The mean of two samples of sizes $$200$$ and $$300$$ were found to be $$25$$, $$10$$ respectively. Their standard deviations were $$3$$ and $$4$$ respectively. The variance of combinedsample of size $$500$$ is

Question

Infinity Learn · Accepted Answer

Combined mean x¯$$=n1x$$¯$$1+n2x$$¯$$2n1+n2$$                            $$=200$$×$$25+300$$×$$10500=16$$  Let                        $$d1=x$$¯$$1$$−x¯$$=25$$−$$16=9$$   and                               $$d2=x$$¯$$2$$−x¯$$=10$$−$$16=$$−$$6Now$$ we know that                             σ$$2=n1$$σ$$12+d12+n2$$σ$$22+d22n1+n2=200(9+81)+300(16+36)500=33600500=67.2$$

The mean of two samples of sizes 200 and 300 were found to be 25, 10 respectively. Their standard deviations were 3 and 4 respectively. The variance of combined
sample of size 500 is

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

The mean of two samples of sizes 200 and 300 were found to be 25, 10 respectively. Their standard deviations were 3 and 4 respectively. The variance of combinedsample of size 500 is

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

The mean of two samples of sizes 200 and 300 were found to be 25, 10 respectively. Their standard deviations were 3 and 4 respectively. The variance of combined
sample of size 500 is