Statistics

1. Where this chapter sits

In Class 9 you classified data into ungrouped and grouped frequency distributions, drew bar graphs, histograms and frequency polygons, and met three measures of central tendency — single numbers that "represent" a whole data set: the mean (average), the median (middle value) and the mode (most frequent value).

This chapter takes one step further: it finds those same three measures when the data is grouped into class intervals (like $10\text{-}25,\ 25\text{-}40,\dots$). When data is large, we condense it into classes — but then we no longer know the individual values, so we need new formulae. We also build the idea of cumulative frequency.

Quick recap of the basic terms:

• Class interval: a range such as $10\text{-}25$; here $10$ is the lower limit, $25$ the upper limit.
• Class size $h$ = upper limit − lower limit (e.g. $25-10=15$).
• Class mark $x_i=\dfrac{\text{upper limit}+\text{lower limit}}{2}$ — the mid-point, used as the representative value of the class.
• Frequency $f_i$ = number of observations in that class.

2. Mean of ungrouped data — the starting point

If observations $x_1,x_2,\dots,x_n$ occur with frequencies $f_1,f_2,\dots,f_n$, the mean is the sum of all values divided by the number of values:

The Greek letter $\sum$ (sigma) just means "add them all up".

3. Mean of grouped data — Direct Method

When data is grouped, we no longer know each value. We assume the frequency of each class is centred at its class mark $x_i$, and treat $x_i$ as the representative value. Then the same formula works:

Converting Example 1 into classes of width $15$ gives Table 13.3. (Note the convention: a value on an upper limit, e.g. $40$, belongs to the next class $40\text{-}55$, not $25\text{-}40$.)

$\bar{x}=\dfrac{1860}{30}=62$. Notice this differs from the ungrouped $59.3$ — that gap is the price of the mid-point assumption ($59.3$ is exact, $62$ is approximate).

4. Mean — Assumed-Mean Method

When the $x_i$ are large, multiplying $f_ix_i$ is tedious. Instead, pick one class mark as the assumed mean $a$ (best taken near the middle), and compute the deviation $d_i=x_i-a$. These are smaller numbers:

For Table 13.3 take $a=47.5$:

$\bar{x}=47.5+\dfrac{435}{30}=47.5+14.5=62$ — same answer, less labour. The mean does not depend on which $a$ you choose.

5. Mean — Step-Deviation Method

If all the deviations $d_i$ share a common factor (usually the class size $h$), divide them by $h$ to get even smaller numbers $u_i=\dfrac{x_i-a}{h}$. Then:

For Table 13.3 with $a=47.5,\ h=15$:

$\bar{x}=47.5+15\times\dfrac{29}{30}=47.5+14.5=62$. All three methods agree — they are just simpler forms of the direct method. Use direct when $x_i,f_i$ are small; assumed-mean or step-deviation when they are large.

6. More worked mean examples

7. Mode of grouped data

The mode is the value occurring most often. For ungrouped data you simply spot the highest frequency. For grouped data you cannot read a single value, so you first find the modal class — the class with the maximum frequency — and then use:

where $l$ = lower limit of the modal class, $h$ = class size, $f_1$ = frequency of the modal class, $f_0$ = frequency of the class before it, $f_2$ = frequency of the class after it.

8. Cumulative frequency

The cumulative frequency (cf) of a class is its frequency plus the frequencies of all classes before it. There are two types:

• "Less than" type: running total going down — e.g. number of students scoring less than $20$, less than $30$, … Uses the upper limits.
• "More than" type: running total subtracted from the top — number scoring more than or equal to $0$, $10$, … Uses the lower limits.

The cumulative frequency column is the key tool for finding the median (next section). Example of a "less than" cf for marks of $53$ students:

9. Median of grouped data

The median is the middle-most value. For ungrouped data of $n$ observations (arranged in order): if $n$ is odd it is the $\left(\dfrac{n+1}{2}\right)$th value; if $n$ is even it is the average of the $\dfrac{n}{2}$th and $\left(\dfrac{n}{2}+1\right)$th values.

For grouped data the middle observation falls inside a class, so we find the median class — the first class whose cumulative frequency is $\ge\dfrac{n}{2}$ — then use:

where $l$ = lower limit of the median class, $n=\sum f_i$, $cf$ = cumulative frequency of the class before the median class, $f$ = frequency of the median class, $h$ = class size.

10. Choosing a measure & the empirical relation

Which measure is best?

