Question

The time taken to complete one cycle of a particular task in a large factory is normally distributed with a mean of 8.8 hours and a standard deviation of 1.7 hours.

The time taken by the 20% of slowest cycles is above how many hours?

Answer 1

Let X be the time taken to complete one cycle of a particular task in a large factory

X ~ N ( 8.8, 1.7)

The time taken by the 20% of slowest cycles is above K hours

So

$P(X \leq K) = 0.2$

$P(Z \leq \frac{K-8.8}{1.7}) = 0.2$

From standard normal distribution

P( Z <= -0.8416 ) = 0.2

So

$\frac{K-8.8}{1.7} = -0.8416 \\ So \\ K = 7.36928$

Note: Please round the value of K as per requirement.

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