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Question
In a photographic process, the developing time of prints may be looked upon as a random variable having the normal distribution with a mean of 16.28 seconds and a standard deviation of 0.12 second. Find the probability that it will take less than 16.35 seconds to develop prints
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Solution
Let x be the random variable have long the normal distribution with a mean of 16.28 seconds and a standard deviation of 0.12 second
µ = 16,28 and σ = 0.12
The standard normal variate
z = `(x - mu)/sigma = (x - 16.28)/0.12` = 1
P(Less than 16.35 seconds) = P(x < 16.35)
When x = 16.35
z = `(16.35 - 16.28)/0.12 = 0.17/0.12` = 0.583
P(X < 16.35) = P(Z < 0.583)
= P(`-oo` < z < 0) + P(0 < z < 0.583)
= 0.5 + 0.2190
= 0.7190
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