• Mean — most used; takes every value into account. But extreme values distort it, so it can mislead for very uneven data.
• Median — best when individual values don't matter and there are outliers (e.g. typical wage), since extremes don't affect the middle value.
• Mode — best for "most popular / most frequent" questions (most-watched TV programme, most-demanded product).

The three are linked by an approximate empirical relationship:

Rearranged: $\text{Mode}=3\,\text{Median}-2\,\text{Mean}$. Given any two of the three, you can estimate the third. Note: before using the mode or median formula, make sure class intervals are continuous (no gaps) — convert if needed.

11. NCERT Exercise 13.1 — fully solved

Q1. Plants per house (20 houses). Classes $0\text{-}2,\dots,12\text{-}14$, $f:1,2,1,5,6,2,3$. Class marks $1,3,5,7,9,11,13$. Direct method (small values): $\sum f_ix_i=1+6+5+35+54+22+39=162$, $\sum f_i=20$. Mean $=\dfrac{162}{20}=8.1$ plants. Direct method chosen because $x_i,f_i$ are small.

Q2. Daily wages of 50 workers. Classes $500\text{-}520,\dots,580\text{-}600$, $f:12,14,8,6,10$. Take $a=550,\ h=20$. $u_i:-2,-1,0,1,2$; $f_iu_i:-24,-14,0,6,20$, $\sum f_iu_i=-12$. Mean $=550+\dfrac{-12}{50}\times20=550-4.8=545.2$ rupees.

Q3. Missing frequency $f$ (mean = 18). Classes $11\text{-}13,\dots,23\text{-}25$, $f:7,6,9,13,f,5,4$. Class marks $12,14,16,18,20,22,24$. $\sum f_ix_i=84+84+144+234+20f+110+96=752+20f$; $\sum f_i=44+f$. Set $\dfrac{752+20f}{44+f}=18\Rightarrow752+20f=792+18f\Rightarrow2f=40\Rightarrow f=20$.

Q4. Heartbeats of 30 women. Classes $65\text{-}68,\dots,83\text{-}86$, $f:2,4,3,8,7,4,2$. Take $a=75.5,\ h=3$. $u_i:-3,-2,-1,0,1,2,3$; $f_iu_i:-6,-8,-3,0,7,8,6=4$. Mean $=75.5+\dfrac{4}{30}\times3=75.5+0.4=75.9$ beats/min.

Q5. Mangoes per box. Classes (continuous: $49.5\text{-}52.5,\dots$), $f:15,110,135,115,25$, $\sum f=400$. Class marks $51,54,57,60,63$. Take $a=57,\ h=3$. $u_i:-2,-1,0,1,2$; $f_iu_i:-30,-110,0,115,50=25$. Mean $=57+\dfrac{25}{400}\times3=57+0.1875\approx57.19$ mangoes. Step-deviation chosen.

Q6. Daily food expenditure (25 households). Classes $100\text{-}150,\dots,300\text{-}350$, $f:4,5,12,2,2$. Take $a=225,\ h=50$. $u_i:-2,-1,0,1,2$; $f_iu_i:-8,-5,0,2,4=-7$. Mean $=225+\dfrac{-7}{25}\times50=225-14=211$ rupees.

Q7. SO$_2$ concentration. Classes $0.00\text{-}0.04,\dots,0.20\text{-}0.24$, $f:4,9,9,2,4,2$, $\sum f=30$. Class marks $0.02,0.06,0.10,0.14,0.18,0.22$. $\sum f_ix_i=0.08+0.54+0.90+0.28+0.72+0.44=2.96$. Mean $=\dfrac{2.96}{30}=0.099$ ppm.

Q8. Days absent (40 students). Classes $0\text{-}6,6\text{-}10,10\text{-}14,14\text{-}20,20\text{-}28,28\text{-}38,38\text{-}40$ (unequal), $f:11,10,7,4,4,3,1$. Class marks $3,8,12,17,24,33,39$. $\sum f_ix_i=33+80+84+68+96+99+39=499$. Mean $=\dfrac{499}{40}=12.475\approx12.48$ days.

Q9. Literacy rate of 35 cities. Classes $45\text{-}55,\dots,85\text{-}95$, $f:3,10,11,8,3$. Take $a=70,\ h=10$. $u_i:-2,-1,0,1,2$; $f_iu_i:-6,-10,0,8,6=-2$. Mean $=70+\dfrac{-2}{35}\times10=70-0.57=69.43\%$.

