/*! 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} Q8 In Exercises 5鈥20, find the ra... [FREE SOLUTION] | 91影视

91影视

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. Herewe find measures of variation.) Then answer the given questions.

What Happens in Vegas . . . Listed below are prices in dollars for one night at different hotels located on Las Vegas Boulevard (the 鈥淪trip鈥). How useful are the measures of variation for someone searching for a room?

212 77 121 104 153 264 195 244

Short Answer

Expert verified

The value of the range of the prices is equal to 187.0 dollars.

The variance of the prices is equal to 4626.2 dollars squared.

The standard deviation of the prices is equal to 68.0 dollars.

The sample has a collection of eight different hotels with different price ranges. Thus, the values can be useful in searching for a room on Las Vegas Boulevard as it provides a fair idea of the available hotels.

Step by step solution

01

Given information

The given data shows the prices for one night at eight different hotels in Las Vegas.

The number of values (n) is 8.

02

Computation of the measures of variation

The measures of dispersion are used for analyzing the spread of a given set of values.

The following are the three highly used measures of variation:

The rangeis the value obtained when the minimum value is subtracted from the maximum value.

Range=MaximumValue-MinimumValue=264-77=187.0dollars

.

Therefore, for the given sample of prices, the range is equal to 187.0 dollars.

Sample variances2is used to determine the inherent variation in the sample. The square of the unit of the data forms the unit of variance. It is calculated using the following formula:

s2=1=1nxi-x2n-1

Here,

x represents the sampled values, and

xis the sample mean.

The sample mean is calculated as

x=1=1nxin=212+77+...+2448=13708171.3

.

Thus, the sample mean is 171.3 million.

The variance of the sample is calculated as

s2=i=1nxi-x2n-1=212-171.33+77-171.33+...+244-171.338-1=32383.57=4626.2

.

Therefore, the sample variance of the prices for a one-night stay is equal to 4626.2dollars2.

Thestandard deviation of the sample also shows the variation in the data. The units of the values in the data are the units of standard deviation. It is calculated using the following formula as

s=s2=4626.268.0dollars

Therefore, the sample standard deviation of the prices for a one-night stay is equal to 68.0 dollars.

03

Interpretation

As the sample consists of a range of eight different hotels with different prices, it can be said that the values of the measures of variation are useful for a person looking for a room on Las Vegas Boulevard.

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

Using the sample data from Exercise 1, find the z score corresponding to the prediction error of 0 min. Is that prediction error significantly low or high? Why or why not?

In Exercises 5鈥8, express all z scores with two decimal places.

Plastic Waste Data Set 31 鈥淕arbage Weight鈥 in Appendix B lists weights (lb) of plastic discarded by households. The highest weight is 5.28 lb, the mean of all of the weights is x = 1.911 lb, and the standard deviation of the weights is s = 1.065 lb.

a. What is the difference between the weight of 5.28 lb and the mean of the weights?

b. How many standard deviations is that [the difference found in part (a)]?

c. Convert the weight of 5.28 lb to a z score.

d. If we consider weights that convert to z scores between -2 and 2 to be neither significantly low nor significantly high, is the weight of 5.28 lb significant?

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.

California Smokers In the California Health Interview Survey, randomly selected adults are interviewed. One of the questions asks how many cigarettes are smoked per day, and results are listed below for 50 randomly selected respondents. How well do the results reflect the smoking behavior of California adults?

9 10 10 20 40 50 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

In Exercises 29鈥32, find the mean of the data summarized in the frequency distribution. Also, compare the computed means to the actual means obtained by using the original list of data values, which are as follows: (Exercise 29) 36.2 years; (Exercise 30) 44.1 years; (Exercise 31) 224.3; (Exercise 32) 255.1..

Blood Platelet Count of Males (1000 cells /渭尝 )

Frequency(f)

100-199

25

200-299

92

300-399

28

400-499

0

500-599

2

Why Divide by n 鈭 1? Let a population consist of the values 9 cigarettes, 10 cigarettes, and 20 cigarettes smoked in a day (based on data from the California Health Interview Survey). Assume that samples of two values are randomly selected with replacement from this population. (That is, a selected value is replaced before the second selection is made.)

a. Find the variance2 of the population {9 cigarettes, 10 cigarettes, 20 cigarettes}.

b. After listing the nine different possible samples of two values selected with replacement, find the sample variance s2 (which includes division by n - 1) for each of them; then find the mean of the nine sample variances s2.

c. For each of the nine different possible samples of two values selected with replacement, find the variance by treating each sample as if it is a population (using the formula for population variance, which includes division by n); then find the mean of those nine population variances.

d. Which approach results in values that are better estimates of2 part (b) or part (c)? Why? When computing variances of samples, should you use division by n or n - 1?

e. The preceding parts show that s2 is an unbiased estimator of 2. Is s an unbiased estimator of ? Explain

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.