Statistics

A class teacher marks thirty scripts and now holds thirty separate numbers — a tangle no one can absorb at a glance. In Class IX you already learned to compress such a tangle into three single numbers: the mean, the median and the mode of ungrouped data. But real data — marks, wages, heartbeats, literacy rates — rarely arrives as a clean list of individual values. It arrives already bucketed into ranges, a table that says "10 to 25 marks: 2 students; 25 to 40 marks: 3 students," because that is how large data is actually recorded. This chapter stretches the three familiar measures of central tendency from a list of individual numbers to a table of grouped, ranged data, and introduces a new bookkeeping device — cumulative frequency — that makes this stretch possible for the median. The chapter's spine has three matching pillars. First, the mean of grouped data, found three ways — the direct method, the assumed mean method and the step-deviation method — all giving the identical answer, so the choice is purely about which one keeps the arithmetic light. Second, the mode, found by locating the single most crowded class interval (the modal class) and reading off a precise value inside it with one formula. Third, the median, found by building a cumulative frequency table and locating the class where the running total first crosses the halfway mark. A short closing result, the empirical relationship $3\,\text{Median} = \text{Mode} + 2\,\text{Mean}$ , ties all three together. By the end you will be able to take the very same table of thirty students' marks and produce three different, complementary answers to the vague question "how did the class do": a mean of $62$ , a modal score of $52$ , and, for a differently sized class, a median that tells you where the middle student stands. Each measure answers a slightly different question, and knowing which one to reach for matters as much as knowing how to compute it.

Key takeaways

The mean of grouped data assumes each class's frequency is concentrated at its class mark $x_i = \dfrac{\text{upper limit} + \text{lower limit}}{2}$ ; the direct method computes $\bar{x} = \dfrac{\Sigma f_i x_i}{\Sigma f_i}$ .
The assumed mean method computes $\bar{x} = a + \dfrac{\Sigma f_i d_i}{\Sigma f_i}$ with $d_i = x_i - a$ ; the step-deviation method computes $\bar{x} = a + h\!\left(\dfrac{\Sigma f_i u_i}{\Sigma f_i}\right)$ with $u_i = \dfrac{x_i - a}{h}$ — all three agree exactly (NCERT Examples 1, 2).
The value chosen for $a$ never changes the final mean; step-deviation is fastest when the deviations $d_i$ share a common factor, usually $h$ .
The mode lies inside the modal class (highest frequency), given by $\text{Mode} = l + \left(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) h$ .
The cumulative frequency of a class is the running total of frequencies up to it ("less than" type) or from it onward ("more than" type).
The median lies inside the median class — the first class whose cumulative frequency is $\geq n/2$ — given by $\text{Median} = l + \left(\dfrac{\tfrac{n}{2} - cf}{f}\right) h$ .
Both the mode and median formulas require continuous class intervals; inclusive intervals must be converted first.
An empirical relationship links all three for moderately skewed data: $3\,\text{Median} = \text{Mode} + 2\,\text{Mean}$ .
Key idea: the mean, median and mode of grouped data are each found by collapsing every class interval to one representative value — the class mark for the mean, the modal-class formula for the mode, and the cumulative-frequency crossing of n over 2 for the median — and the three are tied together by 3 Median = Mode + 2 Mean.

The core insight

Grouped data is summarised by collapsing every class to a single representative point: the mean uses the class mark $x_i$ in $\bar{x} = a + h\!\left(\dfrac{\Sigma f_i u_i}{\Sigma f_i}\right)$ , the mode uses the modal-class formula $l + \left(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\right)h$ , and the median uses a cumulative frequency table and $l + \left(\dfrac{n/2 - cf}{f}\right)h$ , with the three linked by $3\,\text{Median} = \text{Mode} + 2\,\text{Mean}$ .

What this chapter covers

The chapter moves in three matched stages, each extending a Class IX idea from a raw list of numbers to a grouped table. It opens with the mean of grouped data: first the direct method, which assumes each class's frequency sits entirely at its class mark, then two arithmetic shortcuts — the assumed mean method and the step-deviation method — proved algebraically to always agree with the direct method, however large or awkward the numbers. It then turns to the mode of grouped data, defining the modal class as the class of highest frequency and giving a single formula that locates a precise value inside it. The chapter's final and most intricate idea is the median of grouped data, which needs a new tool — the cumulative frequency table, built either as "less than" or "more than" running totals — to locate the median class and then interpolate a precise median value inside it. The chapter closes by comparing all three measures side by side on the same data, discussing when each is the right choice, and stating the empirical relationship that connects them.

