Mean and Median — ICSE Class 9 Maths Important Questions

Answer: (c) $3$3.

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

$\sum f=3+5+4+2=14$Σ f=3+5+4+2=14.

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

$\bar{x}=\dfrac{\sum f x}{\sum f}=\dfrac{165}{14}\approx11.79.$x̄=(Σ f x)/(Σ f)=165/14≈11.79.

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.

(i) Lower limit $20$20, upper limit $29$29. The gap between classes is $1$1, so subtract and add $0.5$0.5: boundaries $19.5$19.5 and $29.5$29.5.

(ii) Class size $=29.5-19.5=10$=29.5-19.5=10. Class mark $=\dfrac{20+29}{2}=24.5$=(20+29)/2=24.5.

(iii) $9.5$9.5–$19.5$19.5, $19.5$19.5–$29.5$29.5, $29.5$29.5–$39.5$39.5.

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.

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

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

Given $\bar{x}=8$x̄=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:

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

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

Class marks: $5, 15, 25, 35, 45$5, 15, 25, 35, 45. Plot the points $(5,4)$(5,4), $(15,7)$(15,7), $(25,12)$(25,12), $(35,9)$(35,9), $(45,5)$(45,5) and join them with straight lines. Close the polygon by joining to the mid-points of the empty classes on either side, $(-5,0)$(-5,0) and $(55,0)$(55,0).

Students scoring $20$20 or more $=12+9+5=26$=12+9+5=26.

(i) Sum $=45+62+38+70+55+48+62+80+40=500$=45+62+38+70+55+48+62+80+40=500.
Mean $=\dfrac{500}{9}\approx55.56$=500/9≈55.56 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.