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

Answer 1

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