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Parin Sir > Quantitative Aptitude (Maths) > Statistical Data > Examples 5 to 8 > Examples 5 to 8

Examples 5 to 8

June 20, 2020
Category: Examples 5 to 8

(5) Find the mode from the data given below :

Marks Obtained	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
Number of Students	2	15	21	7	3

Soln. Modal Class = The Class having Maximum Frequency = 20 – 30

$\therefore$ L = 20, C = 10, $f_{m}$ = 21, $f_{1}$ = 15, $f_{2}$ = 7

Now, Z = $L+ \frac{f_{m}-f_{1}}{2f_{m}-f_{1}-f_{2}} \times C$ = $20+ \frac{21-15}{2(21)-15-7} \times 10$ = 23

(6) Marks obtained by a student in 5 subjects are given below. Maths = 48, Chemistry = 52, Physics = 49, English = 45 and Sanskrit = 38. If the weightage given to these subjects be 5, 4, 4, 1, 1 respectively, then find the weighted mean.

Soln. Weighted Mean = $\frac{\left (48 \times 5 \right ) + \left (52 \times 4 \right ) + \left (49 \times 4 \right ) + \left (45 \times 1 \right ) +\left (38 \times 1 \right )}{5+4+4+1+1}$ = $\frac{727}{15}$ = 48.47

(7) Find the mean weight for the weights of 50 students are given below :

Weight ( in kg. )	40	50	60	70
Number of Students	15	19	9	7

Soln.

Weight ( x )	Freq. $f_{i}$	$f_{i}x_{i}$
40	15	600
50	19	950
60	9	540
70	7	490
Total	n = $\sum f_{i}$ = 50	$\sum f_{i} x_{i}$ = 2580

Now, $\overline{x}= \frac{\sum f_{i}x_{i}}{n} : n= \sum f_{i}$

$= \frac{2580}{50}$ = 51.6 Kg.

(8) Find A.M. from the data given below :

Class	5 – 15	15 – 25	25 – 35	35 – 45	45 – 55
Frequency	5	8	20	10	7

Soln. We have:

Class	Mid-value $x_{i}$	$f_{i}$	$d_{i}=\frac{x_{i}-A}{C}$	$f_{i}d_{i}$
5-15	10	5	– 2	– 10
15-25	20	8	– 1	– 8
25-35	30 = A	20	0	0
35-45	40	10	1	10
45-55	50	7	2	14
Total	––	n = 50	––	$\sum f_{i}d_{i}$ = 6

Now, $\overline{x}= A +\frac{\sum f_{i}d_{i}}{n} \times C$

$=30 + \frac{6}{50} \times 10$

$= 30+1.2$

$=31.2$

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Examples 5 to 8

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