The same procedure once the data arrive already grouped

What to assume they know

• Working it out about the mean and about the median, for a plain list — the four steps for a plain list, about either centre
• Reading a frequency table, discrete and with class intervals
• Cumulative frequency, and building it by running addition
• The mean of a frequency distribution as a weighted total divided by the frequency total
• Class interval, class width, lower limit, mid-point
• Multiplying and adding across a table of several columns without losing place

What they should be able to do

• Compute the mean deviation about the mean of a discrete frequency distribution, using a worked table with the products written out
• Locate the median of a discrete frequency distribution from its cumulative frequencies
• Replace each class of a continuous distribution by its mid-point and say precisely what assumption that makes
• Compute the mean of a continuous distribution by the step-deviation method, and state what the Note on p. 268 restricts that method to
• Locate the median class of a continuous distribution and apply the interpolation formula, explaining what the formula assumes about the interior of that class
• Compute the mean deviation about the median for a continuous distribution
• Convert a distribution with gaps between its classes into a continuous one, as Exercise 13.1 item 12 requires

Where it usually goes wrong

• "The grouped formula is a new definition." It is the same average of distances, with repeated distances collected. Write out one small frequency longhand once and the weighting stops looking like a rule.
• "N is the number of classes." N is the total of the frequencies — 40 in Example 4 and Example 6, 30 in Example 5, 50 in Example 7. The number of classes is the number of rows and appears only as the upper limit of the summation.
• "The mid-point is where the observations actually are." It is where they are assumed to be. In Example 6 the answer 10 is the mean deviation of forty values placed exactly at 15, 25, 35 and so on, and no real measurement guarantees that.
• "Half of N will be one of the cumulative frequencies." Usually it will not. The rule is to take the first cumulative frequency that reaches or passes it; in Example 5 half of 30 is 15 and the cumulative frequencies step from 14 straight to 18.
• "The median is the mid-point of the median class." In Example 7 the median class is 20–30, whose mid-point is 25, and the median is 28. The formula exists precisely because the median is generally not the mid-point.
• "The assumed mean has to be the true mean." It is chosen for convenience — a value near the middle of the table. The formula corrects for whatever was assumed. Table 13.5 happens to assume the true mean and so corrects by zero.
• "The step-deviation table also gives the mean deviation." The Note on p. 268 rules this out. The step-deviation columns feed the mean only.
• "Class intervals like 16–20 and 21–25 are continuous." They have a gap between 20 and 21, and Exercise 13.1 item 12 has to close it before the median formula can be applied.

Questions to check understanding

• Compute the mean deviation about the mean of a discrete frequency distribution
• Locate the median of a discrete distribution from a cumulative frequency row
• Compute the mean deviation about the mean of a continuous distribution using mid-points
• Use the step-deviation method to find the mean of a continuous distribution, then complete the mean deviation
• Identify the median class and apply the interpolation formula
• Convert a distribution with gaps into a continuous one and then find its median
• State the assumption that mid-points make, and what it costs

Examples worth working on the board

Values marked verified are worked out here on data printed in this chapter.

