A bank manager wanted to double-check her claim that recent process improvements have reduced customer wait times to an average of 2.9 minutes, with a standard deviation of 0.8 minutes. To test this, she randomly sampled 25 customers and recorded their wait times. The average wait time of the customers in the sample was 3.3 minutes. Assuming the claim about wait time is true, what is the probability that a random sample of n = 25 customers would have a sample mean as large as or greater than 3.3 minutes? What should the manager conclude about her claim?

the manager claim is therefore rejected.

Step-by-step explanation:

To ascertain the manager claim, we use the normal distribution curve.

z = (x - x') / σ

, where

z is called the normal standard variate,

x is the value of the variable, = 2.9

x' is the mean value of the distribution = 3.3

σ is the standard deviation of the distribution = 0.8

so, z = (2.9 - 3.3) / 0.8 = - 0.5

Using a table of normal distribution to check the partial areas beneath the standardized normal curve, a z-value of - 0.5 corresponds to an area of 0.1915 between the mean value. The negative z-value shows that it lies to the left of the z=0 ordinate.

The total area under the standardized normal curve is unity and since the curve is symmetrical, it follows that the total area to the left of the z=0 ordinate is 0.5000. Thus the area to the left of the z=-0.5 ordinate is 0.5000-0.1915 = 0.3085 of the total area of the curve.

therefore, the probability of the average wait time greater than or equal to 3.3 minutes is 0.3085.

For a group of 25 customers, 25*0.3085 = 7.7, i. e. 8 customers only experience the improvement in wait time.

the manager claim is therefore rejected.