Solved Examples

How to Use This Section

This is your practice bank for Statistics. Keep the four formulas in view: mean $\bar{x}=\dfrac{\sum f_ix_i}{\sum f_i}$ (or the assumed-mean / step-deviation shortcuts), mode $=l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}h$, median $=l+\dfrac{\frac{n}{2}-cf}{f}h$, and the empirical relation $3\,\text{Median}=\text{Mode}+2\,\text{Mean}$. Always build a neat table first.

Example 1: Class marks of $50\text{-}60,60\text{-}70,70\text{-}80$. Solution: $55,65,75$.

Example 2: Mean of marks $0\text{-}20,20\text{-}40,40\text{-}60$ (marks $10,30,50$) with frequencies $3,5,2$. Solution: $\dfrac{30+150+100}{10}=\dfrac{280}{10}=28$.

Example 3 (direct method): Find the mean: $0\text{-}10,10\text{-}20,20\text{-}30,30\text{-}40,40\text{-}50$ with frequencies $8,12,10,6,4$. Solution: marks $5,15,25,35,45$; $f_ix_i=40,180,250,210,180$; total $860$; $\sum f=40$; mean $=860/40=21.5$. Answer: 21.5.

Example 4 (step-deviation): Same data, $a=25$, $h=10$. $u_i=-2,-1,0,1,2$; $f_iu_i=-16,-12,0,6,8$; $\sum f_iu_i=-14$; mean $=25+10\times\dfrac{-14}{40}=25-3.5=21.5$. Answer: 21.5 (matches).

Example 5 (missing frequency, mean): The mean of $0\text{-}20,20\text{-}40,40\text{-}60,60\text{-}80,80\text{-}100$ (marks $10,30,50,70,90$) with frequencies $2,3,f,6,2$ is 50. Find $f$. Solution: $\sum f_ix_i=20+90+50f+420+180=710+50f$; $\sum f=13+f$; $\dfrac{710+50f}{13+f}=50\Rightarrow710+50f=650+50f\Rightarrow710=650$?? So this has no solution — adjust: with mean 50 and symmetric marks, the pull must balance. Using frequencies $2,3,f,6,2$: set $\sum f_i(x_i-50)=0$: $2(-40)+3(-20)+f(0)+6(20)+2(40)=-80-60+0+120+80=60\ne0$. So mean is not 50 for any $f$. Corrected data: frequencies $2,3,f,3,2$ give balance $\Rightarrow$ mean $=50$ for any $f$. Takeaway: the class mark equal to the assumed mean (here 50) contributes 0, so a symmetric table has mean 50 automatically.

Example 6 (mode): Find the mode: $0\text{-}20,20\text{-}40,40\text{-}60,60\text{-}80$ with frequencies $6,9,15,9$. Solution: modal class $40\text{-}60$ ($f_1=15$); $l=40,f_0=9,f_2=9,h=20$; mode $=40+\dfrac{15-9}{30-9-9}\times20=40+\dfrac{6}{12}\times20=40+10=50$. Answer: 50.

Example 7 (mode): $10\text{-}20,20\text{-}30,30\text{-}40$ with frequencies $8,20,12$. modal class $20\text{-}30$; $40+$? $l=20,f_1=20,f_0=8,f_2=12$; mode $=20+\dfrac{20-8}{40-8-12}\times10=20+\dfrac{12}{20}\times10=20+6=26$. Answer: 26.

Example 8 (median): Find the median: $0\text{-}10,10\text{-}20,20\text{-}30,30\text{-}40,40\text{-}50$ with frequencies $5,8,12,10,5$. Solution: cf $5,13,25,35,40$; $n=40$, $\tfrac n2=20$; median class $20\text{-}30$ (cf 25); $l=20,cf=13,f=12,h=10$; median $=20+\dfrac{20-13}{12}\times10=20+5.83=25.83$. Answer: ~25.83.

Example 9 (median): $100\text{-}120,120\text{-}140,140\text{-}160,160\text{-}180,180\text{-}200$ with frequencies $12,14,8,6,10$. cf $12,26,34,40,50$; $n=50$, $\tfrac n2=25$; median class $120\text{-}140$ (cf 26); $l=120,cf=12,f=14,h=20$; median $=120+\dfrac{25-12}{14}\times20=120+18.57=138.57$. Answer: ~138.57.

Example 10 (empirical): Mean $=53$, median $=55$. Estimate mode. $3(55)-2(53)=165-106=59$. Answer: 59.

Example 11 (empirical): Mode $=60$, median $=66$. Estimate mean. $\text{Mean}=\dfrac{3(66)-60}{2}=\dfrac{198-60}{2}=69$. Answer: 69.

Example 12 (all three): For the marks data (frequencies $2,3,7,6,6,6$), state mean, median and mode. Solution: mean $=62$, median $=62.5$, mode $=52$ (from earlier sections).

Example 13 (mean, assumed): $15\text{-}25,25\text{-}35,35\text{-}45,45\text{-}55$ (marks $20,30,40,50$) with frequencies $10,20,30,40$; $a=40$. $d_i=-20,-10,0,10$; $f_id_i=-200,-200,0,400$; $\sum f_id_i=0$; $\sum f=100$; mean $=40+0=40$. Answer: 40.

Example 14 (mode): $0\text{-}10,10\text{-}20,20\text{-}30$ with frequencies $10,8,6$. modal class $0\text{-}10$; $l=0,f_1=10,f_0=0,f_2=8,h=10$; mode $=\dfrac{10}{20-0-8}\times10=\dfrac{10}{12}\times10=8.33$. Answer: ~8.33.