The core ideas

Mean of grouped data: the direct method

Recall from Class IX that for ungrouped data with observations $x_1, x_2, \ldots, x_n$ occurring with frequencies $f_1, f_2, \ldots, f_n$ , the mean is the sum of all values divided by the total count. For grouped data, an individual observation inside a class is not known exactly — only the class it falls into is — so the chapter makes one controlling assumption: the frequency of each class is treated as concentrated at its class mark, the class midpoint.

\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}, \qquad x_i = \frac{\text{upper class limit} + \text{lower class limit}}{2}. \qquad (1)

where: $x_i$ is the class mark of the $i$ -th class, $f_i$ is its frequency, and $\bar{x}$ is the mean. This is the direct method. Because it replaces every observation by the single value $x_i$ , the grouped mean is only an approximation to the true mean of the raw data — for NCERT Example 1, the exact ungrouped mean is $59.3$ , while the same data grouped into six classes gives $62$ .

Reducing the arithmetic: assumed mean and step-deviation methods

When $x_i$ and $f_i$ are numerically large, multiplying every $f_ix_i$ by hand becomes tedious. Two shortcuts fix this without changing the answer. The assumed mean method picks one class mark as a reference point $a$ (near the data's centre) and works with the small deviations $d_i = x_i - a$ :

\bar{x} = a + \frac{\Sigma f_i d_i}{\Sigma f_i}. \qquad (2)

where: $a$ is the assumed mean, any convenient class mark, and $d_i = x_i - a$ . If the $d_i$ share a common factor — typically the class size $h$ — dividing every $d_i$ by $h$ gives even smaller numbers $u_i = \dfrac{x_i - a}{h}$ , the step-deviation method:

\bar{x} = a + h\left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right). \qquad (3)

where: $h$ is the class size and $u_i$ is the step-deviation of the $i$ -th class. Both formulas are proved equivalent to the direct method in the first derivation below — the choice between the three is purely convenience, governed by how large $x_i$ , $f_i$ are and whether $h$ is uniform.

Figure

A bar chart of six mark class intervals with heights equal to the number of students and each bars midpoint marked as its class mark showing how the mean treats every student in a class as located at that single representative point

Mode of grouped data: locating the modal class

For ungrouped data the mode was simply the observation with the highest frequency, found by inspection. For grouped data this is no longer possible directly — individual observations are hidden inside classes — so the chapter defines the modal class as the class interval with the highest frequency, and gives a formula that pinpoints one exact value inside it.

\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) h. \qquad (4)

where: $l$ is the lower limit of the modal class, $h$ is the class size (equal width assumed), $f_1$ is the frequency of the modal class, $f_0$ is the frequency of the preceding class, and $f_2$ is the frequency of the succeeding class. The formula is derived geometrically below. Grouped data can, in principle, be multimodal; this chapter, like NCERT, restricts attention to a single clear modal class.

Median of grouped data: cumulative frequency and the median class

The median of ungrouped data was the middle-most observation once the data was arranged in order — the $\left(\dfrac{n+1}{2}\right)$ -th observation for odd $n$ , or the average of the $\dfrac{n}{2}$ -th and $\left(\dfrac{n}{2}+1\right)$ -th observations for even $n$ . Grouped data hides the individual order, so a running-total tool is needed first: the cumulative frequency of a class is the sum of the frequencies of that class and all classes before it ("less than" type), or of that class and all classes after it ("more than" type). Once a cumulative frequency column ( $cf$ ) is built, the median class is the first class whose cumulative frequency is greater than (and nearest to) $\dfrac{n}{2}$ , where $n = \Sigma f_i$ .

\text{Median} = l + \left(\frac{\tfrac{n}{2} - cf}{f}\right) h. \qquad (5)

where: $l$ is the lower limit of the median class, $n$ is the total number of observations, $cf$ is the cumulative frequency of the class immediately preceding the median class, $f$ is the frequency of the median class, and $h$ is the class size. The formula is derived below using linear interpolation, assuming the median class's observations are spread evenly across its width.

