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In Melanie's Styling Salon, the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes. The slowest quarter of customers will require longer than how many minutes (to the nearest tenth) for a simple haircut?

User Switz
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1 Answer

6 votes

Answer:

Longer than 27.7 minutes.

Explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean
\mu and standard deviation
\sigma, the zscore of a measure X is given by:


Z = (X - \mu)/(\sigma)

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:


\mu = 25, \sigma = 4

The slowest quarter of customers will require longer than how many minutes (to the nearest tenth) for a simple haircut?

Longer than the 100 - 25 = 75th percentile of times, which is X when Z has a pvalue of 0.75. So X when Z = 0.675. So


Z = (X - \mu)/(\sigma)


0.675 = (X - 25)/(4)


X - 25 = 4*0.675


X = 27.7

So

Longer than 27.7 minutes.

User Adeel Raza Azeemi
by
8.0k points
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