Question

select the correct answer from each drop - down menu. the data set gives the number of hours it took each of the 10 students in a cooking class to master a particular technique. {5, 3, 5, 30, 4, 5, 4, 3, 4, 5} the best measure of center for this data set is, and its value is

Explanation:

Step1: Identify the outlier

First, we look at the data set \(\{5, 3, 5, 30, 4, 5, 4, 3, 4, 5\}\). Most of the values are around 3 - 5, but 30 is much larger, so 30 is an outlier.

Step2: Determine the best measure of center

When there is an outlier, the median is a better measure of center than the mean because the mean is affected by the outlier. To find the median, we first sort the data set.

Step3: Sort the data set

Sorting the data set (excluding the outlier for median calculation? No, we sort all data): \(3, 3, 4, 4, 4, 5, 5, 5, 5, 30\). Wait, no, the original data set has 10 values: let's list them properly. The data set is: 5, 3, 5, 30, 4, 5, 4, 3, 4, 5. Let's sort them in ascending order: \(3, 3, 4, 4, 4, 5, 5, 5, 5, 30\).

Step4: Calculate the median

Since there are 10 values (even number), the median is the average of the 5th and 6th values. The 5th value is 4 and the 6th value is 5. So the median is \(\frac{4 + 5}{2}=\frac{9}{2} = 4.5\)? Wait, no, wait the sorted data: let's count again. The data points are: 3, 3, 4, 4, 4, 5, 5, 5, 5, 30. Wait, that's 10 values. The 5th value is the 5th element: index 4 (0 - based) or 5th (1 - based). Wait 1 - based: positions 1:3, 2:3, 3:4, 4:4, 5:4, 6:5, 7:5, 8:5, 9:5, 10:30. So the 5th value is 4 and the 6th is 5. So median is \(\frac{4 + 5}{2}=4.5\)? But wait, maybe I made a mistake in sorting. Wait the original data: 5, 3, 5, 30, 4, 5, 4, 3, 4, 5. Let's count the frequency of each number: 3 appears 2 times, 4 appears 3 times, 5 appears 4 times, 30 appears 1 time. So sorted: 3, 3, 4, 4, 4, 5, 5, 5, 5, 30. Yes. So median is the average of the 5th and 6th terms. 5th term is 4, 6th term is 5. So median is \(\frac{4 + 5}{2}=4.5\)? Wait, but maybe the question is considering that the outlier is 30, so we should use median. Alternatively, maybe the data set was written incorrectly? Wait, maybe the data set is \(\{5, 3, 5, 3, 4, 5, 4, 3, 4, 5\}\) (without 30)? But the original problem has 30. Wait, maybe a typo, but assuming the data is as given.

But the best measure of center when there is an outlier is the median. Let's check the mean to see the difference. Mean is \(\frac{5 + 3 + 5 + 30 + 4 + 5 + 4 + 3 + 4 + 5}{10}=\frac{5 + 3 = 8; 8 + 5 = 13; 13 + 30 = 43; 43 + 4 = 47; 47 + 5 = 52; 52 + 4 = 56; 56 + 3 = 59; 59 + 4 = 63; 63 + 5 = 68}{10}=6.8\). The median is 4.5, which is more representative of the center of the data (excluding the outlier). So the best measure of center is the median.

Now, let's find the median. Wait, maybe I made a mistake in sorting. Let's list all values:

Original data: 5, 3, 5, 30, 4, 5, 4, 3, 4, 5.

Let's list them:

1. 5

2. 3

3. 5

4. 30

5. 4

6. 5

7. 4

8. 3

9. 4

10. 5

Now, sort them:

3 (from 2), 3 (from 8), 4 (from 5), 4 (from 7), 4 (from 9), 5 (from 1), 5 (from 3), 5 (from 6), 5 (from 10), 30 (from 4). So sorted: [3, 3, 4, 4, 4, 5, 5, 5, 5, 30].

Number of values: 10 (even), so median is the average of the 5th and 6th values. 5th value: 4 (index 4, 0 - based: index 4 is the 5th element), 6th value: 5 (index 5). So median is \(\frac{4 + 5}{2}=4.5\)? Wait, but maybe the problem has a typo and 30 is not supposed to be there? Or maybe I misread the data. Wait, the user's data set: \{5, 3, 5, 30, 4, 5, 4, 3, 4, 5\}. Let's count the sum again: 5+3=8, +5=13, +30=43, +4=47, +5=52, +4=56, +3=59, +4=63, +5=68. 68/10=6.8 (mean). Median: as above, 4.5. But maybe the outlier is 30, so the best measure is median, and its value is 4.5? Wait, but let's check the number of 4s and 5s. Wait, the data without 30: \{5, 3, 5, 4, 5, 4, 3, 4, 5\} (wait, no, original data has 10 values, so 30 is included). So the best measure of center is median, and its value is 4.5? Wait, but maybe the data set was supposed to be without 30? Let's check the frequency of 4: 3 times, 5: 4 times, 3: 2 times, 30:1 time. So the median is between the 5th and 6th terms, which are 4 and 5, so median is 4.5.

So the best measure of center for this data set is the median (because of the outlier 30), and its value is 4.5. Wait, but maybe the problem has a mistake, but based on the data, that's the case.

Answer:

The best measure of center for this data set is the median, and its value is 4.5. Wait, but let's confirm. Alternatively, maybe the data set was written as \{5, 3, 5, 3, 4, 5, 4, 3, 4, 5\} (without 30), in which case the mean and median would be different. But with 30, median is better. So the best measure is median, value 4.5.