If for a distribution $\sum(x-5) = 3$, $\sum(x-5)^2 = 43$ and the total number of item is 18, find the mean and standard deviation.

If for a distribution Σ(x-5) = 3, Σ(x-5)² = 43 and the total number of item is 18, find the mean and standard

NCERT Exemplar · Q14

\sum(x-5) = 3

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Working from the deviations $d = x-5$ : mean $=\dfrac{31}{6}\approx 5.17$ and standard deviation $=\dfrac{\sqrt{85}}{6}\approx 1.54$ .

Let $d_i = x_i - 5$ , so $\sum d_i = 3$ , $\sum d_i^2 = 43$ and $n = 18$ .

Mean. The mean of the deviations is $\bar d = \dfrac{\sum d_i}{n} = \dfrac{3}{18} = \dfrac{1}{6}$ . Since $x = d + 5$ ,

\bar x = \bar d + 5 = \frac{1}{6} + 5 = \frac{31}{6} \approx 5.17.

Standard deviation. Adding a constant does not change the spread, so $\sigma_x = \sigma_d$ : …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2023Set ANNUAL4 marks
$Q. Find the standard deviation for the following data: x: 3, 8, 13, 18, 23 f: 7, 10, 15, 10, 6$
›Reveal solutionSolution

Using the step-deviation method (assumed mean $A=13$ , width $h=5$ ), the standard deviation of the given distribution is approximately $6.12$ .

Given data:

$x_i$ 3 8 13 18 23
$f_i$ 7 10 15 10 6

Step 1: Total frequency.

$N = 7+10+15+10+6 = 48$

Step 2: Step-deviation, using $A=13,\ h=5$ .

$d_i = \dfrac{x_i-A}{h}$

$x_i$ 3 8 13 18 23
$d_i$ $-2$ $-1$ $0$ $1$ $2$
$f_i$ 7 10 15 10 6
$f_id_i$ $-14$ $-10$ $0$ $10$ $12$
$f_id_i^2$ 28 10 0 10 24

Step 3: Sum the columns.

$\sum f_id_i = -14-10+0+10+12 = -2$

$\sum f_id_i^2 = 28+10+0+10+24 = 72$

Step 4: Apply the step-deviation variance formula.

…

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Set ANNUAL4 marks
$x_i$
›Reveal solutionSolution

Applying the step-deviation shortcut ( $A=13$ , $h=5$ ) to the frequency data gives a standard deviation of approximately $6.12$ .

The data ( $x_i$ : $3,8,13,18,23$ ; $f_i$ : $7,10,15,10,6$ ) is equally spaced with common difference $h=5$ . Take assumed mean $A=13$ and let $d_i=\dfrac{x_i-A}{h}$ :

$x_i$ $f_i$ $d_i$ $f_id_i$ $f_id_i^2$
3 7 $-2$ $-14$ $28$
8 10 $-1$ $-10$ $10$
13 15 $0$ $0$ $0$
18 10 $1$ $10$ $10$
23 6 $2$ $12$ $24$

$N=\sum f_i=7+10+15+10+6=48,\qquad \sum f_id_i=-2,\qquad \sum f_id_i^2=72.$

The step-deviation formula for standard deviation is …

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$x_i$	3	8	13	18	23
$f_i$	7	10	15	10	6

$x_i$	3	8	13	18	23
$d_i$	$-2$	$-1$	$0$	$1$	$2$
$f_i$	7	10	15	10	6
$f_id_i$	$-14$	$-10$	$0$	$10$	$12$
$f_id_i^2$	28	10	0	10	24

$x_i$	$f_i$	$d_i$	$f_id_i$	$f_id_i^2$
3	7	$-2$	$-14$	$28$
8	10	$-1$	$-10$	$10$
13	15	$0$	$0$	$0$
18	10	$1$	$10$	$10$
23	6	$2$	$12$	$24$

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Q.If for a distribution $\sum(x-5) = 3$ , $\sum(x-5)^2 = 43$ and the total number of item is 18, find the mean and standard deviation.

Concept understanding — Variance and Standard Deviation of Data

Variance and Standard Deviation of Data

Q.If for a distribution ∑(x−5)=3\sum(x-5) = 3∑(x−5)=3, ∑(x−5)2=43\sum(x-5)^2 = 43∑(x−5)2=43 and the total number of item is 18, find the mean and standard deviation.

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Q.If for a distribution $\sum(x-5) = 3$ , $\sum(x-5)^2 = 43$ and the total number of item is 18, find the mean and standard deviation.

Unlock everything free for 14 days