Question

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? ​

Answer 1

Answer:

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.