Name: Variance and Standard Deviation for Ungrouped Distribution
Uploaded: 2019-07-19
Duration: 6 min 21 s
Description: In this video, let us learn to find the variance and standard deviation of ungrouped data. We will also learn the formula to find the discrete frequency dist...

Question

The mean and standard deviation of

100

observations were calculated as

40

and

5.1

respectively by a student who took by mistake

50

instead of

40

for one observation. What are the correct mean and standard deviation?

Accepted Answer

We have given

$n = 100$

Mean $= \bar{X} = 40$

$Standard deviation (σ) = 5.1$

$\frac{1}{n} \sum x_{i} = \bar{X}$

$\Rightarrow \sum x_{i} = 100 \times 40 = 4000$

corrected sum $\sum x_{i} = 4000 - 50 + 40$

$\Rightarrow$ corrected sum $= \frac{Corrected sum}{100}$

$= \frac{3990}{100}$

$= 39.9 . . . (1)$

$\sqrt{V ariance} = σ$

$\Rightarrow variance = σ^{2} = (5.1)^{2}$

$= 26.01$

According to the formula,

$\frac{1}{n} \sum x_{i}^{2} - (\bar{X})^{2} =$ variance

$\Rightarrow \frac{1}{100} \sum x_{i}^{2} - (40)^{2} = 26.01$

$\Rightarrow \frac{1}{100} \sum x_{i}^{2} - 1600 = 26.01$

$\Rightarrow \frac{1}{100} \sum x^{2} = 1626.01$

$\Rightarrow \sum x_{i}^{2} = 162601$ (But this is incorrect due to misread values)

$C o r r e c t e d \sum x_{i}^{2} = 162601 - 50^{2} + 40^{2}$

$= 161701$

Corrected variance $= \frac{1}{100} c o r r e c t e d \underset{}{\sum {x_{i}}^{2} - (c o r r e c t e d m e a n)^{2}}$

$= \frac{161701}{100} - (39.9)^{2}$

$= 1617.01 - 1592.01$

$= 25$

Corrected standard deviation $= \sqrt{Corrected variance}$

$= \sqrt{25}$

$= 5$

Corrected mean $= 39.9$

Corrected standard deviation $= 5$

x;	0	1	2	3	4	5	6	7	8	9	10	11	12
f:	51	203	383	525	532	408	273	139	43	27	10	4	2

The mean and standard deviation of $100$ observations were calculated as $40$ and $5.1$ respectively by a student who took by mistake $50$ instead of $40$ for one observation. What are the correct mean and standard deviation?

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