P(x < −0.5) = P(x > 0.5) = 1 − P(x < 0.5) = 0.3085
(c) Applying the transformation to 5 and 8 gives
5\rightarrow{\frac{5-7}{2}}=-1\;\;\;\;\;\;\;8\rightarrow{\frac{8-7}{2}}=0.5and so we require P(−1 < x < 0.5). Therefore
P(x < 0.5) = 0.6915 P(x < −1) = 0.1587
and then
P(−1 < x < 0.5) = 0.6915 − 0.1587 = 0.5328
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