Question

for homework during a stormy night, all the students in a school were asked to place a cup outside to collect rainwater at their house. the graph displays the amount of rainwater, in cups, that each class gathered.
how many classes participated in the project?
if the principle decided to divide the water collected to give each class a fair share, how much water would each class get?
what is the median amount of rainwater gathered?
what is the range of rainwater gathered?
what is the mode of rainwater gathered?
rainwater gathered
number of classes

Explanation:

Sub - Question 1: How many classes participated in the project?

Step 1: Count the number of 'x's

We count the number of 'x' marks on the line plot. For \(\frac{1}{8}\) there is 1 'x', for \(\frac{1}{4}\) there are 2 'x's, for \(\frac{3}{8}\) there is 1 'x', for \(\frac{1}{2}\) there are 3 'x's, and for 1 there is 1 'x'.

Step 2: Sum the counts

We sum these counts: \(1 + 2+1 + 3+1=8\).

Step 1: Calculate total water

First, we find the total amount of water. For \(\frac{1}{8}\) (1 class): \(1\times\frac{1}{8}=\frac{1}{8}\); for \(\frac{1}{4}\) (2 classes): \(2\times\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\); for \(\frac{3}{8}\) (1 class): \(1\times\frac{3}{8}=\frac{3}{8}\); for \(\frac{1}{2}\) (3 classes): \(3\times\frac{1}{2}=\frac{3}{2}\); for 1 (1 class): \(1\times1 = 1\). Now we convert all to eighths: \(\frac{1}{8}+\frac{4}{8}+\frac{3}{8}+\frac{12}{8}+\frac{8}{8}=\frac{1 + 4+3 + 12+8}{8}=\frac{28}{8}=\frac{7}{2}\).

Step 2: Find the average

There are 8 classes. So the average (fair share) is \(\frac{7}{2}\div8=\frac{7}{2}\times\frac{1}{8}=\frac{7}{16}\)? Wait, no, wait. Wait, let's recalculate the total:

Wait, \(\frac{1}{8}\times1=\frac{1}{8}\); \(\frac{1}{4}\times2=\frac{2}{4}=\frac{1}{2}=\frac{4}{8}\); \(\frac{3}{8}\times1=\frac{3}{8}\); \(\frac{1}{2}\times3=\frac{3}{2}=\frac{12}{8}\); \(1\times1=\frac{8}{8}\). Now sum: \(\frac{1 + 4+3 + 12+8}{8}=\frac{28}{8}=\frac{7}{2}\). Then the average is \(\frac{7}{2}\div8=\frac{7}{16}\)? Wait, no, \(\frac{7}{2}\) divided by 8 is \(\frac{7}{2}\times\frac{1}{8}=\frac{7}{16}\)? Wait, but let's check the counts again. Wait, the number of classes: 1 (for \(\frac{1}{8}\)) + 2 (for \(\frac{1}{4}\))+1 (for \(\frac{3}{8}\)) + 3 (for \(\frac{1}{2}\))+1 (for 1) = 8. The total water: \(\frac{1}{8}+2\times\frac{1}{4}+1\times\frac{3}{8}+3\times\frac{1}{2}+1\times1\). Let's compute:

