Statistics Model Questions Set 1 Practice Questions Answers Test with Solutions & More Shortcuts
Statistics PRACTICE TEST [1 - EXERCISES]
Statistics Model Questions Set 1
Question : 6
Number of credit cards held by an individual can be treated as
a) Discrete data
b) Categorical data
c) Qualitative data
d) None of the above
Answer »Answer: (b)
Categorical data
Question : 7
Consider the following grouped frequency distribution :
x | f |
0-10 | 8 |
10-20 | 12 |
20-30 | 10 |
30-40 | p |
40-50 | 9 |
a) 10
b) 11
c) 9
d) 12
Answer »Answer: (b)
Mean = (sum of ∞) / (sum of f) = (5*5 + 12*15 + 10*25 + p*35 + 9*45) / (8 + 12 + 10 + P + 9) = 25.2 (875 + 35P) / (39 + P) = 25.2 ⇒ P = 11
Question : 8
Square diagrams are drawn to represent the following data:
Country | Pakistan | India | Myanmar | China |
Labour | 36 | 81 | 25 | 100 |
Production(inRs.) |
a) 1.2 cm
b) 1 cm
c) 1.8 cm
d) 2 cm
Answer »Answer: (c)
Rs.25 = 1 $cm^2$
∴ Rs.1 = $1/{25} cm^2$
∴ Rs.81 = ${81}/{25} cm^2$ = Area of square
Side of square = $√{{81}/{25}} = 9/5$ = 1.8 cm
Question : 9
In a pie diagram, there are four slices with angles 150°, 90°, 60° and 60°. A new pie diagram is formed by deleting one of the slices having angle 60° in the given pie diagram. In the new pie diagram
a) The smallest slice has angle 70°
b) The largest slice has angle 180°
c) The largest slice has angle 150°
d) The smallest slice has angle 90
Answer »Answer: (b)
Four slices 150°, 90°, 60°, 60°
when 60° is deleted
Remaining 150°, 90°, 60°
Total = 300°
While making Pie chart where 300° is taken as 100%
${150}/{300}$ × 100 = 50%
${90°}/{300}$ × 100 = 30%
${60°}/{300}$ × 100 = 20%
50% of 360° will be 180°
∴ largest slice as angle – 180°
Question : 10
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 mode?
a) 2
b) 1, 2 and 4
c) 1
d) None of these
Answer »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)
= $3(2) - 2(7/3) = {18 - 14}/3 = 4/3$
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