# Video: KS2-M19 • Paper 2 • Question 22

This graph shows the maximum temperature for five days. For what fraction of the five days was the maximum temperature below 10°C? And then what was the mean maximum temperature, to one decimal place?

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### 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.