The Management at Ohio National Bank does not want its customers to wait in line

Question

The Management at Ohio National Bank does not want its customers to wait in line for service for too long. The manager of a branch of this bank estimated that the customers currently have to wait an average of 8 minutes for service. Assume that the waiting times for all customers at this branch have a normal distribution with a mean of 8 minutes and a standard deviation of 2 minutes.

Find the probability that randomly selected customer will have to wait for less than 4 minutes?

Instructions: Show all steps:

Draw the normal curve and indicate the mean, standard deviation, and the X bar scale

Identify the area of interest (that is shade the area under the curve that you will compute the probability).

Covert the X bar values in Z scores

Look up the Z standardized table for the cumulative area(s).

Now, make your decision ( that a customer will wait for less than 4 minutes)

Explanation / Answer

Here mean=8 and sd=2

We need to find P(x<4)

As this is normal distribution we will convert x to z=x-mean/sd=4-8/2=-2

So we need to find P(z<-2)=0.5+P(0<z<-2)=0.5-0.4772=0.0228=2.28%

Hence customer will not wair for less than 4 minutes as probability is too low

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