# Answer in Statistics and Probability for KCK #249251

**SOLUTION **

I drew a frequency table to represent the data and ease the calculation

(a) Calculate the mean

“Mean(mu)=frac{Sigma{f(x)}}{n}=frac{780}{100}=7.8”

**Answer **“=7.8”

(b) Calculate the variance and standard deviation

“sigma^2=frac{Sigma{f(x)^2}-(Sigma{f(x))^2}}{n}”

“=frac{6440-(780)^2}{100}=3.56\nsigma=sqrt{3.56}=1.8868”

**Answer **“=1.8868”

(c) Calculate the mode

Maximum frequency is 40. The mode class is 7-9.

“L=” lower boundary point of mode class “=7”

“f_1=” frequency of the mode class“=40”

“f_0=” frequency of the preceding class “=30”

“f_2=” frequency of the succeeding class “=20”

“c=” class length of mode class “=2”

“Mode=Z=L+Big(frac{f_1-f_0}{2f_1-f_0-f_2}Big)*c\=7+Big(frac{40-30}{2(40)-30-20}Big)*2=7.6667”

**Answer **“=7.6667”

(d) Calculate the median

value of “(n/2)th” observation “=” value of “(100/2)th” observation “=”

value of “(50)th” observation

From the column of cumulative frequency “cf,” we find that the “(50)th” observation lies in the class 7-9.

The median class is 7-9.

“L=” lower boundary point of median class “=7”

“n=” Total frequency “=100”

“cf=” Cumulative frequency of the class preceding the median class “=35”

“f=” Frequency of the median class “=40”

“c=” class length of median class “=2”

“median=M=L+(dfrac{dfrac{n}{2}-cf}{f})cdot c”

“=7+(dfrac{50-35}{40})cdot 2=7.75”

**Answer **“=7.75”

(e) Calculate the coefficient of variation

coefficient of variation “=dfrac{sigma}{bar{x}}cdot100%=dfrac{1.8868}{7.8}cdot100%approx24.19%”

**Answer **“=24.19%”