Statistics Model Questions Set 1 Practice Questions Answers Test with Solutions & More Shortcuts

Statistics PRACTICE TEST [1 - EXERCISES]

Statistics Model Questions Set 1

Question : 21

Frequency density of a class is computed by the ratio

a) Class frequency to total frequency

b) Class frequency to total number of classes

c) Class frequency to the class width

d) Cumulative frequency up to that class to total frequency

Answer: (c)

Question : 22

The following pairs relate to frequency distribution of a discrete variable and its frequency polygon. Which one of the following pairs is not correctly matched ?

a) Ordinates of the – Class vertices of the polygon frequencies

b) Abscissa of the vertices – Class marks of of the polygon the frequency distribution

c) Base line of the polygon – X-axis

d) Area of the – Total polygon frequency of the distribution

Answer: (d)

Area of the polygon gives sum of $f_i x_i$ not sum of frequency distribution $(f_i)$.

Question : 23

The arithmetic mean of two numbers is 10 and their geometric mean is 8. What are the two numbers?

a) 12, 8

b) 16, 4

c) 15, 5

d) 18, 2

Answer: (b)

Question : 24

In a pi-diagram there are three sectors. If the ratio of the angles of the sectors is 1 : 2 : 3, then what is the angle of the largest sector?

a) 180°

b) 150°

c) 200°

d) 120°

Answer: (b)

Sum of angles of a pie diagram = 360°

∴ θ + 2θ + 3θ = 360°

6θ = 360°

θ = 60°

Hence, angle of Largest sector

6θ = 3 × 60°

= 180°

Question : 25

Read the following information carefully to answer the questions that follow. The arithmetic mean, geometric mean and median of 6 positive numbers a, a, b, b, c, c, where a < b < c are $7/3$, 2, 2, respectively.What is the value of c?

a) 2

b) 3

c) 1

d) 4

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

The value of c is 4.

IMPORTANT quantitative aptitude EXERCISES

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