Figure

A cumulative frequency curve for the marks of fifty three students rising in a smooth increasing line from zero to fifty three with a horizontal line at half of fifty three marking where the curve crosses inside the median class sixty to seventy

Choosing the right measure, and the empirical relationship

The mean uses every observation and is the most common measure, but it is pulled about by extreme values — a single very large or small class frequency distorts it. When a "typical" value matters more than every individual value, the median is often better, since it depends only on how many observations lie on each side, not on how extreme they are. When the most frequently occurring value matters most — the most popular size, the most common wage — the mode is the natural choice. For moderately skewed distributions, the three are tied together by a useful empirical rule:

3\,\text{Median} = \text{Mode} + 2\,\text{Mean}. \qquad (6)

where: all three quantities refer to the same data set. This relationship is empirical — observed to hold well for many real, moderately skewed distributions — not an exact identity for every data set, so it is used for quick estimation, not as a substitute for direct computation when precision matters.

Derivation ADeriving the assumed mean and step-deviation formulas for the mean

The goal is to show algebraically that the assumed mean and step-deviation methods always reproduce exactly the same answer as the direct method $\bar{x} = \Sigma f_i x_i / \Sigma f_i$ , for any choice of the reference value $a$ and any nonzero class size $h$ . This justifies picking whichever method keeps the numbers smallest, with complete confidence the final mean will not change.

Step 1. Start from $d_i = x_i - a$ , so $x_i = a + d_i$ . Substitute into the direct-method formula: $\bar{x} = \dfrac{\Sigma f_i x_i}{\Sigma f_i} = \dfrac{\Sigma f_i (a + d_i)}{\Sigma f_i}$ .

Step 2. Split the sum, since $a$ does not depend on $i$ : $\Sigma f_i(a+d_i) = a\,\Sigma f_i + \Sigma f_i d_i$ , so $\bar{x} = \dfrac{a\,\Sigma f_i}{\Sigma f_i} + \dfrac{\Sigma f_i d_i}{\Sigma f_i}$ .

Step 3. Simplify the first term: $\dfrac{a\,\Sigma f_i}{\Sigma f_i} = a$ , since $\Sigma f_i$ cancels. This leaves the assumed mean formula, $\bar{x} = a + \dfrac{\Sigma f_i d_i}{\Sigma f_i}$ , valid for any $a$ — which is why the result does not depend on which class mark is picked as $a$ .

Step 4. Bring in the step-deviation: since $u_i = \dfrac{x_i-a}{h} = \dfrac{d_i}{h}$ , we have $d_i = hu_i$ . Substituting, $\bar{x} = a + \dfrac{\Sigma f_i(hu_i)}{\Sigma f_i} = a + h\cdot\dfrac{\Sigma f_i u_i}{\Sigma f_i}$ , exactly the step-deviation formula.

Step 5. Confirm on NCERT's numbers: with $a=47.5$ , $h=15$ , $\Sigma f_i d_i = 435$ , $\Sigma f_i = 30$ , so $\bar{x} = 47.5+435/30 = 62$ . The step-deviation gives $\Sigma f_i u_i = 29$ , so $\bar{x} = 47.5+15(29/30) = 62$ — matching the direct method's $1860/30 = 62$ .

Go deeper · Derivation B · 1 minDeriving the mode formula for grouped data

Derivation BDeriving the mode formula for grouped data

The goal is to justify $\text{Mode} = l + \left(\dfrac{f_1-f_0}{2f_1-f_0-f_2}\right)h$ geometrically, using the histogram of the modal class and its two neighbours, rather than quoting it. Two straight lines are drawn across the top of the histogram, and where they cross gives the mode — the point where the modal class's crowding is pulled most strongly toward whichever neighbour has the higher frequency.

Step 1. Set up coordinates: the modal class occupies $[l, l+h]$ with height $f_1$ ; the preceding class occupies $[l-h,l]$ with height $f_0$ ; the succeeding class occupies $[l+h,l+2h]$ with height $f_2$ . The needed real corners are: modal top-left $A=(l,f_1)$ , modal top-right $B=(l+h,f_1)$ , preceding top-right $D=(l,f_0)$ , succeeding top-left $C=(l+h,f_2)$ .

Step 2. Draw line $AC$ (modal top-left to succeeding top-left) and line $BD$ (modal top-right to preceding top-right). Using local coordinate $x'=x-l$ (running $0$ to $h$ ), line $AC$ is $y = f_1+(f_2-f_1)\dfrac{x'}{h}$ and line $BD$ is $y = f_0+(f_1-f_0)\dfrac{x'}{h}$ .

