/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q24 In Exercises 21鈥24, find the c... [FREE SOLUTION] | 91影视

91影视

In Exercises 21鈥24, find the coefficient of variation for each of the two samples; then compare the variation. (The same data were used in Section 3-1.) 21.

Bank Queues Waiting times (in seconds) of customers at the Madison Savings Bank are recorded with two configurations: single customer line; individual customer lines.

Single Line 390 396 402 408 426 438 444 462 462 462

Individual Lines 252 324 348 372 402 462 462 510 558 600

Short Answer

Expert verified

The coefficient of variation for waiting times corresponding to the single customer line is equal to 6.7%.

The coefficient of variation for waiting times corresponding to the individual customer lines is equal to 25.5%.

There is a significantly large difference in the variation of the waiting times of the single customer line and the individual customer lines.

Step by step solution

01

Given information

The waiting times of customers are given under two categories: single customer line and individual customer line.

02

Formula for the coefficient of variation

The coefficient of variation is a percentage of change in the standard deviation measure over the mean of data values.

C.V.=sx100

, where

sis the sample standard deviation;

xis the sample mean.

03

Calculation of sample means 

The mean waiting time for a single customer line is calculated as shown below:

x1=i=1n1x1in1=390+396+...+46210=429.0

Thus, the mean waiting time for a single customer line is 429.0 seconds.

The mean waiting time for individual customer lines is calculated as follows:

x2=i=1n2x2in2=252+324+...+60010=429.0

Thus, the mean waiting time for individual customer lines is 429.0 seconds.

04

Calculations for sample standard deviations

The standard deviation for waiting time for a single customer line is given as follows:

s1=i=1nx1i-x12n1-1=390-429.02+396-429.02+...+462-429.0210-1=28.6

Thus, the standard deviation for waiting time for a single customer line is 28.6 seconds.

The standard deviation for waiting time for individual customer lines is given as follows:

s2=i=1n2x2i-x22n2-1=252-429.02+324-429.02+...+600-429.0210-1=109.3

Thus, the standard deviation for waiting time for individual customer lines is 109.3 seconds.

05

Calculations for sample coefficients of variation

The coefficient of variation for waiting time for a single customer line is computed as follows:

CV1=s1x1100=28.6429100=6.7%

Therefore, the coefficient of variation for waiting time for a single customer line is equal to 6.7%.

The coefficient of variation for waiting time for individual customer lines is computed as follows:

CV2=s2x2100=109.3429100=25.5%

Therefore, the coefficient of variation for waiting time for individual customer lines is equal to 25.5%.

06

Comparison of the two coefficient of variation measure

The coefficient of variation for thewaiting time of individual customer lines isgreater than the coefficient of variation forwaiting time for individual customer lines by a large margin.

Hence, the waiting time for individual customer lines varies more than for a single customer line.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises 37鈥40, refer to the frequency distribution in the given exercise and find the standard deviation by using the formula below, where x represents the class midpoint, f represents the class frequency, and n represents the total number of sample values. Also, compare the computed standard deviations to these standard deviations obtained by using Formula 3-4 with the original list of data values: (Exercise 37) 11.5 years; (Exercise 38) 8.9 years; (Exercise 39) 59.5; (Exercise 40) 65.4.

Standard deviation for frequency distribution

s=nf.x2-f.x2nn-1

Age (yr) of Best Actress When Oscar Was Won

Frequency

20-29

29

30-39

34

40-49

14

50-59

3

60-69

5

70-79

1

80-89

1

Trimmed Mean Because the mean is very sensitive to extreme values, we say that it is not a resistant measure of center. By deleting some low values and high values, the trimmed mean is more resistant. To find the 10% trimmed mean for a data set, first arrange the data in order, then delete the bottom 10% of the values and delete the top 10% of the values, then calculate the mean of the remaining values. Use the axial loads (pounds) of aluminum cans listed below (from Data Set 30 鈥淎luminum Cans鈥 in Appendix B) for cans that are 0.0111 in. thick. An axial load is the force at which the top of a can collapses. Identify any outliers, then compare the median, mean, 10% trimmed mean, and 20% trimmed mean.

247 260 268 273 276 279 281 283 284 285 286 288

289 291 293 295 296 299 310 504

Degrees of Freedom Five pulse rates randomly selected from Data Set 1 鈥淏ody Data鈥 in Appendix B have a mean of 78.0 beats per minute. Four of the pulse rates are 82, 78, 56, and 84.

a. Find the missing value.

b. We need to create a list of n values that have a specific known mean. We are free to select any values we desire for some of the n values. How many of the n values can be freely assigned before the remaining values are determined? (The result is referred to as the number of degreesof freedom.)

In Exercises 5鈥20, find the range, variance, and standard deviation for the given sample data. Include appropriate units (such as 鈥渕inutes鈥) in your results. (The same data were used in Section 3-1, where we found measures of center. Here we find measures of variation.) Then answer the given questions.

TV Prices Listed below are selling prices in dollars of TVs that are 60 inches or larger and rated as a 鈥渂est buy鈥 by Consumer Reports magazine. Are the measures of variation likely to be typical for all TVs that are 60 inches or larger?

1800 1500 1200 1500 1400 1600 1500 950 1600 1150 1500 1750

Weighted Mean A student of the author earned grades of 63, 91, 88, 84, and 79 on her five regular statistics tests. She earned grades of 86 on the final exam and 90 on her class projects. Her combined homework grade was 70. The five regular tests count for 60% of the final grade, the final exam counts for 10%, the project counts for 15%, and homework counts for 15%. What is her weighted mean grade? What letter grade did she earn (A, B, C, D, or F)? Assume that a

mean of 90 or above is an A, a mean of 80 to 89 is a B, and so on

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.