What is the age distribution of promotion-sensitive shoppers? A supermarket super shopper is defined as a shopper for whom at least 70% of the items purchased were on sale or purchased with a coupon. Age range, years Midpoint x Percent of super shoppers 18-28 23 10% 29-39 40-50 51-61 56 62 and over 34 45 67 44% 21% 12% 13% a) Graph the histogram of probability distribution of using the age midpoints x and the percentage of super shoppers. (% of supper shopper) x( midpoints) b) Compute the expected age u of a super shopper. (Round your answer to two decimal places.) (Using u = Erp(x) ) %3D

What is the age distribution of promotion-sensitive shoppers? A supermarket super shopper is defined as a shopper for whom at least 70% of the items purchased were on sale or purchased with a coupon. Age range, years Midpoint x Percent of super shoppers 18-28 23 10% 29-39 40-50 51-61 56 62 and over 34 45 67 44% 21% 12% 13% a) Graph the histogram of probability distribution of using the age midpoints x and the percentage of super shoppers. (% of supper shopper) x( midpoints) b) Compute the expected age u of a super shopper. (Round your answer to two decimal places.) (Using u = Erp(x) ) %3D

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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The accompanying data represent the total travel tax (in dollars) for a 3-day business trip in 8 randomly selected cities. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. Complete parts (a) through (c) below. 68.42 78.57 68.76 84.62 79.17 87.73 101.93 98.62 D Click the icon to view the table of critical t-values. (a) Determine a point estimate for the population mean travel tax. A point estimate for the population mean travel tax is $ (Round to two decimal places as needed.) Clear all Check answer Tutoring Help me solve this Get more help- MacBook Pro G Search or type URL esc & @ 2# $ del 2 3 4 6. 7 P Q W E T Y tab 11 J K .... >
New Survey A survey was conducted two years ago asking college students their top motivations for using a credit card. To determine whether this distribution has changed, you randomly select 425 college students and ask each one what the top motivation is for using a credit card. Can you conclude that there has been a change in the claimed or expected distribution? Use a= 0.05. Complete parts (a) through (d). Old Survey % Frequency, f 112 Response Rewards 27% 23% 97 Low rates Cash back Discounts Other 21% 108 9% 48 20% 60 a. State H, and H, and identify the claim. What is the null hypothesis, Ho? O A. The distribution of motivations is 27% rewards, 23% low rate, 21% cash back, 9% discounts, and 20% other. O B. The distribution of motivations is 112 rewards, 97 low rate, 108 cash back, 48discounts, and 60 other. O C. The distribution of motivations differs from the expected distribution. What is the alternate hypothesis, H2? O A. The distribution of motivations differs from the claimed…
The accompanying data represent the total travel tax (in dollars) for a 3-day business trip in 8 randomly selected cities. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. Complete parts (a) through (c) below. 68.42 78.57 68.76 84.62 79.17 87.73 101.93 98.62 O Click the icon to view the table of critical t-values. ... (a) Determine a point estimate for the population mean travel tax A point estimate for the population mean travel tax is $ 83.48 (Round to two decimal places as needed.) (b) Construct and interpret a 95% confidence interval for the mean tax paid for a three-day business trip. Select the correct choice below and fill in the answer boxes to complete your choice.. (Round to two decimal places as needed.) O A. One can be % confident that the mean travel tax for all cities is between $ and $. O B. One can be % confident that the all cities have a travel tax between $ and $. OC. The…
Full data set D A researcher wanted to determine if carpeted or uncarpeted rooms contain more bacteria. The table shows the results for the number of bacteria per cubic foot for both types of rooms. A normal probability plot and boxplot indicate that the data are approximately normally distributed with no outliers. Do carpeted rooms have more bacteria than uncarpeted rooms at the a = 0.05 level of significance? Carpeted 11.4 10.2 Uncarpeted 10.9 10.9 5.9 10 13 6.9 10.9 8 5.2 6.6 6.4 7.7 7.1 1.2 Click the icon to view the Student t-distribution table. What are the null and alternative hypotheses? Ho: versus H1: Calculate the test statistic, to. to = (Round to two decimal places as needed.) %D Now find the critical value. Select the correct choice below and fill in the answer box within your choice. (Round to three decimal places as needed.) O A. ta = B. ta/2= Do carpeted rooms have more bacteria than uncarpeted rooms at the a = 0.05 level of significance? O A. No, because the test…
