### Video Transcript

The diagram shows information about the shoe sizes of the students in class A. Part a), a student from class A is chosen at random. Calculate the probability that the student’s shoe size is the same as the mode for the class. The second diagram shows information about the shoe sizes of students in class B. For Part b), show that the shoe sizes of the students in class A are more consistent than the shoe sizes of the students in class B. Use both diagrams.

There’s also a part c) that we’ll look at after we finish the first two parts. In part a), we want to know the probability that the student shoe size is the same as the mode for the class. To calculate this probability, we remember that probability is equal to the expected outcome over all possible outcomes. All possible outcomes would be all the students from class A. And the numerator, the expected outcome is the number of students with the mode as their shoe size.

If we look carefully at our diagram, we’ll see that six students have a shoe size of four and that 10 students have a shoe size of 5. 12 students have a shoe size of six, and nine students have a shoe size of seven. This is because the line falls halfway between eight and 10. To find the total number of students, we’ll need to add six plus 10 plus 12 plus nine.

We remember that the mode is the value that is repeated most often. The shoe size six is repeated most often. It’s repeated 12 times. The mode equals six. But that is not what we put as the numerator. The numerator is the number of students with the mode as their shoe size, and there are 12 students with the mode shoe size. In our denominator, six plus 10 plus 12 plus nine equals 37. And so, we say that our probability is 12 out of 37.

In Part b), we need to show that the shoe sizes of the students in class A are more consistent than the shoe sizes of that in class B. Consistency here means how spread out our data is. To compare this, we need to compare the range of both shoe size in class A and shoe size in class B.

To find the range from class A, we’ll take the largest shoe size and subtract the smallest shoe size. The largest shoe size here is seven and the smallest is four. Seven minus four is three. The range of data in class A is three. We’ll follow the same procedure to find the range in class B. The largest shoe size is eight, and the smallest shoe size is three. Eight minus three equals five. Hence, there is greater variety of shoe sizes in class B than class A. And that means class A has a more consistent shoe size.

Part c), Tim is one of the students in class B. His shoe size is the same as the median shoe size for class B. Work out Tim’s shoe size.

To find the median, we’ll first need to know how many students are in class B. One student has a shoe size of three. Five students have a shoe size of four. Six students have a shoe size of five. 12 students have a shoe size of six. 10 students have a shoe size of seven. And three students have a shoe size of eight. One plus five plus six plus 12 plus 10 plus three equals 37. This tells us that there are 37 total students.

The median shoe size is the shoe size that half of the class have a shoe size that is less than or equal to it and the other half of the class has a shoe size greater than or equal to it. Our data is organized from the lowest shoe size to the highest shoe size. If we add the first three shoe sizes together, one plus five plus six equals 12. If we add the last two shoe sizes together, 10 plus three equals 13. And all of the rest of the values are shoe size of six. And so, we can say that the median is going to be equal to six.

There’s also a formula for calculating the median data point. The median will be found by adding 𝑛 plus one and dividing by two where 𝑛 is your total number of data points. For us, that would be 37 plus one divided by two. 38 divided by two equals 19. And so, we say the 19th student, in order from least to greatest or from greatest to least, will have the median. And the 19th student would fall somewhere in the sixes. Both methods show that, in this case, the median is six.