Statistical Measures

We are given a table showing the number of likes Tearrik received for his photos. We want to find the average number of daily likes.

Remember, the mean of a data set is calculated by finding the sum of all values in the set and then dividing by the number of values in the set. We need to add all of the values, and divide them by the number of observations. Here, we have data for 7 days, so we should divide the sum by 7.

Tearrik receives an average of about 34 likes for each photo.

We want to find the median of the given data set. The median is the value that divides the ordered set into two halves. The first thing we have to do is to rearrange the data values from least to greatest. Given Data 25, 30, 28, 36, 40, 36,42 ⇓ Ordered Data 25, 28, 30, 36, 36, 40, 42 Since the number of values in our set is 7, the median is the fourth value in order. Therefore, the median is 36 likes.

The mode is the most common value in the set. Let's take a look at the ordered set one more time! Ordered Data 25, 28, 30, 36, 36, 40, 42 We can see that the most common value in the given data set is 36. This means that the mode of our data set is 36 likes.

We want to find the range of the given data set.

The range is the difference between the greatest and least values in a set of data. We can order the data from least to greatest to find the range. Given Data 85, 72, 90, 68, 78, 85, 91, 79 ⇓ Ordered Data 68, 72, 78, 79, 85, 85, 90, 91 For this exercise, the greatest value is 91 and the least value is 68. Range 91- 68= 23

We want to find the interquartile range of the data set. 68, 72, 78, 79, 85, 85, 90, 91 Since the interquartile range is the difference between the upper quartile and the lower quartile (Q_3-Q_1), we only need to find these two quartiles.

• Lower Quartile (Q_1) is the median of the lower half of the data set.
• Upper Quartile (Q_3) is the median of the upper half of the data set.

We have already sorted the data set from smallest value to largest value in Part A. 68, 72, 78, 79,^(Lower Half) | 85, 85, 90, 91^(Upper Half) The number of values in our data set is 8, so both the lower and upper halves contain four data values. We have two middle values for each half. Thus, we need to calculate the mean of those middle values. Lower Quartile:& 72+ 78/2= 75 [1.1em] Upper Quartile:& 85+ 90/2= 87.5 The last step to calculate the interquartile range is to calculate the difference between the upper and the lower quartile. Let's do it! Interquartile Range 87.5- 75=12.5

We want to find the mean, median, and mode of the given data set. 93, 75, 85, 83, 89, 93, 88, 33, 90 Let's begin by calculating the mean.

Mean

The mean of a data set is the sum of the values divided by the total number of values in the set. There are 9 values in our set, so we have to divide the sum by 9.

The mean is 81. We can continue by finding the median.

Median

When the data are arranged in numerical order, the median is the middle value — or the mean of the two middle values — in a set of data. Let's arrange the given values and find the median. Given Data 93, 75, 85, 83, 89, 93, 88, 33, 90 ⇓ Ordered Data 33, 75, 83, 85, 88, 89, 90, 93, 93 Since the number of values in our set is 9, the median is the value in the middle, 88. The last measure we need is the mode. Let's find it!

Mode

The mode is the value or values that appear most often in a set of data. Arranging the data set from least to greatest makes it easier to see how often each value appears. Let's take a look at the ordered set one more time! Ordered Data 33, 75, 83, 85, 88, 89, 90, 93, 93 The mode is 93.

We have to calculate the interquartile range (IQR) to identify any outliers. Let's start by revisiting some information about quartiles.

• Lower Quartile (Q_1) is the median of the lower half of the data set.
• Second Quartile (Q_2) is the median of the data set. It divides the set of data into two halves.
• Upper Quartile (Q_3) is the median of the upper half of the data set.
• Interquartile Range is the difference between the upper quartile and the lower quartile (Q_3-Q_1).

Now we can identify the quartiles and the interquartile range of the data. We will look at the ordered data set!

The interquartile range of this data set is 12.5. Next, we need to determine the maximum and minimum values for data to be considered an outlier. Outliers are more than 1.5 times the IQR away from the upper and lower quartiles.

Let's start with the minimum value. We substitute 79 for Q_1 and 12.5 for IQR.

The only value less than 60.25 is 33. This means that 33 is an outlier. Now let's check if there are other outliers by calculating the maximum value. We substitute 91.5 for Q_3 and 12.5 for IQR.

Our data set does not contain values greater than 110.25. This means that there are no more outliers than 33.

Let's repeat the process, this time excluding 33 from the data set. 75, 83, 85, 88, 89, 90, 93, 93

Mean

Let's start by calculating the sum of the given values without 33 and divide the sum by 8.

The mean of the data without the outlier is 87.

