Question

The following table gives the grouped data on the weights of all 100 babies born at 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

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} Class & f & \ \ x\ \ & \ f(x)\ & f( x^2) & cf\\ \hline 3-5 & 5 & 4 & 20 & 80 & 5\\ \hdashline 5-7 & 30 & 6 & 180 & 1080 & 35\\ \hdashline 7-9 & 40 & 8 & 320 & 2560 & 75\\ \hdashline 9-11 & 20 & 10 & 200 & 2000 & 95\\ \hdashline 11-13 & 5 & 12 & 60 & 720 & 100\\ \hdashline & n & & \sum f (x) & \sum f (x^2) & & \\ & =100 & & =780 & =6440 & \\ \end{array}$$

(a)

$$mean=\bar{x}=\dfrac{\sum_if_i(x_i)}{n}=\dfrac{780}{100}=7.8$$

(b)

$$\sigma^2=\dfrac{\sum_if_i(x_i)^2-(\sum_if_i(x_i))^2}{n}$$

$$=\dfrac{6440-(780)^2}{100}=3.56$$

$$\sigma=\sqrt{3.56}\approx1.8868$$

(c)

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 succedding class $=20$

$c=$  class length of mode class $=2$

$$mode=Z=L+(\dfrac{f_1-f_0}{2f_1-f_0-f_2})\cdot c$$

$$=7+(\dfrac{40-30}{2(40)-30-20})\cdot2\approx7.6667$$

(d)

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 $=2$

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

$$=7+(\dfrac{50-35}{40})\cdot 2=7.75$$

(e)

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