The average (arithmetic mean) the 100 dimensions is 23 and also the mean of 50 added is 27. What is the typical of the 150 measurements?

Hint: What is the sum of the first \$100\$? What is the amount of the extr \$50\$? What now?

Alternate Hint: We can pretend that the very first \$50\$ measurements and the following \$50\$ dimensions both had actually an average of \$23.\$ What is the mean of \$23,23,\$ and \$27\$? This method is less straightforward, and also requires justification, however it quiet works.

You are watching: The average arithmetic mean of 100 measurements is 23

Hint: You can assume (think why) that all the early \$100\$ dimensions are same to \$23\$ each and all the succeeding \$50\$ measurements are same to \$27\$ each. Does this help?

<(23*100)+(27*50)>/150=24.3333333333

<(23*50)+(23*50)+(27*50)>/150=24.3333333333

Think of it as there being 100 23"s and 50 27"s since it doesn"t matter what the in reality numbers are due to the fact that they will certainly each make up for each various other in a means such that it is tantamount to every number in the collection being that number.

Another example: <(26*50)+(20*50)+(30*25)+(24*25)>/150=24.3333333333

Here you check out (23*100)/100 is equal to <(26*50)+(20*50)>/100

Because 26 is three more than 23 and also 20 is 3 less

26*50 gives you an over count of 150 and also 20*50 offers you one under counting of 150 castle cancel out

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edited Nov 5 "14 at 15:45
answer Nov 5 "14 at 15:33

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