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) 73.5
(b) 78.7
(c) 58.8
(d) 80.6
The correct answers to the above question in:
Answer: (a)
According to the question,
250 = x% of 340 ⇒ x = ${250 × 100}/{340}$ = 73.5%
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Read more model question set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
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 »Answer: (c)
Question : 2
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 »Answer: (d)
Area of the polygon gives sum of $f_i x_i$ not sum of frequency distribution $(f_i)$.
Question : 3
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 »Answer: (b)
Question : 4
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 : 5
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 : 6
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.
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