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Mean: 10.529
Median: 10
Mode: 9 and 10
Range: 7
Standard Deviation: Approximately 1.882
Outliers: none
We want to find the mean, median, mode, range, and standard deviation of the given data set, as well as whether there are any outliers.
The mean of a data set is the sum of the values divided by the total number of values in the set. Let's start by calculating the sum of the given values. 10+15+9+8+11+10+9+8+ 13+9+12+10+13+11+9+12+10 = 179 There are 17 values in our set, so we have to divide the sum by 17. Mean: 179/17 ≈ 10.529
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.
8, 8, 9, 9, 9, 9, 10, 10, 10,
10, 11, 11, 12, 12, 13, 13, 15
The median of this set is 10.
The mode is the value or values that appear most often in a set of data. Let's find the mode of the given values. 10+15+9+8+11+ 10+9+8+ 13+9+12+ 10+13+11+9+12+ 10 Since the data set has two values that appear more often than the other values but equally as often as each other, there are two modes. Mode: 9and 10
The range is the difference between the greatest and least values in a set of data. For our exercise, the greatest value is 15 and the least value is 8. Range: 15-8=7
The standard deviation in a set of data is the average amount by which each individual value deviates or differs from the mean. Standard Deviation sqrt((x_1-μ )^2+(x_2-μ )^2+... +(x_n-μ )^2/n) In the above formula, x_1, ... ,x_n are the values of the set of data, μ is the mean, and n is the number of values. For this exercise, the mean is the sum of the values 179 divided by the number of values 17. μ= 179 17 ≈ 10.529 Let's use this value and apply the formula to each value in the set.
| x_n | x_n-μ | (x_n-μ)^2 |
|---|---|---|
| 10 | 10-10.529=- 0.529 | (- 0.529)^2≈ 0.280 |
| 15 | 15-10.529=4.471 | 4.471^2≈ 19.990 |
| 9 | 9-10.529=- 1.529 | (- 1.529)^2≈ 2.338 |
| 8 | 8-10.529=- 2.529 | (- 2.529)^2≈ 6.396 |
| 11 | 11-10.529=0.471 | 0.471^2≈ 0.222 |
| 10 | 10-10.529=- 0.529 | (- 0.529)^2≈ 0.280 |
| 9 | 9-10.529=- 1.529 | (- 1.529)^2≈ 2.338 |
| 8 | 8-10.529=- 2.529 | (- 2.529)^2≈ 6.396 |
| 13 | 13-10.529=2.471 | 2.471^2≈ 6.106 |
| 9 | 9-10.529=- 1.529 | (- 1.529)^2≈ 2.338 |
| 12 | 12-10.529=1.471 | 1.471^2≈ 2.164 |
| 10 | 10-10.529=- 0.529 | (- 0.529)^2≈ 0.280 |
| 13 | 13-10.529=2.471 | 2.471^2≈ 6.106 |
| 11 | 11-10.529=0.471 | 0.471^2≈ 0.222 |
| 9 | 9-10.529=- 1.529 | (- 1.529)^2≈ 2.338 |
| 12 | 12-10.529=1.471 | 1.471^2≈ 2.164 |
| 10 | 10-10.529=- 0.529 | (- 0.529)^2≈ 0.280 |
| Sum of Values | ≈ 60.238 | |
Finally, we need to divide by 17 and then calculate the square root. Standard Deviation: sqrt(60.238/17)≈ 1.882
To find the outliers, let's start by defining some important characteristics of data sets.
Let's find the upper quartile, the lower quartile, and the interquartile range for the given data set. Do not forget to write the values in numerical order! 8, 8, 9, 9, 9, 9, 10, 10, | 10 | 10, 11, 11, 12, 12, 13, 13, 15 We have two middle values for each half. Thus, we need to calculate the mean of those middle values. Upper Quartile:& 12+ 122=12 Lower Quartile:& 9+ 92=9 Interquartile Range:& 12-9=3 An outlier is an extremely high or extremely low value when compared with the rest of the values in the set. To check for them, we look for data values that are beyond the upper or lower quartiles by more than 1.5 times the interquartile range. Minimum=& Q1-IQR*1.5 Maximum=& Q2+IQR*1.5 Let's find the minimum and maximum for a value not to be an outlier. ccc Minimum& &Maximum 9-1.5* 3=4.5 & & 12+1.5* 3=16.5 Any value between 4.5 and 16.5 is not an outlier. Notice that the data set does not have any outliers.