Step 3. Set the two equal: $f_1-f_0 = \dfrac{x'}{h}\left[(f_1-f_0)-(f_2-f_1)\right] = \dfrac{x'}{h}(2f_1-f_0-f_2)$ .

Step 4. Solve: $x' = h\left(\dfrac{f_1-f_0}{2f_1-f_0-f_2}\right)$ , so $\text{Mode} = l+x' = l+\left(\dfrac{f_1-f_0}{2f_1-f_0-f_2}\right)h$ — the stated formula.

Step 5. Verify against NCERT's Example 6: modal class $40$ – $55$ , $l=40$ , $h=15$ , $f_1=7$ , $f_0=3$ , $f_2=6$ : $x' = 15\left(\dfrac{4}{5}\right)=12$ , so $\text{Mode}=52$ , matching NCERT.

Go deeper · Derivation C · 1 minDeriving the median formula for grouped data by linear interpolation

Derivation CDeriving the median formula for grouped data by linear interpolation

The goal is to justify $\text{Median} = l + \left(\dfrac{n/2 - cf}{f}\right)h$ using one controlling assumption: within the median class, the $f$ observations are spread evenly across its width $h$ , the same representative-point idea used for the mean. This turns "find the middle observation" into a proportion problem.

Step 1. Let the median class be $[l,l+h]$ with frequency $f$ , and $cf$ the cumulative frequency of all classes before it. We want the point $m$ inside $[l,l+h]$ where exactly $n/2$ observations lie at or below $m$ .

Step 2. Since $cf$ observations already lie below $l$ , the additional observations needed from inside the median class are $\dfrac{n}{2}-cf$ .

Step 3. Under the uniform-spread assumption, the density inside this class is $\dfrac{f}{h}$ observations per unit length. To accumulate $\left(\dfrac{n}{2}-cf\right)$ observations from $l$ , we need length $x$ with $\dfrac{f}{h}x = \dfrac{n}{2}-cf$ , so $x = h\left(\dfrac{n/2-cf}{f}\right)$ .

Step 4. Since $x$ is measured from $l$ , $\text{Median} = l+x = l+\left(\dfrac{n/2-cf}{f}\right)h$ — the stated formula.

Step 5. Verify against NCERT's Example 7 (heights of $51$ girls): median class $145$ – $150$ , $l=145$ , $cf=11$ , $f=18$ , $h=5$ , $n/2=25.5$ : $x = 5\left(\dfrac{14.5}{18}\right)\approx4.03$ , so $\text{Median}\approx149.03$ cm, matching NCERT.

Worked examples

Board level - mean percentage of female teachers by all three methods (NCERT Example 2)

The table gives the percentage distribution of female teachers in rural primary schools across states and union territories: classes $15$ – $25$ , $25$ – $35$ , $35$ – $45$ , $45$ – $55$ , $55$ – $65$ , $65$ – $75$ , $75$ – $85$ with frequencies $6, 11, 7, 4, 4, 2, 1$ respectively. Find the mean percentage by the direct, assumed mean, and step-deviation methods, and confirm they agree.

Step 1. Class marks: $x_i = 20, 30, 40, 50, 60, 70, 80$ . Direct method: $\Sigma f_i x_i = 120+330+280+200+240+140+80 = 1390$ , $\Sigma f_i = 35$ , so $\bar{x} = 1390/35 = 39.71$ .

Step 2. Assumed mean method, $a = 50$ : $d_i = -30,-20,-10,0,10,20,30$ , so $f_id_i = -180,-220,-70,0,40,40,30$ , summing to $\Sigma f_i d_i = -360$ . So $\bar{x} = 50 + (-360)/35 = 39.71$ .

Step 3. Step-deviation method, $a=50$ , $h=10$ : $u_i = -3,-2,-1,0,1,2,3$ , so $f_iu_i = -18,-22,-7,0,4,4,3$ , summing to $\Sigma f_i u_i = -36$ . So $\bar{x} = 50 + 10\times(-36/35) = 39.71$ .

Step 4. All three methods agree exactly, as the first derivation above guarantees for any $a$ and $h$ . Answer: the mean percentage of female teachers is $39.71\%$ , by all three methods.

