Question

A particular process at Sparkly Company follows a normal distribution with a standard deviation of 7 minutes. If 95% of the processes require more than 150 minutes, what is the mean time of the processes?

Select one: a. 138.5 b. 143.35 c. 161.5 d. 156.65

Answer 1

Given the, standard deviation = 7 minutes and 95% of the processes require more than 150 minutes. That means, P(X > 150) = 0.95

We want to find, the population mean (μ)

First we find, z-score such that,

P(Z > z) = 0.95

=> 1 - P(Z < z) = 0.95

=> P(Z < z) = 0.05

Using standard normal z-table we get, z-score corresponding probability of 0.05 is z = -1.645

So, for z = -1.645, we find the population mean (μ),

=> Population Mean (μ) = 161.5

Answer : c) 161.5

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