Measures of Central Tendency (Mean, Median, Quartiles and Mode) — Important Questions

Answer: (b) $3$.

$\bar{x}=\dfrac{1+2+3+4+5}{5}=\dfrac{15}{5}=3.$

Answer: (c) $6$.

The value $6$ occurs three times, more often than any other value, so the mode is $6$.

Answer: (a) $7.5$.

In order: $3,4,7,8,10,11$. With $n=6$, median $=$ mean of the $3$rd and $4$th values $=\dfrac{7+8}{2}=7.5.$

Answer: (a) $18$.

Sum of $6$ observations $=6\times17=102$. After removing $12$, sum $=102-12=90$, so new mean $=\dfrac{90}{5}=18.$

Answer: (a) All three methods are algebraically equivalent; the step-deviation method merely simplifies computation and rescales by $h$, giving the identical mean. R correctly explains A.

$\sum f=2+4+3+1=10.$

$\sum fx=(5)(2)+(10)(4)+(15)(3)+(20)(1)=10+40+45+20=115.$

$\bar{x}=\dfrac{\sum fx}{\sum f}=\dfrac{115}{10}=11.5.$

Mean $=\dfrac{x+(x+2)+(x+4)+(x+6)+(x+8)}{5}=\dfrac{5x+20}{5}=x+4.$

Given $x+4=13\Rightarrow x=9.$

The five values are $9,11,13,15,17$; being in order with $n=5$, the median is the $3$rd value $=13.$

Let $A=25$ and $d=x-25$.

$\sum f=30,\ \ \sum fd=-40-30+0+35+30=-5.$

$\bar{x}=A+\dfrac{\sum fd}{\sum f}=25+\dfrac{-5}{30}=25-0.1\overline{6}=24.83\ (\text{approx}).$

Class marks $x$: $5,15,25,35,45$. Let $A=25$, $h=10$, $u=\dfrac{x-25}{10}$.

$\sum f=40,\ \ \sum fu=-12-8+0+9+10=-1.$

$\bar{x}=A+\dfrac{\sum fu}{\sum f}\times h=25+\dfrac{-1}{40}\times10=25-0.25=24.75.$

$\sum f=4+6+10+p+6=26+p.$

$\sum fx=(10)(4)+(30)(6)+(50)(10)+(70)(p)+(90)(6)$

$=40+180+500+70p+540=1260+70p.$

Using $\bar{x}=\dfrac{\sum fx}{\sum f}=54$:

$\dfrac{1260+70p}{26+p}=54.$

Cross-multiplying: $1260+70p=54(26+p)=1404+54p.$

$70p-54p=1404-1260\Rightarrow 16p=144\Rightarrow p=9.$

The missing frequency is $p=9.$

Check: $\sum f=35$, $\sum fx=1260+630=1890$, $\bar{x}=\dfrac{1890}{35}=54.$ Correct.

Mean (step-deviation): class marks $x=15,25,35,45,55$; $A=35$, $h=10$, $u=\dfrac{x-35}{10}$.

$\sum f=40,\ \sum fu=-10-8+0+9+6=-3.$

$\bar{x}=35+\dfrac{-3}{40}\times10=35-0.75=34.25.$

Median: cumulative frequencies: $5,13,28,37,40$. With $n=40$, $\dfrac{n}{2}=20$, which falls in the class $30$–$40$ (median class). Using $\text{median}=l+\dfrac{\frac{n}{2}-cf}{f}\times h$ with $l=30$, $cf=13$, $f=15$, $h=10$:

$\text{median}=30+\dfrac{20-13}{15}\times10=30+\dfrac{7}{15}\times10=30+4.67=34.67.$

Mode: the highest frequency is $15$, so the modal class is $30$–$40$.

So mean $=34.25$, median $\approx34.67$ and the modal class is $30$–$40$.

Cumulative frequency table:

• $0$–$10$: cf $=6$
• $10$–$20$: cf $=16$
• $20$–$30$: cf $=32$
• $30$–$40$: cf $=44$
• $40$–$50$: cf $=50$

Here $n=50$.

Lower quartile $Q_1$: position $\dfrac{n}{4}=12.5$, which lies in class $10$–$20$ ($l=10,\ cf=6,\ f=10,\ h=10$).

$Q_1=10+\dfrac{12.5-6}{10}\times10=10+6.5=16.5.$

Median: position $\dfrac{n}{2}=25$, which lies in class $20$–$30$ ($l=20,\ cf=16,\ f=16,\ h=10$).

$\text{Median}=20+\dfrac{25-16}{16}\times10=20+\dfrac{9}{16}\times10=20+5.625=25.63.$

Upper quartile $Q_3$: position $\dfrac{3n}{4}=37.5$, which lies in class $30$–$40$ ($l=30,\ cf=32,\ f=12,\ h=10$).

$Q_3=30+\dfrac{37.5-32}{12}\times10=30+\dfrac{5.5}{12}\times10=30+4.58=34.58.$

So $Q_1=16.5$, median $\approx25.63$, $Q_3\approx34.58$, and interquartile range $\approx34.58-16.5=18.08.$

(i) Sum $=38+70+48+34+42+55+63=350$. Mean $=\dfrac{350}{7}=50$ runs.

(ii) In order: $34,38,42,48,55,63,70$. With $n=7$ (odd), median $=$ the $\left(\dfrac{7+1}{2}\right)=4$th value $=48$ runs.

(iii) New sum $=350+50=400$ over $8$ innings, so new mean $=\dfrac{400}{8}=50$ runs.

(iv) Range $=$ highest $-$ lowest $=70-34=36$ runs.