Amanda\'s lecture was working on finding the mean of a minor set of numbers. The

Question

Amanda's lecture was working on finding the mean of a minor set of numbers. The lecture was given four scores: 70, 80, 90, and 80 and asked them to find the lecturer then added a fifth score 50, to these scores and asked the lecture to find the new mean. Here is how Amanda labeled her solution method: "50 is 30 fewer than 80, so I divided 30 by five and got 6. And 80 minus 6 is 74 so that is the new mean." Describe why Amanda's method works by using the mean as balance point. Given four numbers a_1, a_2, a_3, and a_4 with mean a bar, if a fifth number, a_5, is added to the set then the mean of the five numbers is given by the expression ________. Prove Part B). Restate this as an arbitrarily large data set. Explain why Part D) is true by generalizing Part A).

Explanation / Answer

Solution

Back-up Theory

If average of n numbers is A, then their total = nA. ……………………………(1)

Now, to work out solution,

Part (a)

The average of the first 4 numbers =(70 + 80 + 90 + 80)/4 = 80. Hence, by (1) above, their total = (4 x 80).

On including one more value, 50, the new total of 5 values = (4 x 80) + 50 = (5 x 80) + 50 – 80

[by adding and subtracting 80 does not change the sum] = (5 x 80) – 30.

Hence, the new average of 5 values = {(5 x 80) – 30}/5 = 80 – (30/5) = 80 – 6 = 74. DONE   

Part (b)

abar – {(abar – a5)}/5} ANSWER

Part (c)

The average of the first 4 numbers =(a1 + a2 + a3 + a4)/4 = abar.. Hence, by (1) above, their total = (4 x abar).

On including one more value, a5, the new total of 5 values = (4 x abar) + a5

= (5 x abar) + a5 – abar[by adding and subtracting abar does not change the sum]

= (5 x abar) – (abar – a5).

Hence, the new average of 5 values = {(5 x abar) – (abar – a5)}/5

= abar – {(abar – a5)/5} DONE

Part (d)

To generalize,

Let A be the average of n values a1, a2, ……., an. Then, their total = (a1+ a2 + ……. + an)

= nA

Now, suppose, k additional values, b1, b2, ……., bk are added to this set. Then, the total of all (n+ k) values

= nA + (b1+ b2 + ……., + bk) and their average

= {nA + (b1+ b2 + ……., + bk)}/(n + k).

Adding to and subtracting from the numerator of the above, kA, we have,

New total = {(nA + kA) – kA + (b1+ b2 + ……., + bk)} and hence the new average

= {(nA + kA) – kA + (b1+ b2 + ……., + bk)}/(n + k)

= [A(n + k) – {kA - (b1+ b2 + ……., + bk)}]/(n + k)

= A – [{kA - (b1+ b2 + ……., + bk)}/(n + k)]

= A – [{(A – b1) + (A – b2) + ….. + (A – bk)}/(n + k)]

= A – [{[1,k](A – bi)}/(n + k)] ANSWER

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