Statistics Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (a)

Number of Type A pencil = ${50}/1$ = 50

Number of Type B pencil = $x/{1.50}$

Number of Type C pencil = ${20}/2$ = 10

Average = ${\text"Total money spent"}/{\text"totalno.of pencil"}$ = 1.25

= ${x + 50 + 20}/{50 + 10 + {x/{1.50}}}$ = 1.25

= 70 + x = 1.25 $(60 + x/{1.50})$

70 + x = 75.00 + ${1.25}/{1.50}$x

x - ${125}/{150}$ x = 5

= ${25}/{150}$ x = 5

x = 30

Answer: (b)

Number of total bulbs = 100

∴ 7 + x + 40 + y + 10 + 2 = 100

⇒ x + y = 41 ... (i)

The median life is 20 h, so median interval will be (18-23).

Here, l = 18, $N/2$ = 50

c = 7 + x, f = 40, h = 5

∴ Median = l + ${(N/2 - C)}/f$ × h

⇒ 20 = 18 + ${(50 – 7 – x)}/{40}$ × 5

⇒ 2 = ${50 – 7 – x}/8$

⇒ 16 = 50 – 7 – x

⇒ x = 43 – 16

⇒ x = 27

Missing frequency 'y' is 14.

Answer: (a)

Difference in the number of boys studying Mathematics and Physics = 180 – 150 = 30

Answer: (d)

Since, intelligence of students is an attribute, arithmetic mean is not suitable method.

Answer: (b)

Mode of frequency distribution of series II is 46.

Answer: (d)

a < b < c

Total numbers = 6

Increasing order a, a, b, b, c, c

∴ Median = ${\text"(6/2)th term + (6/2 + 1)th term"}/2$

= ${\text"3rd term + 4th term"}/2$

2 = ${b + b}/2$ = b

Arithmetic mean = ${a + a + b + b + c + c}/6$

⇒ $7/3 = {a + b + c}/3$

⇒ a + b + c = 7

⇒ a + c = 7 – 2 = 5 ... (i)

Geometric mean = $(a^2 × b^2 × c^2)^{1/6}$

⇒ 2 = $(abc)^{1/3}$

⇒ abc = 8

⇒ ac = $8/2$ = 4 ... (ii)

⇒ c = $4/a$

From equation (i),

a + $4/a$ = 5

⇒ ${a^2 + 4}/a$ = 5

⇒ $a^2$ – 5a + 4 = 0

⇒ $a^2$ – 4a – a + 4 = 0

⇒ a(a – 4) – 1(a – 4) = 0

⇒ (a – 4) (a – 1) = 0

if a = 1 then c = 4

a = 4 then c = 1

a = 1, c = 4 and b = 2

Mode = 3 (Median) – 2 (Mean)

= $3(2) - 2(7/3) = {18 - 14}/3 = 4/3$