Statistics Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 1 EXERCISES
The following question based on Statistics topic of quantitative aptitude
(a) 30
(b) 60
(c) 20
(d) 80
The correct answers to the above question in:
Answer: (a)
Difference in the number of boys studying Mathematics and Physics = 180 – 150 = 30
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Read more model question set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
Read the following information carefully to answer the questions that follow. The average age of 6 persons living in a house is 23.5 years. Three of them are majors and their average age is 42 years. The difference in ages of the three minor children is same.What is the median of the ages of minor children?
a) 5 years
b) 7 years
c) 3 years
d) Cannot be determined
Answer »Answer: (a)
Total age of six persons = 23.5 × 6 = 141 years
Total age of three major persons = 42 × 3 = 126 years
∴ Total age of three minor children = 141 – 126 = 15 years
The difference in ages of the three minor children is same.
Therefore, we take ages may be:
5, 5, 5; 3, 5, 7; 2, 5, 8 and 1, 5, 9
In all the cases, median will be 5 years.
Median age of minor children = 5 years.
Question : 2
Consider the following statements related to cumulative frequency polygon of a frequency distribution, the frequencies being cumulated from the lower end of the range :
1. The cumulative frequency polygon gives an equivalent representation of frequency distribution table.
2. The cumulative frequency polygon is a closed polygon with one horizontal and one vertical side. The other sides have non–negative slope.
Which of the above statements is / are correct ?
a) Only 2
b) Both 1 and 2
c) Only 1
d) Neither 1 nor 2
Answer »Answer: (c)
Here, Statement 1 is correct but Statement 2 is not correct.
Question : 3
Let 1 $\ov{x_1}$ and $\ov{x_2}$ (where $\ov{x_2} > \ov{x_1}$) be the means of two sets comprising $n_1 and n_2$ (where $n_2 < n_1$) observations respectively. If $\ov{x}$ is the mean when they are pooled, then which one of the following is correct?
a) $\ov{x} > \ov{x_2}$
b) $\ov{x} < \ov{x_1}$
c) $\ov{x_1} < \ov{x} < \ov{x_2}$
d) $(\ov{x_1} - \ov{x}) + (\ov{x_2} - \ov{x}) = 0$
Answer »Answer: (c)
Let if possible $\ov{x}_1 < \ov{x} < \ov{x}_2$
Consider
$\ov{x}_1 < \ov{x}$
$\ov{x}_1 < {n_1 \ov{x}_1 + n_2 \ov{x}_2}/{n_1 + n_2}$
$n_1 \ov{x}_1 + n_2\ov{x}_1 < n_1\ov{x}_1 + n_2 \ov{x}_2$
⇒ $\ov{x}_1 < \ov{x}_2$
[it is true as given in Question]
Hence $\ov{x}_1 < \ov{x}$ ...(1)
Consider $\ov{x} < \ov{x}_2$
${n_1\ov{x}_1 + n_2 \ov{x}_2}/{n_1 + n_2} < \ov{x}_2$
⇒ $\ov{x}_1 < \ov{x}_2$
[True as given in Question]
Hence $\ov{x} < \ov{x}_2$
From (1) & (2)
$\ov{x}_1 < \ov{x} < \ov{x}_2$
Question : 4
The following table gives the frequency distribution of life length in hours of 100 electric bulbs having median life 20 h.
| Life of bulbs(in hours) | Number of bulbs |
| 8-13 | 7 |
| 13-18 | x |
| 18-23 | 40 |
| 23-28 | y |
| 28-33 | 10 |
| 33-38 | 2 |
a) 24
b) 14
c) 27
d) 11
Answer »Answer: (b)
Number of total bulbs = 100
∴ 7 + x + 40 + y + 10 + 2 = 100
⇒ x + y = 41 ... (i)
| Life of bulbs (in hours) | Number of bulbs | Cumulative Frequency |
| 8 - 13 | 7 | 7 |
| 13 - 18 | x | 7 + x |
| 18 - 23 | 40 | 47 + x |
| 23 - 28 | y | 47 + x + y |
| 28 - 33 | 10 | 57 + x + y |
| 33 - 38 | 2 | 59 + x + y |
| N = 100 |
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.
Question : 5
An individual purchases three qualities of pencils. The relevant data is given below :
| Quality | Price per Pencil (in Rs.) | Money spent (in Rs.) |
| A | 1.00 | 50 |
| B | 1.50 | x |
| C | 2.00 | 20 |
a) Rs. 30
b) Rs.40
c) Rs.10
d) Rs.60
Answer »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
Question : 6
If each of n numbers xi = i (i = 1, 2, 3,..... n) is replaced by (i + 1) $x_i$ , then the new mean is
a) ${n(n + 1)}/2$
b) ${(n + 1)(n + 2)}/{3n}$
c) ${n + 3}/2$
d) ${(n + 1)(n + 2)}/3$
Answer »Answer: (d)
$(i + 1)x^i = (i + 1)x^i$ where i = 1, 2, 3, ------- n
$Σ↖{n}↙{i = 1}(i + 1)$ = 1.2 + 2.3 + 3.4 + 4.5 + -------meters
= $Σ↖{n}↙{n = 1} T_n$
= Σ n (n + 1)
= $Σn^2 + Σn$
= ${(n + 1)n(2n + 1)}/6 + {n(n + 1)}/2$
Mean = $1/n [{(n + 1) n (2n + 1)}/6 + {n(n + 1)}/2]$
= ${(n + 1)}/2 [{2n + 1}/3 + 1]$
= ${(n + 1)}/2 {(2n + 4)}/3$
= ${(n + 1)(n + 2)}/3$
So, option (d) is correct.
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