Reference no: EM131646597

Based on all the data that has been collected so far the Bank Manager believes the mean (m) number of minutes to process a customer transaction is 8.0 minutes with a standard deviation (sigma) equal to 2.4 minutes.

Part One: Compute how often it will require 6 minutes or less to process a customer transaction using the normal distribution and the Normal Table (Table N).

x = 6

mean = 8.0

sigma = 2.4

Z = (x – mean) / sigma

Z =

The Tail Value from the Normal Table =

Part Two: Clearly explain the answer in Part One above for the Bank Manager.

Part Three: Compute how often it will require 6 minutes or more to process a customer transaction using the normal distribution and the Normal Table (Table N).

Since the total probability (percentage) needs to equal 1 (or 100%) this answer can be easily determined by subtracting the Tail Value found in Part One above from the number 1.

1 – Tail =

Part Four: Clearly explain the answer in Part Three above for the Bank Manager.

Part Five: Compute how often it will require 11 minutes or more to process a customer transaction using the normal distribution and the Normal Table (Table N).

x = 11

mean = 8.0

sigma = 2.4

Z = (x – mean) / sigma

Z =

From the Normal Table the Tail value =

Part Six: Clearly explain the answer in Part Five above for the Bank Manager.

Part Seven: Compute how often it will require 11 minutes or less to process a customer transaction using the normal distribution and the Normal Table (Table N).

Since the total probability (percentage) needs to equal 1 (or 100%) this answer can be easily determined by subtracting the Tail Value found in Part Five above from the number 1.

1 – Tail =

Part Eight: Clearly explain the answer in Part Seven above for the Bank Manager.