### Bat Wings

Two students collected some data on the wingspan of bats, but each lost a measurement. Can you find the missing information?

### Kate's Date

When Kate ate a giant date, the average weight of the dates decreased. What was the weight of the date that Kate ate?

### Balancing the Books

How many visitors does a tourist attraction need next week in order to break even?

# Average Temperature

##### Age 11 to 14 Short Challenge Level:

Finding the sum of the temperatures
The mean of the first $6$ cities is the sum of all of the temperatures and then dividing by $6$.
So the sum of the temperatures, before dividing by $6$, must have been $6\times5=30^\circ$C.

When $12^\circ$C and $-6^\circ$C are added to the list, the new sum will be $30 + 12 - 6 = 36^\circ$C.
There are now $8$ cities on the list, so the new average temperature can be found by dividing by $8$.

$36^\circ$C$\div8 = 4.5^\circ$C so the new average temperature is $4.5^\circ$C

Finding the distances from the mean
The mean temperature is balanced between the individual temperatues, so that the total distance from the mean to the temperatures above it is equal to the total distance to the temperatures above it.
$12^\circ$C is $7^\circ$C above the mean, so it will pull the mean up.
$-6^\circ$C is $11^\circ$C below the mean, so it will pull the mean down.

If there was only $1$ city in the sample, then the mean would be its temperature.
If there were $2$ cities in the sample, then the mean would be halfway between their temperatures.
If there were $3$ cities in the sample, then each would have a $\frac13$ impact on the mean.
So since there are now $8$ cities in the sample, each will have a $\frac18$ impact on the mean.

So the city where it is $12^\circ$C accounts for a $\frac78 ^\circ$C increase in the mean, and the city where it is $-6^\circ$C accounts for a $\frac{11}8^\circ$C decrease in the mean.
So the total change will be $\frac78-\frac{11}8=-\frac48=-\frac12$
So the new mean will be $5^\circ$C$-\frac12^\circ$C=$4\frac12^\circ$C.

Pairs of cities
Both of the methods above can also be applied to pairs of cities. Instead of thinking of adding 2 cities to a sample of 6, we can think of adding 1 pair to a sample of 3, where the mean of the new pair is $(12+-6)\div2=3^\circ$C.

Finding the sum of the mean temperatures of the pairs
The mean temperature of the first $3$ means in the sample is $5^\circ$C, so the sum of the first $3$ means in the sample must have been $3\times5^\circ$C$=15^\circ$C.

So including the mean of the new pair, the sum is $15^\circ$C$+3^\circ$C$=18^\circ$C.

So the new mean is $18^\circ$C$\div4=4.5^\circ$C.

Finding the distance from the mean
This pair has mean $3^\circ$C which is $2^\circ$C less than the mean of the first $3$ pairs.
There are now $4$ pairs, so each pair's temperature has a $\frac14$ impact on the mean.

So this pair decreases the mean by $\frac14\times 2^\circ$C$= \frac12^\circ$C.
So the new mean is $5^\circ$C$-\frac12^\circ$C$=4\frac12^\circ$C.
You can find more short problems, arranged by curriculum topic, in our short problems collection.