Question

Automatic banking machine (ABM) customers can perform a number of transactions quickly and efficiently. A banking consultant has noted that the times to complete a transaction are normally distributed with a mean of 34 seconds and a standard deviation of 4 seconds. What percentage of ABM customers take more than 31 seconds to do their banking?

Answer 1

Answer:

77.34% of ABM customers take more than 31 seconds to do their banking.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

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

$Z = \frac{X - \mu}{\sigma}$

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 problem, we have that:

$\mu = 34, \sigma = 4$

What percentage of ABM customers take more than 31 seconds to do their banking?

This is 1 subtracted by the pvalue of Z when X = 31.

So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{31 - 34}{4}$

$Z = -0.75$

$Z = -0.75$ has a pvalue of 0.2266.

So 1-0.2266 = 0.7734 = 77.34% of ABM customers take more than 31 seconds to do their banking.