Answer:
1. The range = 17
2. The mean = 29.053846153846
3. The median = 29.5
4.i) Q1 = 28
ii) Q3 = 32
5. The IQR = 4
6. The variance = 19.7626923
7. The standard deviation = 4.445524975
8. Identify any outliers = 17.5
Step-by-step explanation:
We are given this random samples of data
{32.3, 28.8, 29.1, 29.5, 30.8, 31.0, 17.5, 34.5, 23.5, 33.0, 31.7, 28.6, 27.4}
First step before solving any of the question is we rearrange the numbers from lowest to highest
17.5, 23.5, 27.4, 28.6, 28.8, 29.1, 29.5, 30.8, 31.0, 31.7, 32.3, 33.0, 34.5
1. The range
This is the difference between the Maximum value and the Minimum value in a given data set.
Maximum value = 34.5
Minimum value = 17.5
Range = 34.5 - 17.5
= 17
2. The mean
Mean = Sum of terms/Number of terms
= 17.5 + 23.5 + 27.4 + 28.6 + 28.8 + 29.1 +29.5 + 30.8 + 31.0 +31.7 +32.3 +33.0 +34.5/13
= 377.7/13
= 29.053846153846
3. The median
The formula for Median = 1/2(n + 1)th value
n = 13
= 1/2(13 + 1)th value
= 1/2(14)th value
= 7th value.
The 7th value for the data set = 29.5
4.i) Q1 means the First Quartile
The formula for First Quartile
= 1/4(n + 1)th value
= 1/4(13 + 1)th value
= 14/4 th value
= 3.5 th value
This means, the first quartile is between the 3rd and 4th value
3rd value = 27.4
4th value = 28.6
Q1 = 27.4 +28.6/2
= 56/2
= 28
ii) Q3
The Formula for Third Quartile
= 3/4(n + 1)th value
= 3/4(13 + 1)th value
= 42/4 th value
= 10.5th value
This means, the first quartile is between the 10th and 11th value
10th value = 31.7
11th value = 32.3
Q1 = 31.7 +32.3/2
= 64/2
= 32
5. The IQR
IQR = Interquartile range. It is the difference between Q3 and Q1
= 32 - 28
= 4
6. The sample variance
The formula = (x -mean)²/n - 1
= [(17.5 -29.053846153846)² + (23.5 -29.053846153846)² + (27.4 -29.053846153846)² + (28.6 -29.053846153846)² + (28.8 -29.053846153846)² + (29.1 -29.053846153846)² + (29.5 -29.053846153846)² +(30.8 -29.053846153846)² + (31.0 -29.053846153846)² + (31.7 -29.053846153846)² +(32.3 -29.053846153846)² +(33.0 -29.053846153846)² +(34.5 -29.053846153846)² ] /13 - 1
= [133.4913609+ 30.8452071 +2.735207101 +0.2059763313 + 0.0644378698 + 0.002130177514 + 0.1990532545 + 3.049053254 + 3.787514792 + 7.002130177 + 10.53751479 + 15.57213018 +29.66059171] /12
= 237.1523076/12
= 19.7626923
7. The standard deviation =
= √Variance
= √ 19.7626923
= 4.445524975
8. Identify any outliers
Outliers are number in a data set that varies different from the other data. It can be a extremely large or extremely small.
The outliers in the data set = 17.5