Median

Recall the ordered set with 33 excluded. 75, 83, 85, 88, | 89, 90, 93, 93 This time the number of values in our set is 8. This is why the median is the mean of the middle values. 88+89/2 = 88.5 The median is 88.5.

Mode

We can see that the most common value in the given data set is still 93. 75, 83, 85, 88, 89, 90, 93, 93 This is the mode of our data set.

Let's summarize our findings in the table below so it is easier to compare the results.

We can see that removing the outlier changed the mean and the median. They both increased but the mean increased more than the median. Removing the outlier did not change the mode of the data. The answer is mean.

We want to find the range and the interquartile range of the given data. 28, 27, 33, 29, 25, 43, 29, 28, 23, 33

Range

Let's start by finding the range. The range of a data set is one of the measures of variation. It is the difference of the greatest value and the least value. Let's order the data values from least to greatest. 28, 27, 33, 29, 25, 43, 29, 28, 23, 33 ⇓ 23, 25, 27, 28, 28, 29, 29, 33, 33, 43 The least value in the data set is 23. The greatest value in the data set is 43. Let's find the range by subtracting the least value from the greatest value. Range 43 - 23 = 20 The range of the data values is 20.

Interquartile Range

Next, we will find the interquartile range or IQR. It is the difference of the third quartile and the first quartile. Let's identify the quartiles.

We can see that Q_2 = 28.5. The first quartile, Q_1 is the median of the lower half of the data. The third quartile, Q_3 is the median of the upper half. This means that Q_1 = 27, and Q_3 = 33. Their difference is 6. Interquartile Range 33- 27=6 The interquartile range is 6.

In Part A, we found that the interquartile range of the given data is 6. We also found that the first quartile is Q_1 = 27 and the third quartile is Q_3 = 33. Now, we can identify the outliers. We know that any value less than Q_1 - 1.5(IQR) or greater than Q_3 + 1.5(IQR) is an outlier.

Let's find the exact values of the expressions.

Therefore, the values less than 18 and greater than 42 are outliers. In our data set, there are no values below 18, but there is only one value above 42, namely 43. 23, 25, 27, 28, 28, 29, 29, 33, 33, 43 Therefore, 43 is an outlier.

Let's repeat the process, this time excluding 43 from the data set. 23, 25, 27, 28, 28, 29, 29, 33, 33

Range Without Outlier

We subtract the least value from the greatest. Range Without Outlier 33-23 = 10 The range decreased to 10.

Interquartile Range Without Outlier

Next, we will find the IQR of the new data set. Let's divide the data set into the lower half and the upper half.

We can see that Q_2 = 28. The first quartile, Q_1 is the median of the lower half of the data. The third quartile, Q_3 is the median of the upper half. This means that Q_1 = 26, and Q_3 = 31. Their difference is 5. Interquartile Range 31- 26=5 The interquartile range without the outlier is 5.

Let's summarize our findings in a table below to more easily compare the results.

In this example, removing the outlier reduced both the range and the interquartile range. That said, the range decreased more than the interquartile range. This suggests that outliers have a greater effect on the range than on the interquartile range.

The mean absolute deviation (MAD) is the average of the absolute values of the differences between the mean and each value in the data set. Let's see the given data set. 54, 46, 55, 59, 48, 50 We will start by calculating the mean of the given set of numbers. We can see that there are 6 values in our data set. Let's calculate the mean!

We found that the mean of the given data set is 52. We are ready to calculate the MAD. As previously stated, the MAD of a set of data is the average of the absolute values of the differences between the mean and each value in the data set. Let's use a table to find the sum of the absolute values of the differences.

Finally, we add the absolute values found in the table and then divide the sum by the number of values 6.

The mean absolute deviation of the data is 4. This number indicates that the data, on average, are 4 units away from the mean of 52.

We are given a data set consisting of the number of pages read by 10 different students. 40, 42, 45, 48, 49, 50, 52, 66, 67, 74 We are asked to describe the number of pages that are within one standard deviation — 11 pages — of the mean. First, we need to find the mean. We add all values and divide the sum by the number of the values.

The mean of the data set is 53.5. Now we can find the range of values that are within one standard deviation of the mean! To do that, we need to add one standard deviation to the mean and subtract one standard deviation from the mean.

The data values that are between 42.3 and 64.3 are within one standard deviation of the mean. 40, 42, 45, 48, 49, 50, 52, 66, 67, 74 ↓ 42.3 2.35cm ↓ 64.3 1.3cm The values that are between 43.3 and 64.3 are 45, 48, 49, 50, and 52. Therefore, the values within the range of one standard deviation from the mean are 45, 48, 49, 50, and 52.