Example 15 (median class): frequencies $3,9,15,30,18,5$ ($n=80$). cf $3,12,27,57,75,80$; $\tfrac n2=40$; median class is the one with cf first past 40 → cf 57, the 4th class. Answer: 4th class.

Example 16 (ogive points): frequencies $4,7,18,11,6,5$ (classes below 140, 140-145, …). Less-than cf: $4,11,29,40,46,51$; points $(140,4),(145,11),(150,29),(155,40),(160,46),(165,51)$.

Example 17 (more-than cf): frequencies $5,8,12,10,5$ ($n=40$). More-than-or-equal cf (from the top of each class): $40,35,27,15,5$. Points (lower limit, cf): $(0,40),(10,35),(20,27),(30,15),(40,5)$.

Example 18 (median from ogive): For Example 17 data, $\tfrac n2=20$; on the less-than ogive (cf $5,13,25,35,40$) the height 20 falls in class $20\text{-}30$; median $=20+\dfrac{20-13}{12}\times10=25.83$. Answer: ~25.83.

Example 19 (mean, step-deviation, big numbers): marks $100,300,500,700,900$ with frequencies $4,6,10,6,4$; $a=500$, $h=200$. $u_i=-2,-1,0,1,2$; $f_iu_i=-8,-6,0,6,8=0$; mean $=500$. Answer: 500.

Example 20 (mode): $25\text{-}35,35\text{-}45,45\text{-}55,55\text{-}65$ with frequencies $7,31,33,17$. modal class $45\text{-}55$; $l=45,f_1=33,f_0=31,f_2=17,h=10$; mode $=45+\dfrac{33-31}{66-31-17}\times10=45+\dfrac{2}{18}\times10=45+1.11=46.11$. Answer: ~46.11.

Example 21 (median): $1500\text{-}2000,2000\text{-}2500,2500\text{-}3000,3000\text{-}3500$ with frequencies $24,40,33,28$. cf $24,64,97,125$; $n=125$, $\tfrac n2=62.5$; median class $2000\text{-}2500$ (cf 64); $l=2000,cf=24,f=40,h=500$; median $=2000+\dfrac{62.5-24}{40}\times500=2000+481.25=2481.25$. Answer: ~2481.25.

Example 22 (empirical): Median $=28.5$, mode $=27$. Mean $=\dfrac{3(28.5)-27}{2}=\dfrac{85.5-27}{2}=29.25$. Answer: 29.25.

Example 23 (missing freq, median): $n=60$, median $=28.5$, classes $0\text{-}10,\dots,50\text{-}60$ frequencies $5,x,20,15,y,5$, median class $20\text{-}30$. From $\sum f=60$: $x+y=15$. Median: $28.5=20+\dfrac{30-(5+x)}{20}\times10\Rightarrow8.5=\dfrac{25-x}{2}\Rightarrow x=8$, $y=7$. Answer: $x=8,y=7$.

Example 24 (mean): $0\text{-}6,6\text{-}12,12\text{-}18,18\text{-}24,24\text{-}30$ (marks $3,9,15,21,27$) with frequencies $6,8,10,9,7$. $f_ix_i=18,72,150,189,189$; total $618$; $\sum f=40$; mean $=618/40=15.45$. Answer: 15.45.

Example 25 (mode = median case): frequencies $12,12,12$ across three equal classes — is there a unique mode? All equal frequencies mean no single modal class; the data is multimodal / has no unique mode. Takeaway: the mode formula needs a single maximum-frequency class.

Example 26 (median): $0\text{-}100,100\text{-}200,200\text{-}300,300\text{-}400,400\text{-}500$ with frequencies $2,5,x,12,6$ and $n=40$. Then $x=40-25=15$. cf $2,7,22,34,40$; $\tfrac n2=20$; median class $200\text{-}300$ (cf 22); $l=200,cf=7,f=15,h=100$; median $=200+\dfrac{20-7}{15}\times100=200+86.67=286.67$. Answer: ~286.67.

Example 27 (mean, direct): ages $0\text{-}15,15\text{-}30,30\text{-}45,45\text{-}60$ (marks $7.5,22.5,37.5,52.5$) with frequencies $5,10,8,7$. $f_ix_i=37.5,225,300,367.5$; total $930$; $\sum f=30$; mean $=31$. Answer: 31.

Example 28 (mode): $0\text{-}5,5\text{-}10,10\text{-}15$ with frequencies $10,15,5$. modal class $5\text{-}10$; $l=5,f_1=15,f_0=10,f_2=5,h=5$; mode $=5+\dfrac{15-10}{30-10-5}\times5=5+\dfrac{5}{15}\times5=5+1.67=6.67$. Answer: ~6.67.

Example 29 (empirical, estimate median): Mean $=45.5$, mode $=40.6$. Median $=\dfrac{\text{Mode}+2\,\text{Mean}}{3}=\dfrac{40.6+91}{3}=\dfrac{131.6}{3}=43.87$. Answer: ~43.87.

Example 30 (full): For frequencies $5,10,20,7,8$ (classes $0\text{-}10,\dots,40\text{-}50$):

1. cf $5,15,35,42,50$; $n=50$, $\tfrac n2=25$; median class $20\text{-}30$ (cf 35); median $=20+\dfrac{25-15}{20}\times10=25$.
2. modal class $20\text{-}30$ ($f_1=20$); mode $=20+\dfrac{20-10}{40-10-7}\times10=20+\dfrac{10}{23}\times10=24.35$.
3. mean: marks $5,15,25,35,45$; $f_ix_i=25,150,500,245,360=1280$; mean $=1280/50=25.6$. Final Answer: mean $=25.6$, median $=25$, mode $\approx24.35$. Takeaway: build one table, then read off all three measures.