Question

The median and the mode of the following distribution are known to be 27 and 26 respectively.

Class 0-10 10-20 20-30 30-40 40-50

Frequency 3 a 20 12 b

(i) Determine the values of a and b.                              (5 Marks)

(ii) Compute the arithmetic mean and the standard deviation of the distribution.

Answer 1

(i) Given median = 27 and mode = 26

So, N= 3 + a+ 20 + 12 + b

= 35 + a + b

Therefore Mode lies in the 20 – 30 class and its frequency will be 20

So, Mode = L +$\dfrac{(f_m-f_1)h}{(f_m-f_1)(f_m-f_2)}$

   

$26= 20+ \dfrac{(20-a)10}{(20-a)+(20-12)}$

      $6=\dfrac{200-10a}{28-a}$

     200 – 10 a = 168 – 6a

      Or 4a = 200 – 168

      Or a = 8

Now Median = $l + (\dfrac{\frac{N}{2}-c.f.}{f})h$

Now class interval for median = 20 – 30, so l = 20

So, $\dfrac{N}{2}= \dfrac{(35+a+b)}{2}$

  c.f.= F = 3 + a=11

 f = 20 and h = 10

Now,

Median = $l + (\dfrac{\frac{N}{2}-c.f.}{f})h$

$27 = 20 +\dfrac{ (\dfrac{35+a+b}{2}-11)10}{20}$

$14 = \dfrac{43+b}{2}-11\\43+b=50\\\Rightarrow b=7$

(ii)

Mean $\bar x =A + \dfrac{\sum fd}{n}\times h$

=25+1250⋅10

=25+0.24⋅10

=25+2.4

Mean =27.4

Standard deviation $S=\sqrt{\dfrac{\sum f\times d^2-\frac{(\sum fd)^2}{n}}{n-1}}\times h$

S= $\sqrt{\dfrac{60-\frac{(12)^2}{50}}{49}}⋅10$

S =10.7968

Standard Deviation = 10.7968