12. NCERT Exercise 13.2 — fully solved

Q1. Ages of patients — mode and mean. Classes $5\text{-}15,\dots,55\text{-}65$, $f:6,11,21,23,14,5$. Modal class $35\text{-}45$ (max $f=23$): $l=35,f_1=23,f_0=21,f_2=14,h=10$. $\text{Mode}=35+\dfrac{23-21}{46-21-14}\times10=35+\dfrac{20}{11}=36.8$ years. Mean: $a=30,h=10$, $u_i:-2,-1,0,1,2,3$; $f_iu_i:-12,-11,0,23,28,15=43$, $\sum f=80$. Mean $=30+\dfrac{43}{80}\times10=30+5.37=35.37$ years. Maximum patients are around age $36.8$; the average age is $35.37$.

Q2. Modal lifetime of 225 components. Classes $0\text{-}20,\dots,100\text{-}120$, $f:10,35,52,61,38,29$. Modal class $60\text{-}80$ (max $f=61$): $l=60,f_1=61,f_0=52,f_2=38,h=20$. $\text{Mode}=60+\dfrac{61-52}{122-52-38}\times20=60+\dfrac{9}{32}\times20=60+5.625=65.625$ hours.

Q3. Monthly expenditure of 200 families — mode and mean. Modal class $1500\text{-}2000$ (max $f=40$): $l=1500,f_1=40,f_0=24,f_2=33,h=500$. $\text{Mode}=1500+\dfrac{40-24}{80-24-33}\times500=1500+\dfrac{16}{23}\times500=1500+347.83=1847.83$ rupees. Mean: $a=2750,h=500$, $\sum f_iu_i=-235$ (from $u_i:-3,-2,-1,0,1,2,3,4$ with $f:24,40,33,28,30,22,16,7$). Mean $=2750+\dfrac{-235}{200}\times500=2750-587.5=2662.5$ rupees.

Q4. Teacher-student ratio — mode and mean. Classes $15\text{-}20,\dots,50\text{-}55$, $f:3,8,9,10,3,0,0,2$. Modal class $30\text{-}35$ (max $f=10$): $l=30,f_1=10,f_0=9,f_2=3,h=5$. $\text{Mode}=30+\dfrac{10-9}{20-9-3}\times5=30+\dfrac{5}{8}=30.6$. Mean: $a=32.5,h=5$, $\sum f=35$, $\sum f_iu_i=-23$. Mean $=32.5+\dfrac{-23}{35}\times5=32.5-3.29=29.2$. Most states have ratio $\approx30.6$; the average is $29.2$.

Q5. Runs by batsmen — mode. Modal class $4000\text{-}5000$ (max $f=18$): $l=4000,f_1=18,f_0=4,f_2=9,h=1000$. $\text{Mode}=4000+\dfrac{18-4}{36-4-9}\times1000=4000+\dfrac{14}{23}\times1000=4608.7$ runs.

Q6. Cars per period — mode. Classes $0\text{-}10,\dots,70\text{-}80$, $f:7,14,13,12,20,11,15,8$. Modal class $40\text{-}50$ (max $f=20$): $l=40,f_1=20,f_0=12,f_2=11,h=10$. $\text{Mode}=40+\dfrac{20-12}{40-12-11}\times10=40+\dfrac{80}{17}=44.7$ cars.

13. NCERT Exercise 13.3 — fully solved

Q1. Electricity consumption (68 consumers) — median, mean, mode. Classes $65\text{-}85,\dots,185\text{-}205$, $f:4,5,13,20,14,8,4$. Mean: $a=135,h=20$, $\sum f_iu_i=7$; mean $=135+\dfrac{7}{68}\times20=137.06$ units. Median: $\dfrac{n}{2}=34$; cf: $4,9,22,42,\dots$ — median class $125\text{-}145$ ($l=125,cf=22,f=20,h=20$). $\text{Median}=125+\dfrac{34-22}{20}\times20=125+12=137$ units. Mode: modal class $125\text{-}145$: $\text{Mode}=125+\dfrac{20-13}{40-13-14}\times20=125+\dfrac{140}{13}=135.77$ units. The three measures are close ($\approx135\text{-}137$).

Q2. Missing $x,y$ (median = 28.5, total = 60). Classes $0\text{-}10,\dots,50\text{-}60$, $f:5,x,20,15,y,5$. Sum $=45+x+y=60\Rightarrow x+y=15$ … (1). Median $28.5$ lies in $20\text{-}30$: $l=20,f=20,h=10,cf=5+x$, $\dfrac{n}{2}=30$. $28.5=20+\dfrac{30-(5+x)}{20}\times10\Rightarrow8.5=\dfrac{25-x}{2}\Rightarrow17=25-x\Rightarrow x=8$. From (1): $y=7$.

