# Statistics - Pooled Variance (r)

Pooled Variance/Change is the weighted normal for assessing the fluctuations of two autonomous variables where the mean can differ between tests however the genuine difference continues as before.

### Example

Problem Statement:

Compute the Pooled Variance of the numbers 1, 2, 3, 4 and 5.

Solution:

#### Step 1

Decide the normal (mean) of the given arrangement of information by including every one of the numbers then gap it by the aggregate include of numbers given the information set.

${Mean = \frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3 }$

#### Step 2

At that point, subtract the mean worth with the given numbers in the information set.

${\Rightarrow (1 - 3), (2 - 3), (3 - 3), (4 - 3), (5 - 3) \Rightarrow - 2, - 1, 0, 1, 2 }$

#### Step 3

Square every period's deviation to dodge the negative numbers.

${\Rightarrow (- 2)^2, (- 1)^2, (0)^2, (1)^2, (2)^2 \Rightarrow 4, 1, 0, 1, 4 }$

#### Step 4

Now discover Standard Deviation utilizing the underneath equation

${S = \sqrt{\frac{\sum{X-M}^2}{n-1}}}$

Standard Deviation = ${\frac{\sqrt 10}{\sqrt 4} = 1.58113 }$

#### Step 5

${Pooled\ Variance\ (r)\ = \frac{((aggregate\ check\ of\ numbers\ - 1) \times Var)}{(aggregate\ tally\ of\ numbers - 1)} , \\[7pt] \ (r) = (5 - 1) \times \frac{2.5}{(5 - 1)}, \\[7pt] \ = \frac{(4 \times 2.5)}{4} = 2.5}$

Hence, Pooled Variance (r) =2.5

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