Cumulative Frequency

A cumulative frequency includes all earlier groups as well as the current group.

1. The cumulative frequency at 30 minutes is the number of times up to and including 30 minutes: every group so far.
2. 4+7+8=19.

Subtract the earlier cumulative total from the later cumulative total.

1. The later total includes the earlier group.
2. The interval frequency is 19-6=13.

Use an upper class boundary and the total frequency up to that boundary.

1. At the upper boundary 10, the cumulative frequency is 6+9=15.
2. Therefore (10,15) is a point on the graph.

Use half the total frequency to choose the height at which to read across.

1. Half of 40 is 20, so use cumulative frequency 20.
2. Read across from cumulative frequency 20 and down to the time axis.

3. The estimated median is 30 minutes.

Read the cumulative total above the given time.

1. Read the graph at 20 minutes.
2. The cumulative frequency is about 10, so about 10 journeys took no more than 20 minutes.

The upper quartile has three quarters of the observations at or below it.

1. Three quarters of 80 is 60.
2. At cumulative frequency 60, the graph gives about 40 minutes.
3. Use the highlighted reading on the supplied graph.

Read both quartiles from their cumulative frequency positions before subtracting.

1. The lower and upper quartiles use cumulative frequencies 20 and 60.
2. These give about 20 and 40 minutes respectively.
3. Read both boundaries before subtracting.

4. The interquartile range is about 40-20=20 minutes.

The cumulative frequency counts the journeys up to the threshold, so subtract it from the total.

1. The graph gives about 32 journeys taking no more than 40 minutes.
2. The number above 40 minutes is about 40-32=8.

Locate half the total frequency between two consecutive cumulative totals.

1. There are 40 observations, so locate cumulative frequency 20.
2. The cumulative total is 8 at 10 minutes and 22 at 20 minutes. The median lies in 10<t\le20.

Think about what cumulative frequency counts, rather than one class frequency.

1. Cumulative frequency is a running total of frequencies, and a frequency is never negative.
2. A smaller class frequency may make the graph less steep, but cannot make the running total fall.

Subtract the cumulative count at the threshold from the total before finding the percentage.

1. About 60 students scored no more than 60.
2. About 100-60=40 scored above 60.
3. Out of 100 students, this is about 40\%.

Convert the percentage to a cumulative frequency, then read across and down.

1. 90\% of 100 is 90.
2. Read at cumulative frequency 90 to get an estimated score of 80.
3. Use the highlighted reading on the supplied graph.

Compare medians at half the total and interquartile ranges between the two quartiles.

1. Group A has median about 30 minutes, while Group B has median about 40 minutes.
2. Both groups have 60 people, so each median uses cumulative frequency 30.

3. For A, the quartiles are about 20 and 40; for B they are about 30 and 50. Both interquartile ranges are about 20 minutes.
4. Each group uses cumulative frequencies 15 and 45 for its quartiles.

5. A has the lower median journey time, with approximately the same interquartile range.

Estimate the number above the threshold, then divide by the total.

1. At 35 minutes, the cumulative frequency is about 50.
2. About 80-50=30 waiting times exceed 35 minutes.
3. The estimated probability is \frac{30}{80}=\frac38.
4. Use the highlighted reading on the supplied graph.

Read the proportion in the sample and apply it to the larger group.

1. The sample has about 30 scores no more than 40, a proportion of \frac{30}{100}.
2. Applying this proportion to 200 gives an estimate of 200\times\frac{30}{100}=60 students.

Read the cumulative frequency at both times; the difference is how many lie between them.

1. Read up from 15 minutes to the graph and across: about 20 times are 15 minutes or less.
2. Read up from 35 minutes: about 80 times are 35 minutes or less.
3. Read both boundaries before subtracting.

4. So about 80-20=60 times are more than 15 but no more than 35 minutes.
5. The estimated probability is \frac{60}{100}=\frac{3}{5} (or 0.6).

The 30 removed are all at the top, so where in the original order is the middle of the other 70?

1. The median of 70 people is halfway through them: the 35th time (strictly between the 35th and 36th, but on a graph use 35).
2. The 30 people removed had the longest times, so the 35th shortest time of the remaining 70 is also the 35th shortest time of the original 100.
3. So read across from cumulative frequency 35 (not 50, which is the median of all 100) to the graph and down to the time axis.
4. The retained group ends at cumulative frequency 70, so its halfway point is cumulative frequency 35.

5. This gives about 21 minutes (anything from about 19 to 23.5 minutes is a fair reading).

Locate the class containing half the observations and work out how far through it that position lies.

1. The cumulative frequencies are 5, 15, 45 and 50 at 10, 20, 40 and 50 minutes, so the points are (10,5), (20,15), (40,45) and (50,50).
2. The median is at cumulative frequency \frac{50}{2}=25, which lies on the straight line from (20,15) to (40,45).
3. Along that line the cumulative frequency rises by 30 over 20 minutes. We need to rise 25-15=10, which is \frac{10}{30}=\frac{1}{3} of the way.
4. The median lies one third of the way along the segment from (20,15) to (40,45).

5. So the median is 20+\frac{1}{3}\times20=26.66\ldots minutes, which is 26.7 minutes to 1 decimal place.

The cutoff has the other eighty percent at or below it.

1. The slowest 20\% are above this time, so 80\% of the people are at or below it.
2. 80\% of 50 is 0.8\times50=40, so read across from cumulative frequency 40.
3. The graph reaches 40 at about 33 minutes, so the time is about 33 minutes.
4. Use the highlighted reading on the supplied graph.

Find each count above the threshold, convert each to a percentage, and subtract.

1. Read up from 40 minutes: about 45 people in Group A and about 30 in Group B took 40 minutes or less.
2. The two curves give different counts at the same time of 40 minutes.

3. So 60-45=15 in Group A and 60-30=30 in Group B took more than 40 minutes.
4. As percentages of 60: Group A \frac{15}{60}\times100=25\%, Group B \frac{30}{60}\times100=50\%.
5. The Group B percentage is 50-25=25 percentage points bigger.