\(\frac{1}{8}+\frac{2}{4}+\frac{3}{8}+\frac{3}{2}+1=\frac{1 + 4+3 + 12+8}{8}=\frac{28}{8}=\frac{7}{2}\). Then the average is \(\frac{7}{2}\div8=\frac{7}{16}\)? Wait, no, \(\frac{7}{2}\) is 3.5, and 3.5 divided by 8 is 0.4375 which is \(\frac{7}{16}\). Wait, but maybe I made a mistake in the total. Wait, \(\frac{1}{2}\) is 4/8, so 3 times \(\frac{1}{2}\) is 12/8. 1 is 8/8. So 1 (1/8) + 2 (2/4 = 4/8) + 1 (3/8) + 3 (12/8) + 1 (8/8) = 1+4 + 3+12 + 8 = 28, over 8. So 28/8 = 7/2. Then 7/2 divided by 8 is 7/16. Wait, but maybe the graph has different counts? Wait, the graph: at \(\frac{1}{8}\): 1 x; at \(\frac{1}{4}\): 2 x's; at \(\frac{3}{8}\): 1 x; at \(\frac{1}{2}\): 3 x's; at 1: 1 x. So total x's: 1 + 2+1 + 3+1 = 8. So total water: \(\frac{1}{8}\times1+\frac{1}{4}\times2+\frac{3}{8}\times1+\frac{1}{2}\times3 + 1\times1=\frac{1}{8}+\frac{1}{2}+\frac{3}{8}+\frac{3}{2}+1\). Convert to eighths: \(\frac{1 + 4+3 + 12+8}{8}=\frac{28}{8}=\frac{7}{2}\). Then average is \(\frac{7}{2}\div8=\frac{7}{16}\)? Wait, no, \(\frac{7}{2}\) divided by 8 is \(\frac{7}{16}\). Wait, but maybe I messed up the total. Wait, \(\frac{1}{2}\) is 0.5, 3 times 0.5 is 1.5. \(\frac{1}{4}\) is 0.25, 2 times 0.25 is 0.5. \(\frac{1}{8}\) is 0.125, 1 times 0.125 is 0.125. \(\frac{3}{8}\) is 0.375, 1 times 0.375 is 0.375. 1 times 1 is 1. Now sum: 0.125+0.5 + 0.375+1.5+1 = (0.125 + 0.375)+(0.5 + 1.5)+1 = 0.5 + 2+1 = 3.5. 3.5 divided by 8 is 0.4375, which is 7/16. So the average is 7/16 cups? Wait, but maybe the graph has different values. Wait, the x - axis is marked as 0, 1/8, 1/4, 3/8, 1/2, 5/8, 3/4, 7/8, 1. So the points with x's are at 1/8 (1 x), 1/4 (2 x's), 3/8 (1 x), 1/2 (3 x's), 1 (1 x). So total classes: 1 + 2+1 + 3+1 = 8. Total water: 1(1/8)+2(1/4)+1(3/8)+3(1/2)+1*(1)=1/8 + 2/4+3/8 + 3/2+1. Let's compute:

1/8+3/8 = 4/8 = 1/2; 2/4+3/2 = 1/2 + 3/2 = 2; then 1/2+2 + 1 = 3.5. So total water is 3.5 cups. Then average is 3.5 / 8 = 0.4375 cups or 7/16 cups.

Step 1: Order the data

We have 8 data points. The data points (in cups) are: \(\frac{1}{8}\), \(\frac{1}{4}\), \(\frac{1}{4}\), \(\frac{3}{8}\), \(\frac{1}{2}\), \(\frac{1}{2}\), \(\frac{1}{2}\), 1.

Step 2: Find the median

For 8 data points, the median is the average of the 4th and 5th values. The 4th value is \(\frac{3}{8}\) and the 5th value is \(\frac{1}{2}\). The average of \(\frac{3}{8}\) and \(\frac{1}{2}\) is \(\frac{\frac{3}{8}+\frac{4}{8}}{2}=\frac{\frac{7}{8}}{2}=\frac{7}{16}\)? Wait, no, \(\frac{3}{8}+\frac{1}{2}=\frac{3 + 4}{8}=\frac{7}{8}\), then divide by 2: \(\frac{7}{16}\)? Wait, no, the data points in order:

First, list all the values:

1. \(\frac{1}{8}\)
2. \(\frac{1}{4}\)
3. \(\frac{1}{4}\)
4. \(\frac{3}{8}\)
5. \(\frac{1}{2}\)
6. \(\frac{1}{2}\)
7. \(\frac{1}{2}\)
8. 1

So the 4th term is \(\frac{3}{8}\) and the 5th term is \(\frac{1}{2}\). The median is \(\frac{\frac{3}{8}+\frac{1}{2}}{2}=\frac{\frac{3 + 4}{8}}{2}=\frac{\frac{7}{8}}{2}=\frac{7}{16}\)? Wait, no, \(\frac{3}{8}+\frac{1}{2}=\frac{3 + 4}{8}=\frac{7}{8}\), then the median is \(\frac{7}{8}\div2=\frac{7}{16}\)? Wait, no, the median for an even number of data points is the average of the two middle numbers. The two middle numbers are the 4th and 5th. So \(\frac{3}{8}\) (4th) and \(\frac{1}{2}\) (5th). \(\frac{3}{8}+\frac{1}{2}=\frac{3 + 4}{8}=\frac{7}{8}\), then divide by 2: \(\frac{7}{16}\). Wait, but \(\frac{1}{2}\) is 4/8, so \(\frac{3}{8}\) and \(\frac{4}{8}\) average to \(\frac{7}{16}\)? Wait, no, \(\frac{3}{8}+\frac{4}{8}=\frac{7}{8}\), and \(\frac{7}{8}\) divided by 2 is \(\frac{7}{16}\). Yes.

Answer:

8

Sub - Question 2: If the principal decided to divide the water collected to give each class a fair share, how much water would each class get?