Board level - mode and mean of examination marks compared (NCERT Examples 1, 3 and 6)

Using the grouped marks data from Table 13.3 (classes $10$ – $25$ , $25$ – $40$ , $40$ – $55$ , $55$ – $70$ , $70$ – $85$ , $85$ – $100$ with frequencies $2, 3, 7, 6, 6, 6$ ), find the mode of the data, and compare it with the mean.

Step 1. Identify the modal class. The highest frequency is $7$ , in class $40$ – $55$ : $l=40$ , $h=15$ , $f_1=7$ , $f_0=3$ (preceding class $25$ – $40$ ), $f_2=6$ (succeeding class $55$ – $70$ ).

Step 2. Apply the mode formula: $\text{Mode} = 40 + \left(\dfrac{7-3}{2(7)-3-6}\right)15 = 40 + \left(\dfrac{4}{5}\right)15 = 52$ .

Step 3. Recall the mean by the direct method: class marks $17.5,32.5,47.5,62.5,77.5,92.5$ give $\Sigma f_ix_i = 1860$ , $\Sigma f_i = 30$ , so $\bar{x} = 1860/30 = 62$ .

Step 4. Compare and interpret: the mode ( $52$ ) is noticeably less than the mean ( $62$ ) — a cluster of higher-scoring students pulls the mean above the modal value. "What mark did most students get" is answered by the mode; "what is the class average" by the mean. Answer: mode $= 52$ , mean $= 62$ ; the two measures answer different questions about the same data.

Board HOTS / foundation JEE-NEET - recovering two missing frequencies from a given median (NCERT Example 8)

The median of the following distribution is $525$ , and the total frequency is $100$ . Classes $0$ – $100, 100$ – $200, \ldots, 900$ – $1000$ have frequencies $2, 5, x, 12, 17, 20, y, 9, 7, 4$ respectively (with $x$ in class $200$ – $300$ and $y$ in class $600$ – $700$ ). Find $x$ and $y$ .

Step 1. Use the total-frequency condition first: $2+5+x+12+17+20+y+9+7+4 = 76+x+y = 100$ , so $x+y = 24$ . Call this equation (1).

Step 2. Build the cumulative frequency column in terms of $x$ : up to $100$ : $7$ ; up to $200$ : $7+x$ ; up to $300$ : $19+x$ ; up to $400$ : $36+x$ ; up to $500$ : $56+x$ .

Step 3. Locate the median class: since $525$ lies between $500$ and $600$ , that is the median class, so $l=500$ , $f=20$ , $h=100$ , and $cf$ (cumulative frequency of the class before it) $=36+x$ .

Step 4. Apply the median formula and solve: $525 = 500 + \left(\dfrac{50 - (36+x)}{20}\right)100$ , so $25 = (14-x)\times 5$ , giving $x = 9$ .

Step 5. Substitute into equation (1): $9 + y = 24$ , so $y = 15$ . Answer: $x = 9$ and $y = 15$ . Timing note: aim for under five minutes. The two unknowns need two equations — the total-frequency condition and the median formula — both must be used. The common slip is the cumulative frequency of the median class's preceding class: it must include $x$ (an earlier class) but never $y$ (the later class $600$ – $700$ ); mixing this up produces a wrong equation in Step 4.

Common traps

Using the class limit instead of the class mark. The mean formula needs $x_i$ , the midpoint of each class, not its upper or lower limit.
Believing the choice of assumed mean $a$ changes the answer. It never does; a poor $a$ only makes the arithmetic heavier, never the final mean wrong.
Forgetting to multiply back by $h$ in the step-deviation method. Leaving out the final $\times h$ in $\bar{x} = a + h(\Sigma f_iu_i/\Sigma f_i)$ gives a number close to, but not equal to, the correct mean.
Mixing up $f_0$ and $f_2$ in the mode formula. $f_0$ is always the frequency of the class preceding the modal class, and $f_2$ the class succeeding it — swapping them gives a wildly different mode.
Reading $cf$ as the cumulative frequency of the median class itself. It is the cumulative frequency of the class immediately before the median class.
Choosing the wrong median class. It is the first class whose cumulative frequency is greater than or equal to $n/2$ — not the class with the largest frequency (that is the modal class, an entirely different idea).
Applying the mode or median formula to inclusive (gapped) class intervals. Both assume continuous classes; intervals like $118$ – $126$ , $127$ – $135$ must first be converted to continuous form ( $117.5$ – $126.5$ , and so on).
Treating the grouped mean as an exact value. It is a controlled approximation from the midpoint assumption; the true ungrouped mean can differ slightly, as Example 1's $59.3$ versus $62$ shows.

