Respuesta :
Answer/Step-by-step explanation:
a. To construct a stem-and-leaf plot for the given sample data, 7, 13, 28, 12, 16, 23, 13, 49, 18, 11, 9, 27, 36, 18, 15, 11, 32, 22, 19, 21, first, order the data from the least to the greatest.
7, 9, 11, 11, 12, 13, 13, 15, 16, 18, 18, 19, 21, 22, 23, 27, 28, 32, 36, 49
The plot would like the one shown below:
Stem | Leaf
0 | 7, 9
1 | 1, 1, 2, 3, 3, 5, 6, 8, 8, 9
2 | 2, 3, 7, 8
3 | 2, 36
4 | 9
Key:
13 = 1 | 3
b.
i. Mean = (7 + 9 + 11 + 11 + 12 + 13 + 13 + 15 + 16 + 18 + 18 + 19 + 21 + 22 + 23 + 27 + 28 + 32 + 36 + 49)/20
Mean = 400/20
Mean = 20
ii. Median = average of the 10th and 11th data value
Median = (18 + 18)/2 = 36/2
Median = 18
iii. Range = max value - min value
Range = 49 - 7
Range = 42
iv. Standard deviation for sample data is given as [tex] s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \overline{x})^2} [/tex]
Where,
[tex] \overline{x} = sample mean = 20 [/tex]
[tex] x_i = each value of the sample data [/tex]
N = total number of values in the sample = 20
Let's solve:
First, calculate [tex] (x_i - \overline{x})^2 [/tex] for each data value:
7 => (7 - 20)² = (-13)² = 169
9 => (9 - 20)² = (-11)² = 121
11 => (11 - 20)² = (-9)² = 81
11 => (11 - 20)² = (-9)² = 81
12 => (12 - 20)² = (-8)² = 64
13 => (13 - 20)² = (-7)² = 49
13 => (13 - 20)² = (-7)² = 49
15 => (15 - 20)² = (-5)² = 25
16 => (16 - 20)² = (-4)² = 16
18 => (18 - 20)² = (-2)² = 4
18 => (18 - 20)² = (-2)² = 4
19 => (19 - 20)² = (-1)² = 1
21 => (21 - 20)² = (1)² = 1
22 => (22 - 20)² = (2)² = 4
23 => (23 - 20)² = (3)² = 9
27 => (27 - 20)² = (7)² = 49
28 => (28 - 20)² = (8)² = 64
32 => (32 - 20)² = (12)² = 144
36 => (36 - 20)² = (16)² = 256
49 => (49 - 20)² = (29)² = 841
Next, calculate [tex] \sum_{i=1}^N (x_i - \overline{x})^2 [/tex] by summing all results gotten above:
= 169 + 121 + 81+ 81 + 64 + 49 + 49 + 25 + 16 + 4 + 4 + 1 + 1 + 4 + 9 + 49 + 64 + 144 + 256 + 841
= 2,032
Next, find [tex] \frac{1}{N-1} \sum_{i=1}^N (x_i - \overline{x})^2 [/tex] by dividing the result you have above by (N - 1):
[tex] \frac{2,032}{N - 1} = \frac{2,032}{20 - 1} [/tex]
[tex] \frac{2,032}{19} = 106.947368 [/tex]
Next, calculate [tex] s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \overline{x})^2} [/tex] by finding the square root of the result you have above:
[tex] s = \sqrt{106.947} [/tex]
[tex] s = 10.34 [/tex] (to 2 d.p)
v. Interquartile Range (IQR) = Third Quarter (Q3) - First Quartile (Q1)
Q1 = (12 + 13)/2 = 25/2 = 12.5
Q3 = (23 + 27)/2 = 50/2 = 25
IQR = 25 - 12.5
IQR = 12.5
c. The distribution of the number of horses nominated for each race in the sample is slightly down-skewed. This indicates that most of lower number of horses nominated for each race in the sample chosen are more common. Most of the data values are below the mean value of 20. Also, the median shows the median number of horses selected to be 18, indicating a lower number.
With the distribution having a range value of 42, extreme value such as 49, seem to be an outlier, as it does not reflect the typical number of horses selected in each race, which are mostly lower.