Mathematics, 18.06.2020 23:57 booo1144

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

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Answer from: trentvw1174

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?

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

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

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%

Answer from: Quest

40/5 = 8you just divide the number of gloves by the number of teams

Answer from: Quest

rounding to the nearest thousandth would be 452,000

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