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A recent survey asked students, "On average, how many times in a week do you go to a restaurant for dinner?" Of the 570 respondents, 84 said they do not go out for dinner, 290 said once, 100 said twice, 46 said thrice, 30 said 4 times, 13 said 5 times, 5 said 6 times, and 2 said 7 times. a. Display the data in a table. Explain why the median is \(1 .\) b. Show that the mean is 1.5 . c. Suppose the 84 students who said that they did not go out for dinner had answered 7 times instead. Show that the median would still be 1 . (The mean would increase to 2.54 . The mean uses the numerical values of the observations, not just their ordering.)

Short Answer

Expert verified
a. Median is 1; b. Mean is 1.5; c. Median stays 1 even if all 0 times move to 7 times.

Step by step solution

01

Organize Data into a Frequency Table

To present the survey data in an organized manner, create a frequency table listing the number of times students go out for dinner against the number of respondents. This table is structured as follows: | Number of Times | Frequency | |-----------------|-----------| | 0 | 84 | | 1 | 290 | | 2 | 100 | | 3 | 46 | | 4 | 30 | | 5 | 13 | | 6 | 5 | | 7 | 2 | This table clearly represents the distribution of responses.
02

Explain the Median

The median is the middle value of ordered data. To find it, list out responses from least to greatest based on frequency. There are 570 total responses. To find the median position, use the formula \( \frac{n + 1}{2} \) where \(n\) is the total number of data points. Therefore, the median position is \( \frac{570 + 1}{2} = 285.5 \), which suggests we take the average of the 285th and 286th data points. Both data points are in the '1 time' category, confirming that the median is 1.
03

Calculate the Mean

The mean is found by multiplying each number of times by its frequency, summing these products, and dividing by the total number of respondents. The calculation becomes:\[\frac{(0 \times 84) + (1 \times 290) + (2 \times 100) + (3 \times 46) + (4 \times 30) + (5 \times 13) + (6 \times 5) + (7 \times 2)}{570}\]= \(\frac{0 + 290 + 200 + 138 + 120 + 65 + 30 + 14}{570} = \frac{857}{570} \approx 1.5 \).This shows that the mean is indeed 1.5.
04

Adjust the Data and Explain New Median

If the 84 students who said 0 changed to 7 times, recalculate to see the position for the median which remains the 285th and 286th positions after the change. The new table would redistribute frequencies, but the 285th and 286th responses still fall inside the '1 time' category, as the number for '0' is moved higher and does not affect these positions. Thus the median remains 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Median
The median is a fundamental concept in descriptive statistics that helps us find the middle value of a dataset. When you have an ordered set of numbers, the median is the point that divides your data into two equal halves.
If you were to line up all the responses to the question on dining frequency, from the least to the greatest, the middle point is known as the median. In this exercise, the total number of responses given was 570. To locate the median, you use the formula \( \frac{n + 1}{2} \) to determine the position in the data. Here, that is \( \frac{570 + 1}{2} = 285.5 \).
Since we cannot have a fraction of a response, this suggests we need to find the average of the 285th and 286th responses. Both of these fall in the '1 time' category. Thus, the median is 1. Interestingly, even changing the initial '0 times' responses to '7 times' would not shift the 285th and 286th positions out of the '1 time' category, securely maintaining our median at 1 time per week.
Mean
The mean is another essential aspect of descriptive statistics, often called the average. To calculate the mean, every response is accounted for, and then we determine the average frequency.
The calculation for the mean is done by multiplying each category of response by its frequency, adding all these results together, and then dividing by the total number of responses. So, in this case:
  • 0 times: 84 students, contributing 0 to the sum
  • 1 time: 290 students, contributing 290
  • 2 times: 100 students, contributing 200
  • 3 times: 46 students, contributing 138
  • 4 times: 30 students, contributing 120
  • 5 times: 13 students, contributing 65
  • 6 times: 5 students, contributing 30
  • 7 times: 2 students, contributing 14
The sum of these products is 857. Dividing this by the total of 570 responses gives us approximately 1.5. Thus, the mean frequency of dining out is 1.5 times per week. If we hypothetically change the 84 '0 times' responses to '7 times', the mean would rise significantly to 2.54.
Frequency Table
A frequency table is a useful tool in descriptive statistics for organizing and presenting data in a clear format. It displays each unique value in your data along with the count, or frequency, of how often it occurs.
In this exercise, a frequency table was used to organize how students responded to the survey question about dining frequency per week. Each row of the table represented one of the possible answers – from '0' times to '7' times – and the corresponding frequency, which is the number of students selecting that response.
For example, the table showed that 290 students reported dining out once per week, while 84 students stated they do not dine out at all. This makes it easy to visualize the distribution and helps in calculating descriptive statistics like the median and mean. Using a frequency table simplifies spotting trends, making calculations more intuitive, and enhances understanding of the dataset.

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