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Question 571178: After finding the average of 35 scores, a student carelessly included the average with the 35 scores and found the average of these 36 numbers. The ratio of the second average to the true average was
(A) 1 : 1 (B) 35 : 36 (C) 36 : 35 (D) 2 : 1 (E) None of these
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i believe the answer is 1:1 which would be selection A.
there's a couple of ways you can prove this.
the first way is to take any average number and any number of data points that make up that average.
assume you have 556 data points whose average is equal to 375.
these numbers were chosen at random.
if we let x equal the total of the numbers, then we get:
x / 556 = 375
if we multiply both sides of this equation by 556, then we get:
x = 375 * 556 = 208500
we have a total of 208500 for 556 data points whose average is 375.
we add the average to this set of data points to get a new total of 208500 + 375.
this is equal to 208875/
we divide this by 557 data points (556 of the original plus the additional data point of the average) to get:
208875 / 557 = 375
this is the exact same average that we original got which makes the ratio of the second average to the true average equal to 1:1.
we can also prove this by formula as follows:
the original average is equal to T/N
T is the total of the data points and N is the number of data points.
the second average is equal to T plus the original average divided by N+1
we know that N is equal to 35 which makes N+1 equal to 37.
our original average becomes T/35
our second average becomes (T + the original average)/37
since the original average is equal to T/35, the second average becomes:
(T + T/35)/36
we can simplify this expression to get:
(36T/35)/36
this is because T is equivalent to 35T/35 which, when added to T/35, becomes 36T/35.
our second average is therefore equal to (36T/35)/36
this is the same as 36T/35 * 1/36 which is the same as 36T/(35*36)
since the 36 in the numerator cancels out the 36 in the denominator, we are left with:
second average = T/35.
this is the same as the original average, so the ratio is 1:1.
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