Respuesta :
Solution :
Number Values (X) [tex]$\sum (X_i - \bar X)^2$[/tex]
1 121.5 193494.4144
2 138.9 178489.3504
3 155.7 164576.2624
4 172 151616.7844
5 172.9 150916.7104
6 196.5 133137.4144
7 202.3 128938.4464
8 202.7 128651.3424
9 334.5 51474.5344
10 504 3292.4644
11 563.7 5.3824
12 574.1 161.7984
13 935 139591.9044
14 952.6 153053.0884
15 3194.3 6932267.726
Total 8420.7 8509667.624
Mean [tex]$\bar X = \sum \frac{X_i}{n}$[/tex]
[tex]$=\frac{8420.7}{15}$[/tex]
= 561.38
Median :
[tex]$Q_2 = \left(\frac{1}{2}n+\frac{1}{2}\right)^{th}$[/tex]value
[tex]$ = \left(\frac{1}{2}\times 15+\frac{1}{2}\right)^{th}$[/tex]value
= 202.7
Percentile = [tex]$\left(\frac{P}{100} \times n + \frac{1}{2}\right)^{th}$[/tex] value
= [tex]$\left(\frac{50}{100} \times 15 + \frac{1}{2}\right)^{th}$[/tex] value
= 202.7
Percentile = [tex]$\left(\frac{P}{100} \times n + \frac{1}{2}\right)^{th}$[/tex] value
= [tex]$\left(\frac{40}{100} \times 15 + \frac{1}{2}\right)^{th}$[/tex] value
= 198.82
Percentile = [tex]$\left(\frac{P}{100} \times n + \frac{1}{2}\right)^{th}$[/tex] value
= [tex]$\left(\frac{90}{100} \times 15 + \frac{1}{2}\right)^{th}$[/tex] value
= 1849.28
[tex]$Q_1 = \left(\frac{1}{4}n+\frac{1}{4}\right)^{th}$[/tex]value
[tex]$ = \left(\frac{1}{4}\times 15+\frac{1}{4}\right)^{th}$[/tex]value
= 172
[tex]$Q_3 = \left(\frac{3}{4}n+\frac{3}{4}\right)^{th}$[/tex]value
[tex]$ = \left(\frac{3}{4}\times 15+\frac{3}{4}\right)^{th}$[/tex]value
= 574.1
Therefore, the range = highest value - lowest value
= 3194.3 - 121.5
= 3072.8
Now the interquartile range = [tex]$Q_3-Q_1$[/tex]
= 574.1 - 172
= 402.1
