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__

__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=1+2+3+4+55=155=3Mean=1+2+3+4+55=155=3

#### STEP 2

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

⇒(1−3),(2−3),(3−3),(4−3),(5−3)⇒−2,−1,0,1,2⇒(1−3),(2−3),(3−3),(4−3),(5−3)⇒−2,−1,0,1,2

#### STEP 3

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

⇒(−2)2,(−1)2,(0)2,(1)2,(2)2⇒4,1,0,1,4⇒(−2)2,(−1)2,(0)2,(1)2,(2)2⇒4,1,0,1,4

#### STEP 4

Now discover Standard Deviation utilizing the underneath equation

S=∑X−M2n−1−−−−−−−√S=∑X−M2n−1

Standard Deviation = 1√04√=1.58113104=1.58113

#### STEP 5

Pooled Variance (r) =((aggregate check of numbers −1)×Var)(aggregate tally of numbers−1), (r)=(5−1)×2.5(5−1), =(4×2.5)4=2.5Pooled Variance (r) =((aggregate check of numbers −1)×Var)(aggregate tally of numbers−1), (r)=(5−1)×2.5(5−1), =(4×2.5)4=2.5

Hence, Pooled Variance (r) =2.5