Answer to Question #194657 in Statistics and Probability for ndinoo

Question

The following table is a frequency table of the scores obtained in a competition. Use

the table answer the questions below.

Classes Frequency(f)10 - 13 413 - 16 616 - 19 1219 - 22 1422 - 25 4Total 40a. Find the mean, median and mode of the score. [2,2,2]

b. Find the range, variance, and standard deviation. [1,3,1]

c. Find the coefficient of variation. [2]

d. Compute the interquartile range.

Answer 1

a.

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c} Class & f & m & d & f\cdot d & f\cdot d^2 & cf\\ \hline 10-13 & 4 & 11.5 &-2 &- 8 & 16 & 4 \\ 13-16 & 6 & 14.5 & -1 & -6 & 6 & 10 \\ 16-19 & 12 & 17.5 & 0 & 0 & 0 & 22 \\ 19-22 & 14 & 20.5 & 1& 14 & 14 & 36 \\ 22-25 & 4 & 23.5 & 2 & 8 & 16 & 40 \\ \end{array}$

$A=17.5$

$d=\dfrac{x-A}{h}, h=3$

$Mean=\bar{x}=A+\dfrac{\sum f d}{n}\cdot h$ $=17.5+\dfrac{8}{40}\cdot 3=18.1$

The median class is 16-19.

$L=16, n=40$

Cumulative frequency of the class preceding the median class $cf=10$

Frequency of the median class $f=12$

Class length of median class $c=3$

$Median\ M=L+\dfrac{\dfrac{n}{2}- cf}{f}\cdot c$ $=16+\dfrac{\dfrac{40}{2}- 10}{12}\cdot 3=18.5$

Maximum frequency is14.

The mode class is 19-22.

$L=19$

Cumulative frequency of the class preceding the median class $cf=10$

Frequency of the mode class $f_1=14$

Frequency of the preceding class $f_0=12$

Frequency of the succeeding class $f_2=4$

Class length of mode class $c=3$

$Z=L+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\cdot c$ $=19+\dfrac{14-12}{2(14)-12-4}\cdot 3=19.5$

b.

$Range=\dfrac{22+25}{2}-\dfrac{10+13}{2}=12$ $S^2=\dfrac{\sum f\cdot d^2-\dfrac{(\sum f\cdot d)^2}{n}}{n-1}\cdot h^2$ $=\dfrac{52-\dfrac{(8)^2}{40}}{40-1}\cdot 3^2=\dfrac{151.2}{13}\approx11.63077$ $S=\sqrt{S^2}=\sqrt{\dfrac{151.2}{13}}\approx3.4104$

c.

$Coefficient \ of \ Variation=\dfrac{S}{\bar{x}}\cdot 100\%$ $=\dfrac{\sqrt{\dfrac{151.2}{13}}}{18.1}\cdot 100\%\approx18.842\%$

d.

Class with $(\dfrac{n}{4})^{th}$  value of the observation in $cf$ column

$=(\dfrac{40}{4})^{th}$ value of the observation in $cf$ column

$=(10)^{th}$ value of the observation in $cf$ column

and it lies in the class 13-16.

$Q_1$ class: $13-16$

$L=13$

$Q_1=L+\dfrac{\dfrac{n}{4}- cf}{f}\cdot c$ $=13+\dfrac{\dfrac{40}{4}- 4}{6}\cdot 3=16$

Class with $(\dfrac{3n}{4})^{th}$  value of the observation in $cf$ column

$=(\dfrac{3\cdot 40}{4})^{th}$ value of the observation in $cf$ column

$=(30)^{th}$ value of the observation in $cf$ column

and it lies in the class 19-22.

$Q_3$ class: $19-22$

$L=19$

$Q_3=L+\dfrac{\dfrac{3n}{4}- cf}{f}\cdot c$ $=19+\dfrac{\dfrac{3\cdot 40}{4}- 22}{14}\cdot 3\approx20.7143$ $IQR=Q_3-Q_1\approx20.7143-16\approx4.7143$