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\\\\\n\\sigma=\\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\\%"
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