• The weighted forms (§13.4.2, p. 263). The mean is the total of the products of each value with its frequency, divided by N. The mean deviation about either centre is the total of the products of each frequency with the distance of its value from that centre, again divided by N. Nothing new is being defined: a frequency of 10 means ten identical distances, and multiplying is faster than writing them out ten times. Say that out loud — students treat the weighted formula as a second definition.
• Example 4 (§13.4.2, pp. 263–264, Table 13.1). Values 2, 5, 6, 8, 10, 12 with frequencies 2, 8, 10, 7, 8, 5. Verified: N = 40, the products total 300, so the mean is 7.5; the distances from 7.5 are 5.5, 2.5, 1.5, 0.5, 2.5, 4.5, their weighted total is 92, and the mean deviation is 2.3. Note that the mean, 7.5, is not one of the six listed values — a mid-point of the data, not a member of it.
• The discrete median rule (§13.4.2, p. 263). Sort the values, run the cumulative frequencies down the table, and take the value at which the cumulative frequency first equals or passes half of N.
• Example 5 (§13.4.2, pp. 264–265, Tables 13.2 and 13.3). Values 3, 6, 9, 12, 13, 15, 21, 22 with frequencies 3, 4, 5, 2, 4, 5, 4, 3. Verified: the cumulative frequencies are 3, 7, 12, 14, 18, 23, 27, 30, so N = 30; half of 30 is 15, and both the fifteenth and the sixteenth observations fall inside the cumulative frequency 18, whose value is 13, so the median is 13. Verified: the distances from 13 are 10, 7, 4, 1, 0, 2, 8, 9; weighted they total 149; the mean deviation is 149 ÷ 30, printed to two places as 4.97. Two things to point at: the data was handed over already ordered, so the sorting step vanishes, and the median here is one of the listed values, unlike Example 4's mean.
• The continuous example the chapter opens with (§13.4.2, p. 265): marks 0–10 through 50–60 for 100 students, with 12, 18, 27, 20, 17 and 6 students. It is printed to show the shape of a continuous table and is not worked.
• The mid-point assumption (§13.4.2, p. 265). Each class's whole frequency is treated as sitting at the class mid-point. The consequence is worth stating plainly: two data sets with quite different arrangements inside a class produce the same table and therefore the same answer, and the answer is exact only for data that really does sit at the mid-points.
• Example 6 (§13.4.2, p. 266, Table 13.4). Classes 10–20 through 70–80 with frequencies 2, 3, 8, 14, 8, 3, 2. Verified: N = 40; mid-points 15, 25, 35, 45, 55, 65, 75; the products total 1800, so the mean is 45; the distances from 45 are 30, 20, 10, 0, 10, 20, 30, their weighted total is 400, and the mean deviation is 10. The distribution is symmetric about its middle class, which is why every number here comes out whole.
• Fig 13.3 (§13.4.2, p. 267), measured on a close-up taken wide enough to catch the side captions and the arrow underneath. One horizontal axis, not two: a single double-headed arrow carrying one row of ticks and one row of dots, and labelled twice over — the deviation scale printed along its top edge, the original scale along its bottom edge, with a caption at each end saying which is which. The arrow beneath the axis points at the one tick that reads 60 on the lower scale and 0 on the upper. The lower one is the original scale, ticked 0 to 120 by tens, with an arrow marking the assumed mean at 60; the upper one is the same points relabelled −60 to 60 by tens. The figure is a general illustration of moving the origin, not a picture of Example 6's data — its assumed mean is 60, whereas Example 6 uses 45. Do not caption it as belonging to the example.
• Fig 13.4 (§13.4.2, p. 267). Three rows: the original scale 0 to 120, the deviations from the assumed mean −60 to 60, and above them the step-deviations −6 to 6. The common factor being divided out here is 10.
• Table 13.5 (§13.4.2, p. 268). Example 6 redone with assumed mean 45 and common factor 10, so the step-deviations are −3, −2, −1, 0, 1, 2, 3. Verified: their weighted total is 0, so the mean returns 45 exactly. The zero is not luck — 45 is the true mean, and the step-deviations weighted by frequency must then vanish, which is the identity from the very start of §13.4 seen through a change of variable.
• The Note on p. 268. The step-deviation route is used to get the mean and nothing else; the distance column is still built from the true mid-points and the true mean. A student who tries to read a mean deviation straight off the step-deviation column will be out by the common factor.
• The median of a continuous distribution (§13.4.2, pp. 268–269). Find the class whose cumulative frequency first reaches or passes half of N. Then start at that class's lower limit and move into it by the shortfall — half of N less the cumulative frequency of the class before — as a fraction of that class's own frequency, scaled by the class width. The reason it works: inside the median class the frequency is assumed spread evenly across the width, so each observation occupies the same slice of the interval and the position is a straight-line guess. The chapter gives the formula and not the reason.
• Example 7 (§13.4.2, pp. 269–270, Table 13.6). Classes 0–10 through 50–60 with frequencies 6, 7, 15, 16, 4, 2. Verified: the cumulative frequencies are 6, 13, 28, 44, 48, 50, so N = 50 and the twenty-fifth observation falls in 20–30; with lower limit 20, preceding cumulative frequency 13, class frequency 15 and width 10, the median is 28. Verified: the mid-points are 5, 15, 25, 35, 45, 55, their distances from 28 are 23, 13, 3, 7, 17, 27, the weighted total is 508, and the mean deviation is 10.16. Note that the median, 28, is not a mid-point of anything — it sits three units above the mid-point of its own class.
• Exercise 13.1, the grouped items (pp. 270–271). About the mean: item 5, values 5, 10, 15, 20, 25 with frequencies 7, 4, 6, 3, 5; item 6, values 10, 30, 50, 70, 90 with frequencies 4, 24, 28, 16, 8; item 9, daily income in rupees in classes 0–100 through 700–800 with 4, 8, 9, 10, 7, 5, 4, 3 persons; item 10, heights in cm in classes 95–105 through 145–155 with 9, 13, 26, 30, 12, 10 boys. About the median: item 7, values 5, 7, 9, 10, 12, 15 with frequencies 8, 6, 2, 2, 2, 6; item 8, values 15, 21, 27, 30, 35 with frequencies 3, 5, 6, 7, 8; item 11, marks in classes 0–10 through 50–60 with 6, 8, 14, 16, 4, 2 girls; item 12, ages of 100 persons in classes 16–20, 21–25 and so on to 51–55, with 5, 6, 12, 14, 26, 12, 16, 9 persons, together with a printed hint to close the gaps by taking half a unit off each lower limit and adding half a unit to each upper limit. Verified, by working added here: item 5 has mean 14 and mean deviation 6.32; item 6 has mean 50 and mean deviation 16; item 7 has median 7 and mean deviation about 3.23; item 8 has median 30 and mean deviation about 5.10; item 11 has median class 20–30, median 20 + 110/14, and mean deviation about 10.34; item 12, after the classes are closed, has median 38 and mean deviation 7.35. Item 12 is the only one in the exercise where the given classes have gaps, and the hint exists because the median formula needs a lower limit that touches the class below it.

Figures to have open

• Fig 13.3 and Fig 13.4 redrawn as stacked, aligned number lines. These are the chapter's own figures (p. 267) and their whole content is that the scales line up. Redraw as schematics; do not reproduce the printed art. Caption them as general illustrations, not as Example 6.
• An exploded view of one class interval for section 6: the class as a bar, real observations scattered inside it, then all of them sliding onto the mid-point. An added figure; the chapter states the assumption in prose only.
• A single median-class bar for section 11, with the lower limit, the class width, the preceding cumulative frequency and the shortfall all marked, and the straight-line position of the median inside it. An added figure.
• Working tables matching Tables 13.1 to 13.6 in column structure. These are the chapter's own tables.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 13 "Statistics", §13.4.2 "Mean deviation for grouped data", printed pp. 262–270, covering the discrete case (pp. 262–265), the continuous case (pp. 265–270), the shortcut method (pp. 266–268) and the median formula (pp. 268–269).
• Tables 13.1 to 13.6 and Figures 13.3 and 13.4, all inside those pages.
• Exercise 13.1 items 5 to 12, printed pp. 270–271, including the printed hint to item 12.
• The Note restricting the step-deviation method is on p. 268.