The accompanying data represent the total travel tax (in dollars) for a 3-day business trip in 8 randomly selected cities. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. Complete parts (a) through (c) below. 68.09 79.99 70.42 83.11 79.33 85.94 101.02 97.39 O E Click the icon to view the table of critical t-values. (a) Determine point estimate for the population mean travel tax. A point estimate for the population mean travel tax is $ 83.16 (Round to two decimal places as needed.) (b) Construct and interpret a 95% confidence interval for the mean tax paid for a three-day business trip. Select the correct choice below and fill in the answer boxes to complete your choice. (Round to two decimal places as needed.) O A. One can be % confident that the mean travel tax for all cities is between $ and $. O B. There is a % probability that the mean travel tax for all cities is between $ and $. O C.…
Hospital visits: According to a health agency, there were 409,689 hosital visits for asthma-related illnesses in a recent year. The age distribution was as follows. Round your answers to four decimal places if necessary. Age Range Less than 1 year 1-17 18-44 45-64 65-84 85 and up Total Send data to Excel Part 1 of 3 (a) What is the probability that an asthma patient is between 18 and 44 years old? The probability that an asthma patient is between 18 and 44 years old is 0.1944 Part 2 of 3 (b) What is the probability than an asthma patient is less than 1 year old? The probability that an asthma patient is less than 1 year old is 0.0192 Part: 2/3 Number 7862 103,032 79,653 121,731 80,658 16,753 409,689 Part 3 of 3 (c) Using a cutoff of 0.05, is it unusual for an asthma patient to be 85 years old or older? Based on a cutoff of 0.05, it (Choose one) unusual for an asthma patient to be 85 years old or older. is is not X
The data given below show the number of overtime hours worked in one week per employee. Use the data to complete parts (a) and (b). Overtime hours 0 1 2 3 4 5 6 Employees 4 14 25 54 44 28 18 (a) Construct a probability distribution. (b) Graph the probability distribution using a histogram and describe its shape. (a) Construct the probability distribution by completing the table below. x 0 1 2 3 4 5 6 P(x) enter your response here enter your response here enter your response here enter your response here enter your response here enter your response here enter your response here (Round to three decimal places as needed.) (b) Choose the correct graph of the probability distribution. A. 012345600.10.20.30.4Overtime HoursProbability A histogram has a horizontal axis labeled Overtime Hours from 0 to 6 in increments of 1…
The data given below show the number of overtime hours worked in one week per employee. Use the data to complete parts (a) and (b). Overtime hours 0 1 2 3 4 5 6 Employees 4 14 35 53 46 26 17 (a) Construct a probability distribution. (b) Graph the probability distribution using a histogram and describe its shape. (a) Construct the probability distribution by completing the table below. x 0 1 2 3 4 5 6 P(x) nothing nothing nothing nothing nothing nothing nothing (Round to three decimal places as needed.)
A distribution of scores is positively skewed. Which is the most probable order, from smallest to largest, for the three measures of central tendency? O Mean, median, mode O Mode, median, mean Mean, mode, median O Median, mean, mode
Historical data show that customers who download music from a popular Web service spend approximately $22 per month, with a standard deviation of $3. Assume the spending follows the normal probability distribution. Find the probability that a customer will spend at least $19 per month. How much (or more) do the top 9% of customers spend? What is the probability that a customer will spend at least $19 per month? (Round to four decimal places as needed.) How much do the top 9% of customers spend? (Round to the nearest cent as needed.)
Provide an appropriate response. A lab orders a shipment of 100 rats a week, 52 weeks a year, from a rat supplier for experiments that the lab conducts. Prices for each weekly shipment of rats follow the distribution below: $10.00 $12.50 Price Probability 0.25 0.3 $15.00 0.45 Suppose the mean cost of the rats turned out to be $13.00 per week. Interpret this value. O The average cost for all weekly rat purchases is $13.00. O The median cost for the distribution of rat costs is $13.00. O The rat cost that occurs more often than any other is $13.00. O Most of the weeks resulted in rat costs of $13.00.
Test Prep. A GRE prep company claims that average score improvement of their students is equal to 90 points. We want to see if it is different than this. We randomly sample students for their average improvement. The data we collected are below. GRE scores are known to be normally distributed. Points 90.4 84 77 75.2 81.5 78.4 96.5 97.3 101.5 85.3 69.5 88.6 92.4 Evaluate the company's claim using a hypothesis test. You wish to test the following claim (Ha) at a significance level of a = 0.05. (a) Write the hypothesis for the test. O Ho: H= 90 Ha: p = 90 O H;: H = 90 Ha: p 90 O Ho: H = 85.969 Ha: p = 85.969 Ο H : μ- 85. 969 H3: p 85.969 (b) What is the p-value for this test (round to 4 decimal place) (c) Based on the sample data, we sould conclude that ... O There is enough evidence at the a = 0.05 level to say that the average improvement is not 90 points. O There is not enough evidence at the a = 0.05 level to say that the average improvement is not 90. Question Help: O Message…