Answer to Question #237194 in Statistics and Probability for Steve

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