Question

Waiting times​ (in minutes) of customers at a bank where all customers enter a single waiting line and a bank where customers wait in individual lines at three different teller windows are listed below. Find the coefficient of variation for each of the two sets of​ data, then compare the variation. Bank A​ (single line): 6.5 nbsp 6.7 nbsp 6.7 nbsp 6.8 nbsp 7.1 nbsp 7.3 nbsp 7.4 nbsp 7.7 nbsp 7.7 nbsp 7.7 Bank B​ (individual lines): 4.3 nbsp 5.4

Answer 1

Complete Question

1. Waiting times​ (in minutes) of customers at a bank where all customers enter a single waiting line and a bank where customers wait in individual lines at three different teller windows are listed below. Find the coefficient of variation for each of the two sets of​ data, then compare the variation.

Bank A (single lines): 6.5, 6.6, 6.7, 6.8, 7.2, 7.3, 7.4, 7.6, 7.6, 7.7

Bank B (individual lines): 4.1, 5.4, 5.8, 6.3, 6.8, 7.8, 7.8, 8.6, 9.3, 9.7

- The coefficient of variation for the waiting times at Bank A is ----- %?

- The coefficient of variation for the waiting times at the Bank B is ----- ​%?

- Is there a difference in variation between the two data​ sets?

Answer:

a

The coefficient of variation for the waiting times at Bank A is $l =$6.3%

b

The coefficient of variation for the waiting times at Bank B is $l_1 =$25.116%

c

The waiting time of Bank B has a considerable higher variation than that of Bank A

Step-by-step explanation:

From the question we are told that

  For Bank A  : 6.5, 6.6, 6.7, 6.8, 7.2, 7.3, 7.4, 7.6, 7.6, 7.7

 For  Bank B : 4.1, 5.4, 5.8, 6.3, 6.8, 7.8, 7.8, 8.6, 9.3, 9.7

The sample size is  n =10

The mean for Bank A is

          $\mu_A = \frac{6.5+ 6.6+ 6.7+ 6.8+ 7.2+ 7.3+ 7.4+ 7.6+ 7.6+ 7.7}{10}$

          $\mu_A = 7.14$

The standard deviation is mathematically represented as

      $\sigma = \sqrt{\frac{\sum|x- \mu|}{n} }$

        $k = \sum |x- \mu | ^2 = 6.5 -7.14|^2 + |6.6-7.14|^2+ |6.7-7.14|^2+ |6.8-7.14|^2 + |7.2-7.14|^2+ |7.3-7.14|^2, |7.4-7.14|^2+ |7.6-7.14|^2+|7.6-7.14|^2+|7.7-7.14|^2$

$k = 2.42655$

    $\sigma = \sqrt{\frac{2.42655}{10} }$

    $\sigma = 0.493$

The coefficient of variation for the waiting times at Bank A is  mathematically represented as  

        $l = \frac{\sigma}{\mu} *100$

        $l = \frac{0.493}{7.14} *100$

       $l =$6.3%

Considering Bank B

     The mean for Bank B is

               $\mu_1 = \frac{4.1+ 5.4+ 5.8+ 6.3+6.8+ 7.8+ 7.8+ 8.6+ 9.3+ 9.7}{10}$

              $\mu_1 = 7.16$

The standard deviation is mathematically represented as      

       $\sigma_1 = \sqrt{\frac{\sum|x- \mu_1|}{n} }$

    $\sum |x- \mu_1 | ^2 =4.1-7.16|^2 +| 5.4-7.16|^2+ |5.8-7.16|^2 + | 6.3-7.16|^2 + |6.8-7.16|^2 + | 7.8-7.16|^2 +|7.8-7.16|^2 +|8.6-7.16|^2 + |9.3-7.16|^2 +|9.7-7.16|^2$

$\sum |x- \mu_1 | ^2 =32.34$

    $\sigma_1 = \sqrt{\frac{32.34}{10} }$

     $\sigma_1 = 1.7983$

The coefficient of variation for the waiting times at Bank B is  mathematically represented as  

        $l_1 = \frac{\sigma }{\mu} *100$

        $l_1 = \frac{1.7983 }{7.16} *100$

        $l_1 =$25.116%