Video Transcript
This graph shows the maximum
temperature for five days. Then we’re shown the graph that
shows temperature in degrees C on the 𝑦-axis and the days of the week along the
𝑥-axis. And there are two parts to the
problem. For what fraction of the five days
was the maximum temperature below 10 degrees C? And then what was the mean maximum
temperature to one decimal place?
Before we start, let’s have a look
at the graph. As we’ve said already, the 𝑦-axis
shows the temperature in degrees C. And we can see that the scale goes
up in two degrees: two, four, six, and so on. So each of the horizontal lines
that go across the graph go up in two degrees each time.
Along the bottom, we have the days
of the week from Monday to Friday. And we can see five temperatures
that have been plotted by using crosses. So on Monday, the maximum
temperature was 8.1 degrees. On Tuesday, it was 9.3 degrees. On Wednesday, it was 11.9
degrees. On Thursday, it was 11.8
degrees. And the maximum temperature on
Friday was 12.4 degrees.
The first part of the problem asked
us for what fraction of the five days was the maximum temperature below 10
degrees. Here’s where 10 degrees is on our
graph. We can see that only on two days
was the maximum temperature below 10 degrees. On Monday, it was 8.1. And on Tuesday, it was 9.3. But we can’t give the answer two
days because we’re asked for what fraction of the five days. The temperature was less than 10
degrees on two out of five days. And we can write that as a fraction
as two-fifths, two out of our possible five.
The second part of the problem asks
us to find the mean maximum temperature. The mean is a type of average. How can we find the mean out of
this set of numbers? We find the mean by adding all of
the numbers together to find the overall total. We then divide by the number of
numbers that there are.
Let’s look for ways to combine the
numbers quickly. We can add Monday and Wednesday to
start with. We know that nine-tenths and
one-tenth make one whole. So really this is the same as
saying 11 plus eight plus one, which equals 20.
Now let’s add the other three
values. Three-tenths plus eight-tenths plus
four-tenths equals fifteen-tenths, which is the same as one whole and
five-tenths. In the ones column, we have nine,
one, and two, plus the one that we exchanged. This comes to 13 ones, which is the
same as three ones and one ten.
Finally, in the tens column, two
plus one plus one equals four tens, plus the one that we’ve exchanged equals five
tens. The total of all the temperatures
over the five days is 53.5 degrees.
To find the mean, we just need to
divide this number now by the number of numbers. There were five temperatures that
we added together. So we need to divide by five. How many fives are there in
five? There’s one lot of five in
five. How many fives in three? Well, there are zero fives in
three. So we need to move the three digit
to the next column. How many fives are in 35? Seven times five equals 35. And so the mean temperature is 10.7
degrees.
We’re asked to calculate the mean
maximum temperature to one decimal place. Well, 10.7 is to one decimal
place. And so the mean maximum temperature
is 10.7 degrees. The fraction of the five days where
the maximum temperature was below 10 degrees was two-fifths. And the mean maximum temperature
over the five days was 10.7 degrees.
