Question

From frequency distribution table given below, find-
i) mean
ii) median
iii) mode
iv) variance
v) standard deviation
L1 — L2 50–52 53–55 56–58 59–61 62–64
f 5 10 21 8 6

Answer 1

(i):

$$L_x =(L_2+L_1)/2$$ $$f_x=f \cdot L_x$$ $$Mean= \frac {\sum f_x} {\sum f}=2850/50=57$$

(ii):

$$\text {F - cumulative frequency}$$ $$\frac {\sum f} {2} = 25 \rightarrow \text { class median: 56-58}$$ $$L_m=55.5 \text { - the lower boundary of the class median}$$ $$f_m=21 \text { - the frequency of the class median}$$ $$F_m=15 \text { - the cumulative frequency before class median}$$ $$w=3 \text { - the class width}$$ $$Median=L_m + \frac {\frac {\sum f} {2} - F_m} {f_m} w = 55.5 + \frac {25-15} {21} 3 = 56.929$$

(iii):

$$f_{max} = 21 \rightarrow \text { modal group: 56-58}$$ $$L_{md}=55.5 \text { - the lower boundary of the modal group}$$ $$f_{md} = 21 \text { - the frequency of the modal group}$$ $$Mode = L_{md} + \frac {f_{md} - f_{md-1}} {(f_{md} - f_{md-1})+(f_{md} - f_{md+1})} w=$$ $$=55.5+\frac {21-10} {21-10+21-8} 3=56.875$$

(iv):

$$Variance=\sigma ^2 = \frac {\sum f_x ^2} {\sum f} - (\frac {\sum f_x } {\sum f})^2 = \frac {2162718} {50} - 57^2 = 40005.36$$

(v):

$$\text {Standard deviation} = \sqrt{\sigma^2}=200.013$$