Vardaan Learning Institute
Chapter Solutions: Statistics
SECTION A: OBJECTIVE TYPE QUESTIONS (1 Mark Each)
1. Class mark of 10-25:
Class Mark = (Upper + Lower)/2 = (10+25)/2 = 35/2 = 17.5.
Correct Option: (b) 17.5
2. Mean 12, Median 15, Mode?:
Mode = 3(Median) - 2(Mean) = 3(15) - 2(12) = 45 - 24 = 21.
Correct Option: (b) 21
3. Cannot be determined graphically?
Median (Ogive), Mode (Histogram) possible. Mean (Arithmetic) not graphically.
Correct Option: (a) Mean
4. Sum of lower limits of median and modal class.
Data: 0-5(10), 5-10(15), 10-15(12), 15-20(20), 20-25(9).
N=66. N/2=33. CF: 10, 25, 37, 57, 66.
Median Class: 33 falls in 10-15 (since 25 to 37). Lower limit = 10.
Modal Class: Max freq 20 is 15-20. Lower limit = 15.
Sum = 10 + 15 = 25.
Correct Option: (b) 25
5. Intersection of Ogives gives?
Intersection gives Median.
Correct Option: (b) Median
6. Value of $\Sigma(f_i x_i - \bar{x})$.
$\Sigma(f_i x_i) - \Sigma(f_i \bar{x}) = N\bar{x} - \bar{x}(N) = 0$.
Correct Option: (a) 0
7. In $\bar{x} = a + ...$, $d_i$ are deviations from a of:
$d_i = x_i - a$. $x_i$ are Mid-points.
Correct Option: (c) Mid-points of the classes
8. Assumption about frequencies.
We assume frequencies are centred at class marks.
Correct Option: (b) Centred at the classmarks of the classes
9. Less than 14.6 seconds.
Sum freq below 14.6: 2 + 4 + 5 + 71 = 82.
Correct Option: (c) 82
10. Assertion: Mode of 2,3,5,2,4,3,2,1 is 2.
Freqs: 2(3 times), 3(2 times). Mode is 2. Correct.
Reason: Mode is most frequent observation. Correct and explains A.
Correct Option: (a) Both A and R are true and R is the correct explanation of A.
SECTION B: SHORT ANSWER TYPE QUESTIONS (2 Marks Each)
11. Find Mean.
$x_i$: 5, 15, 25, 35, 45
$f_i x_i$: 15, 75, 225, 175, 135. Sum=625. N=25.
Mean = 25.
25
12. Find Mode.
Modal Class 60-80 (Freq 61).
$l=60, f_1=61, f_0=52, f_2=38, h=20$.
Mode = 65.625.
65.625
13. Mean literacy rate.
Mean = 69.43%
69.43 %
14. Mean 50. Find p.
$\Sigma f_i = 92+p$. $\Sigma f_ix_i = 5160 + 30p$.
$50(92+p) = 5160 + 30p \Rightarrow 4600 + 50p = 5160 + 30p$.
$20p = 560 \Rightarrow p=28$.
p = 28
SECTION C: SHORT ANSWER TYPE II QUESTIONS (3 Marks Each)
15. Median 28.5, find x, y.
$x=8, y=7$.
x = 8, y = 7
16. Mean 18, find f.
$18 = \frac{752+20f}{44+f}$.
$792+18f = 752+20f \Rightarrow 40 = 2f \Rightarrow f=20$.
f = 20
17. Median Height.
Median = 152.86 cm.
152.86 cm
18. Mean 25. Find p.
$\Sigma f_i = 44+p$. $\Sigma f_ix_i = 940 + 35p$.
$25(44+p) = 940+35p \Rightarrow 1100 + 25p = 940 + 35p \Rightarrow 160 = 10p \Rightarrow p=16$.
p = 16
19. Compute Mode.
Modal Class 12-16 (Freq 17).
$l=12, f_1=17, f_0=9, f_2=12, h=4$.
$Mode = 12 + \frac{17-9}{34-9-12} \times 4$
$= 12 + \frac{8}{13} \times 4 = 12 + \frac{32}{13} = 12 + 2.46 = 14.46$.
14.46
20. Convert CF to F and find Modal Class.
Marks: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60.
Freq: 6, 9(15-6), 14(29-15), 12(41-29), 19(60-41), 10(70-60).
Max freq is 19. Modal Class is 40-50.
Modal Class: 40-50
SECTION D: LONG ANSWER TYPE QUESTIONS (5 Marks Each)
21. Find Mean, Median, Mode of Daily Income.
Data: 100-120(12), 120-140(14), 140-160(8), 160-180(6), 180-200(10). N=50.
Mean: $\Sigma f_i u_i$ method (a=150, h=20).
$u_i$: -2, -1, 0, 1, 2. $f_iu_i$: -24, -14, 0, 6, 20. Sum = -12.
Mean = $150 + (-12/50)20 = 150 - 4.8 = 145.2$.
Median: N/2=25. Class 120-140 (CF 26). $l=120, cf=12, f=14, h=20$.
Median = $120 + \frac{25-12}{14} \times 20 = 120 + \frac{13 \times 10}{7} = 120 + 18.57 = 138.57$.
Mode: Class 120-140. $f_1=14, f_0=12, f_2=8$.
Mode = $120 + \frac{14-12}{28-12-8} \times 20 = 120 + \frac{2}{8} \times 20 = 120 + 5 = 125$.
Mean=145.2, Median=138.57, Mode=125
22. Median 525. Find x, y (N=100).
$2+5+x+12+17+20+y+9+7+4 = 100 \Rightarrow 76+x+y=100 \Rightarrow x+y=24$.
Median 525 lies in 500-600. $l=500, h=100, f=20$.
CF before: $2+5+x+12+17 = 36+x$.
$525 = 500 + \frac{50-(36+x)}{20} \times 100$
$25 = (14-x) \times 5 \Rightarrow 5 = 14-x \Rightarrow x=9$.
$y = 24-9 = 15$.
x = 9, y = 15
SECTION E: CASE STUDY ANSWERS
23. Case Study: Heights
(i) Freqs: 4, 7, 18, 11, 6, 5.
(ii) Median = 149.03 cm.
(i) 4,7,18,11,6,5 (ii) 149.03 cm