Mean and Median — ICSE Class 9 Maths Important Questions

Answer: (c) $3$3.

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

Answer: (c) $7$7.

Arranged: $3, 5, 7, 9, 11$3, 5, 7, 9, 11. With $n=5$n=5 (odd), median $=\left(\dfrac{5+1}{2}\right)$=(5+1/2)th $=3$=3rd term $=7$=7.

Answer: (c) $19$19.

Increasing every observation by $4$4 increases the mean by $4$4: new mean $=15+4=19$=15+4=19.

Answer: (b) $30$30.

Sum of $6$6 numbers $=6\times20=120$=6×20=120. Sum of remaining $5$5 numbers $=5\times18=90$=5×18=90. Removed number $=120-90=30$=120-90=30.

Answer: (a) With $n=4$n=4 (even), median $=$= average of the 2nd and 3rd terms $=\dfrac{6+8}{2}=7$=6+8/2=7. R correctly explains A.

$\sum f=3+5+4+2=14$f=3+5+4+2=14.

$\sum f x=5(3)+10(5)+15(4)+20(2)=15+50+60+40=165$f x=5(3)+10(5)+15(4)+20(2)=15+50+60+40=165.

$\bar{x}=\dfrac{\sum f x}{\sum f}=\dfrac{165}{14}\approx11.79.$x= f x/ f=165/1411.79.

With $n=5$n=5 (odd), the median is the 3rd term $=x+3$=x+3.

$x+3=18\ \Rightarrow\ x=15.$x+3=18 x=15.

The values are $13, 15, 18, 20, 22$13, 15, 18, 20, 22. Mean $=\dfrac{13+15+18+20+22}{5}=\dfrac{88}{5}=17.6$=13+15+18+20+22/5=88/5=17.6.

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

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

$\bar{x}=A+\dfrac{\sum fd}{\sum f}=25+\dfrac{-5}{30}=25-0.1667\approx24.83.$x=A+ fd/ f=25+-5/30=25-0.166724.83.

Build the cumulative frequency (cf):

$N=22$N=22 (even). The two middle terms are the $\dfrac{N}{2}=11$N/2=11th and $12$12th terms.

From the cf column, both the 11th and 12th terms fall in the value $x=14$x=14 (cf jumps from 8 to 16).

Median $=\dfrac{14+14}{2}=14$=14+14/2=14.

$\sum f=6+8+p+4=18+p$f=6+8+p+4=18+p.

$\sum f x=5(6)+7(8)+9(p)+11(4)=30+56+9p+44=130+9p$f x=5(6)+7(8)+9(p)+11(4)=30+56+9p+44=130+9p.

Given $\bar{x}=8$x=8:
$8=\dfrac{130+9p}{18+p}.$8=130+9p/18+p.

$8(18+p)=130+9p\ \Rightarrow\ 144+8p=130+9p\ \Rightarrow\ 144-130=9p-8p\ \Rightarrow\ p=14.$8(18+p)=130+9p 144+8p=130+9p 144-130=9p-8p p=14.

The missing frequency is $p=14$p=14.

(i) Direct method:

$\sum f=40$f=40, $\sum fx=60+180+360+320+250=1170$fx=60+180+360+320+250=1170.

$\bar{x}=\dfrac{\sum fx}{\sum f}=\dfrac{1170}{40}=29.25.$x= fx/ f=1170/40=29.25.

(ii) $N=40$N=40 (even). Middle terms are the $20$20th and $21$21st.
From the cf column, cf reaches 15 at $x=20$x=20 and 27 at $x=30$x=30, so both the 20th and 21st terms correspond to $x=30$x=30.

Median $=\dfrac{30+30}{2}=30$=30+30/2=30.

Total frequency: $24+a+28+b+16=120\Rightarrow a+b=52$24+a+28+b+16=120 a+b=52. \quad(1)

$\sum fx=30(24)+40a+50(28)+60b+70(16)$fx=30(24)+40a+50(28)+60b+70(16)
$=720+40a+1400+60b+1120=3240+40a+60b$=720+40a+1400+60b+1120=3240+40a+60b.

Mean $=50$=50:
$50=\dfrac{3240+40a+60b}{120}\ \Rightarrow\ 6000=3240+40a+60b,$50=3240+40a+60b/120 6000=3240+40a+60b,
$40a+60b=2760\ \Rightarrow\ 2a+3b=138. \quad(2)$40a+60b=2760 2a+3b=138. (2)

From (1), $a=52-b$a=52-b. Substitute into (2):
$2(52-b)+3b=138\ \Rightarrow\ 104-2b+3b=138\ \Rightarrow\ b=34.$2(52-b)+3b=138 104-2b+3b=138 b=34.

Then $a=52-34=18$a=52-34=18.

So the missing frequencies are $a=18$a=18 and $b=34$b=34.

(i) Sum $=45+62+38+70+55+48+62+80+40=500$=45+62+38+70+55+48+62+80+40=500.
Mean $=\dfrac{500}{9}\approx55.56$=500/955.56 runs.

(ii) Arranged: $38, 40, 45, 48, 55, 62, 62, 70, 80$38, 40, 45, 48, 55, 62, 62, 70, 80. With $n=9$n=9 (odd), median $=\left(\dfrac{9+1}{2}\right)$=(9+1/2)th $=5$=5th term $=55$=55 runs.

(iii) New sum $=500+100=600$=500+100=600 over 10 innings. New mean $=\dfrac{600}{10}=60$=600/10=60 runs.

(iv) With 100 added, arranged data has $n=10$n=10 (even); median $=$= average of 5th and 6th terms $=\dfrac{55+62}{2}=58.5$=55+62/2=58.5 runs. Yes, the median changes from $55$55 to $58.5$58.5 because the number of observations becomes even and the middle position shifts.