Answer to Question #249251 in Statistics and Probability for KCK

Question #249251
The following table give the grouped data on the weights of all 100 babies born at Matero level 1
clinic last year.
Weight (Units) Number of
babies
3 to less than 5 5
5 to less than 7 30
7 to less than 9 40
9 to less than 11 20
11 to less than 13 5
(a) Calculate the mean
(b) Calculate the variance and standard deviation
(c) Calculate the mode
(d) Calculate the median
(e) Calculate the coefficient of variation
1
Expert's answer
2021-10-14T06:46:02-0400

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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