Test Index

Statistics Part 2

Section: Mathematics

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Question : 13 of 50

Marks: +1, -0

If the mean and the variance of the data

Class	$4-8$	$8-12$	$12-16$	$16-20$
Frequency	3	$\lambda$	4	7

\mu

and 19 respectively, then the value of

\lambda + \mu

[23 jan 2026 Shift 2]

21
19
18
20

Solution:

Concept:

For grouped data, mean

\mu = \frac{\sum f_i x_i}{\sum f_i}

and variance

\sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - \mu^2

, where

x_i

are class marks.

Explanation:

Class marks:

x_i = 6, 10, 14, 18

for classes

4-8, 8-12, 12-16, 16-20

respectively.

Construct table:

Class	$f_i$	$x_i$	$f_i x_i$	$f_i x_i^2$
$4-8$	$3$	$6$	$18$	$108$
$8-12$	$\lambda$	$10$	$10\lambda$	$100\lambda$
$12-16$	$4$	$14$	$56$	$784$
$16-20$	$7$	$18$	$126$	$2268$
Total	$14+\lambda$		$200+10\lambda$	$3160+100\lambda$

Mean:

\mu = \frac{200+10\lambda}{14+\lambda}

.

Variance:

19 = \frac{3160+100\lambda}{14+\lambda} - \mu^2

.

Substitute

\mu

:

19 = \frac{3160+100\lambda}{14+\lambda} - \left(\frac{200+10\lambda}{14+\lambda}\right)^2

.

Multiply by

(14+\lambda)^2

:

19(14+\lambda)^2 = (3160+100\lambda)(14+\lambda) - (200+10\lambda)^2

.

Expand:

19(196+28\lambda+\lambda^2) = (44240+4560\lambda+100\lambda^2) - (40000+4000\lambda+100\lambda^2)

.

Simplify:

3724+532\lambda+19\lambda^2 = 4240+560\lambda

.

Thus

19\lambda^2 -28\lambda -516 =0

.

Solve:

\lambda = \frac{28 \pm \sqrt{28^2-4(19)(-516)}}{38} = \frac{28 \pm 200}{38}

.

Positive

\lambda = \frac{228}{38}=6

.

Then

\mu = \frac{200+10(6)}{14+6} = \frac{260}{20}=13

.

So

\lambda + \mu = 6+13 = 19

.

Answer:

Option B: 19

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