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.