Common trapBoth the mode and median formulas require continuous class intervals — always convert inclusive intervals first

The single biggest misconception in this chapter is plugging the mode or median formula directly into data given as inclusive class intervals, such as $1$ – $10$ , $11$ – $20$ , $21$ – $30$ , where there is a gap of one unit between one class's upper limit and the next class's lower limit. Both formulas assume the class boundaries are continuous — the upper limit of one class must equal the lower limit of the next — because $l$ , $h$ , and the cumulative frequencies only make sense as unbroken measurements along a number line. Before applying either formula to inclusive data, subtract $0.5$ from every lower limit and add $0.5$ to every upper limit (so $1$ – $10$ becomes $0.5$ – $10.5$ , $11$ – $20$ becomes $10.5$ – $20.5$ ), which closes every gap without changing which observations fall in which class. Skipping this conversion produces a mode or median systematically off by half a class-width — avoidable by checking, before you start, whether the class boundaries already touch.

You should be able to…

Convert a raw list of observations into a grouped frequency distribution table and identify each class's class mark.
Compute the mean of grouped data by the direct method, and cross-check it using the assumed mean or step-deviation method.
Choose a sensible assumed mean $a$ and use the class size $h$ as the step-deviation divisor to minimise arithmetic.
Identify the modal class of a grouped distribution and apply the mode formula correctly, keeping $f_0$ , $f_1$ , $f_2$ straight.
Build a cumulative frequency table, of either the "less than" or "more than" type.
Identify the median class using $n/2$ and the cumulative frequency column, and apply the median formula.
Convert inclusive class intervals into continuous class intervals before applying the mode or median formula.
Use $3\,\text{Median} = \text{Mode} + 2\,\text{Mean}$ to estimate an unknown measure from the other two.
Interpret which of the mean, median or mode is the most appropriate measure for a given real-world question.

Ask the tutor: if you are unsure which method to use for a mean, how to pick $f_0$ and $f_2$ for the mode, or how to build a cumulative frequency table for your own data set, ask the tutor to work through it with you step by step.

Why is the grouped mean 62 different from the exact ungrouped mean 59.3 for the very same marks data?

The grouped mean is a controlled approximation, not an error. When raw marks are bucketed into class intervals, the exact individual values are discarded and every observation in a class is assumed to sit at that class's midpoint. This assumption is not exactly true for any real class, so the grouped mean shifts slightly from the true mean computed from the raw list. The gap between $59.3$ and $62$ is the price paid for condensing thirty individual numbers into a six-row table; it shrinks further as the class width $h$ is made smaller, giving the midpoint assumption less room to be wrong.

Does it matter which class mark I choose as the assumed mean a in the assumed mean method?

No — the final computed mean comes out identical no matter which value you pick for $a$ , as the algebra in the first derivation shows: the $a$ terms cancel out exactly, leaving only $\Sigma f_id_i/\Sigma f_i$ to add back. What does change with a poor choice is how much arithmetic you do. Picking $a$ close to the centre of the data keeps the deviations $d_i$ small and balanced between positive and negative, which keeps the running sums manageable by hand.

My class intervals have gaps, like 118 to 126 then 127 to 135 — what do I do before finding the mode or median?

Convert the intervals to continuous form first, because both formulas assume unbroken class boundaries. Take half the gap between one class's upper limit and the next class's lower limit — here the gap is $127-126=1$ , so half is $0.5$ — and subtract it from every lower limit while adding it to every upper limit. So $118$ – $126$ becomes $117.5$ – $126.5$ , $127$ – $135$ becomes $126.5$ – $135.5$ , and so on down the table. This does not change which observations belong to which class; it only removes the gap so that $l$ , $h$ , and the cumulative frequencies behave correctly inside the formulas.

How do I decide whether to report the mean, median or mode for a given data set?

Use the mean when every observation should count and you want an overall average — but it is pulled toward any unusually large or small class. Use the median when you want a "typical" value that is not distorted by a few extreme classes, such as a typical wage. Use the mode when the question is really about the single most common or popular value, such as the most frequently sold shirt size. All three are valid summaries of the same data; the right one depends on which question you are actually being asked.