StatisticsMean of Grouped Data

The plank balanced at 5, and the arithmetic agrees. Riya's list repeats some scores — 2 appears three times and 10 twice — so it is quicker to write it as a table of values and their frequencies :

Instead of adding 2 three times, multiply: . Doing this for every column gives the total number of pani-puris, 553651. Adding the frequencies counts the friends: 11810. So the mean is , exactly where the plank balanced.

Now a real class. Thirty students sat a 100-mark mathematics paper.

Example 1 : The marks obtained by 30 students of Class X of a school in a mathematics paper of 100 marks are given in the table below. Find the mean of the marks obtained by the students.

Solution : We need each mark multiplied by its frequency , so we add an column, as in Table 13.1.

Table 13.1

So 59.35.93593

The mean mark is 59.3. Keep this number in mind: in a moment the very same marks will give a slightly different answer.

From a List to Classes

Thirty marks fit into one table. Three thousand would not. With large data sets we sort the values into class intervals and record only how many fall into each one. Let's group the marks of Example 1 into classes of width 15: 10 – 25, 25 – 40, 40 – 55, and so on up to 85 – 100.

One snag: four students scored exactly 40, and 40 is where 25 – 40 ends and 40 – 55 begins. Each student must be counted in exactly one class, so we follow a fixed rule: a value equal to the upper limit of a class goes into the next class. A mark of 40 therefore counts in 40 – 55, not in 25 – 40.

Drag each mark into its class.

Table 13.3

Adding the last column gives , so

6259.360

Using the class marks as they are, like this, is called the direct method.

But wait. These are the same 30 students as in Example 1, and there the mean was 59.3. Which answer is right? Each grey dot below is one student, placed at their exact mark; the shaded bands are the six classes. The blue line is where the dots balance — their mean. Squash every class onto its class mark and watch the blue line.

Grouping moved the students: a 40 now counts as 47.5, a 70 as 77.5, a 36 as 32.5. On balance the moves were upwards, so the balance point drifted from 59.3 to 62. So 59.3 is the exact mean, and 62 is an approximateexactincorrect mean — the price of the mid-point assumption. Once the raw marks are thrown away, 62 is the best we can do, and for large data sets it is usually very close to the truth.

The direct method is quick here. But when the and are large, the products get long and error-prone. There is a way to shrink the numbers first.

The Assumed Mean Method

We cannot change the frequencies — they are what they are. But we can make every smaller by subtractingaddingmultiplying the same number from each of them.

Step 1. Choose one of the as the assumed mean and call it . A value near the middle of the list keeps the new numbers small, so for Table 13.3 we could take or . Let's take .

Step 2. Find the deviation of each from , that is, the differencesumproduct .

Step 3. Multiply each by its frequency , and add up all the . Table 13.4 shows the working.

From Table 13.4, , and , so

6261.547.5

The same 62 as the direct method, with much smaller multiplications. This is the assumed mean method.

Activity 1 : Would the answer change if we picked a different ? Choose class marks from Table 13.3 as — at least three different ones — and watch the deviations and the mean.

The mean comes out the samedifferentsmaller for every choice of . That makes sense: raising by some amount lowers every — and so — by exactly the same amount, and stays put. So choose whichever makes your arithmetic easiest.

The Step-Deviation Method

Look again at the column of Table 13.4: −30, −15, 0, 15, 30, 45. Every one of them is a multiple of 15, which is the class size . Dividing by makes them smaller still, so let

With and , the deviation −30 becomes −2−30−15, and the deviation 45 becomes 34515. Table 13.5 finishes the working.

This is the step-deviation method. Three methods, one answer:

Let's use all three methods on the same data.

Example 2 : The table below gives the percentage distribution of female teachers in the primary schools of rural areas of various states and union territories (U.T.) of India. Find the mean percentage of female teachers by all the three methods discussed in this section.

Source : Seventh All India School Education Survey

Solution : First, the class marks (Table 13.6):

Table 13.6

Take and , so that and . Table 13.7 has every column we need.

Table 13.7

From the totals, 353634, 139013501425, −360360−324 and .

Direct method: 39.7139.740.29

Assumed mean method: 39.715089.71

Step-deviation method: 39.715048.97

So the mean percentage of female teachers in these schools is 39.71 — the same by every method.

Solution : The class sizes are unequal here, and the are large. Step deviation still works: take and . (One comes out as a decimal, −3.75, and that is fine — the formula holds for any .) This gives Table 13.8.

Table 13.8

So , and

152.89200173.89

On average, each of these 45 bowlers took about 153 wickets in one-day cricket.

Speed Round: Class Marks and Deviations

Every mean calculation leans on two quick skills: finding a class mark, and turning into . Do them in your head.

Get five right to move on.