Q3. Ages of 100 policy holders — median (below-type data). Convert to frequencies: classes $18\text{-}20,20\text{-}25,\dots,55\text{-}60$, $f:2,4,18,21,33,11,3,6,2$. cf: $2,6,24,45,78,\dots$. $\dfrac{n}{2}=50$; median class $35\text{-}40$ ($l=35,cf=45,f=33,h=5$). $\text{Median}=35+\dfrac{50-45}{33}\times5=35+\dfrac{25}{33}=35.76$ years.

Q4. Lengths of 40 leaves — median. Make classes continuous: $117.5\text{-}126.5,\dots,171.5\text{-}180.5$, $f:3,5,9,12,5,4,2$. cf: $3,8,17,29,\dots$. $\dfrac{n}{2}=20$; median class $144.5\text{-}153.5$ ($l=144.5,cf=17,f=12,h=9$). $\text{Median}=144.5+\dfrac{20-17}{12}\times9=144.5+2.25=146.75$ mm.

Q5. Lifetime of 400 lamps — median. Classes $1500\text{-}2000,\dots,4500\text{-}5000$, $f:14,56,60,86,74,62,48$. cf: $14,70,130,216,\dots$. $\dfrac{n}{2}=200$; median class $3000\text{-}3500$ ($l=3000,cf=130,f=86,h=500$). $\text{Median}=3000+\dfrac{200-130}{86}\times500=3000+406.98=3406.98$ hours.

Q6. Letters in 100 surnames — median, mean, mode. Classes $1\text{-}4,\dots,16\text{-}19$, $f:6,30,40,16,4,4$. Median: cf $6,36,76,\dots$, $\dfrac{n}{2}=50$; median class $7\text{-}10$ ($l=7,cf=36,f=40,h=3$). $\text{Median}=7+\dfrac{50-36}{40}\times3=7+1.05=8.05$. Mean: $a=11.5,h=3$, class marks $2.5,5.5,8.5,11.5,14.5,17.5$; $\sum f_iu_i=-106$; mean $=11.5+\dfrac{-106}{100}\times3=11.5-3.18=8.32$. Mode: modal class $7\text{-}10$: $\text{Mode}=7+\dfrac{40-30}{80-30-16}\times3=7+\dfrac{30}{34}=7.88$.

Q7. Weights of 30 students — median. Classes $40\text{-}45,\dots,70\text{-}75$, $f:2,3,8,6,6,3,2$. cf: $2,5,13,19,\dots$. $\dfrac{n}{2}=15$; median class $55\text{-}60$ ($l=55,cf=13,f=6,h=5$). $\text{Median}=55+\dfrac{15-13}{6}\times5=55+1.67=56.67$ kg.

14. Common mistakes to avoid

• Comparing the wrong column to $\dfrac{n}{2}$ — always compare the cumulative frequency, and use the lower limit of the chosen class as $l$.
• For median, taking $cf$ of the median class instead of the class before it.
• Forgetting to make class intervals continuous before applying mode/median formulae (e.g. $1\text{-}4,5\text{-}8\dots$ must become $0.5\text{-}4.5,\dots$).
• Mixing up $f_0$ (class before) and $f_2$ (class after) in the mode formula.
• In the mean, using the class limits instead of the class mark $x_i$.
• Writing the empirical relation as $3\,\text{Mode}=\dots$ — it is $3\,\text{Median}=\text{Mode}+2\,\text{Mean}$.

15. Quick revision checklist

• Class mark $x_i=\dfrac{\text{upper}+\text{lower}}{2}$; class size $h=$ upper − lower.
• Mean: Direct $\bar{x}=\dfrac{\sum f_ix_i}{\sum f_i}$; Assumed $\bar{x}=a+\dfrac{\sum f_id_i}{\sum f_i}$; Step $\bar{x}=a+h\dfrac{\sum f_iu_i}{\sum f_i}$.
• Mode $=l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times h$ (modal class = max frequency).
• Median $=l+\dfrac{\frac{n}{2}-cf}{f}\times h$ (median class = first cf $\ge\frac{n}{2}$).
• Empirical relation: $3\,\text{Median}=\text{Mode}+2\,\text{Mean}$.
• Always build a cf column for median; ensure classes are continuous for mode/median.