Quantitative Aptitude

Statistics

606 Questions
All Ungrouped data dispersion and consistency 209 Unknown terms from given average 124 Grouped data central tendencies 101 Merging means and variances of groups 53 Replacement change in average 27 Alligation and weighted average 21 Data interpretation from tables and charts 20 Net successive percentage change 19 Grouped data variance via step deviation 18 Combination of multiple ratios 1

Statistics Questions

Multiple choice

The mean and standard deviation of $1,2,3,4,5,6$ are
  1. $\dfrac{7}{2},\sqrt{\dfrac{35}{12}}$
  2. $\dfrac{7}{3} ,\sqrt{3}$
  3. $3,3$
  4. $3,35/12$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mean = (1+2+3+4+5+6)/6 = 21/6 = 3.5 or 7/2. Variance = sum(x-mean)^2 / n = (6.25 + 2.25 + 0.25 + 0.25 + 2.25 + 6.25) / 6 = 17.5 / 6 = 35/12. Standard deviation is sqrt(35/12).

Multiple choice

In a series of $100$ observations, half of them is equal to $p$ and remaining half equal to $-p$ if the standard deviation of these observations is $10$ then $|p|=$__
  1. $10$
  2. $\dfrac {1}{10}$
  3. $\dfrac {1}{\sqrt {10}}$
  4. $\sqrt {10}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a set of 100 observations where 50 are p and 50 are -p, the mean is 0. The variance is the average of the squared deviations from the mean: (50 * p^2 + 50 * (-p)^2) / 100 = (50p^2 + 50p^2) / 100 = 100p^2 / 100 = p^2. Since standard deviation is 10, variance is 100, so p^2 = 100 and |p| = 10.

Multiple choice

For $(2n+1)$ observation ${ x }{ 1 },-{ x }{ 1 },{ x }{ 2 },-{ x }{ 2 },.....{ x }{ n },-{ x }{ n }$ and $0$, where ${ { x }_{ i } }^{ \displaystyle 8 }$ are distinct, non-zero numbers. Let S.D. and M.D. denote the standard deviation and median respectively. then, which one of the following is always true?

  1. S.D.<M.D.

  2. S.D.>M.D.
  3. S.D.$=$M.D.

  4. Nothing can be said in general about the relationship of S.D. and M.D.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a symmetric distribution around 0, the median is 0. The standard deviation is non-zero because the values are distinct and non-zero. Thus, S.D. > M.D.

Multiple choice

Let ${ x }{ 1 },{ x }{ 2 },.....{ x }_{ n }$ be $n$ observations and $\overline { X } $ be their arithmetic mean. The formula for the standard deviation is given by
  1. $\sum _{ i=1 }^{ n }{ { { (x }_{ i }-\overline { X } ) }^{ 2 } }$
  2. $\cfrac { 1 }{ n } \sum _{ i=1 }^{ n }{ { { (x }_{ i }-\overline { X } ) }^{ 2 } }$
  3. $\sqrt { \cfrac { 1 }{ n } \sum _{ i=1 }^{ n }{ { { (x }_{ i }-\overline { X } ) }^{ 2 } } }$
  4. $\sqrt { \cfrac { 1 }{ n } \sum _{ i=1 }^{ n }{ { { x }_{ i } }^{ 2 }+{ \overline { X } }^{ 2 } } } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The standard deviation is defined as the square root of the variance. The variance is the average of the squared differences from the mean, calculated as (1/n) * sum((xi - mean)^2).

Multiple choice

A data consists of $n$ observation: $x_1,x_2,........,x_n.$ If $\displaystyle \sum_{i=1}^{n}(x_i+1)^2=9n$ and $\displaystyle \sum_{i=1}^n(x_i-1)^2=5n$, then the standard deviation of this data is:
  1. $5$
  2. $\sqrt{5}$
  3. $\sqrt{7}$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Variance = (1/n) * sum(x_i^2) - mean^2. Given sum(x_i+1)^2 = 9n and sum(x_i-1)^2 = 5n. Expanding: sum(x_i^2 + 2x_i + 1) = 9n and sum(x_i^2 - 2x_i + 1) = 5n. Adding them: 2*sum(x_i^2) + 2n = 14n => sum(x_i^2) = 6n. Subtracting them: 4*sum(x_i) = 4n => mean = 1. Variance = (6n/n) - 1^2 = 5. Standard deviation = sqrt(5).

Multiple choice

Mean deviation for $n$ observations ${ x }{ 1 },{ x }{ 2 },.....{ x }_{ n }$ from their mean $\bar { X } $ is given by:
  1. $\sum _{ i=1 }^{ n }{ ( { x }_{ i }-\bar { X }) }$
  2. $\cfrac { 1 }{ n } \sum _{ i=1 }^{ n }{ \left| { x }_{ i }-\bar { X } \right| } $
  3. $\sum _{ i=1 }^{ n }{ { { (x }_{ i }-\bar { X } ) }^{ 2 } }$
  4. $\cfrac { 1 }{ n } \sum _{ i=1 }^{ n }{ { { (x }_{ i }-\bar { X } ) }^{ 2 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Mean deviation from the mean is defined as the average of the absolute differences between each observation and the mean. This is represented by (1/n) * sum(|xi - mean|).

Multiple choice

The marks of a student in $6$ tests are $41, 45, 54, 57, 43$ and x. If the mean marks of these tests is $48$, then standard deviation of these tests is?
  1. $\dfrac{10}{\sqrt{3}}$
  2. $\dfrac{10}{\sqrt{2}}$
  3. $\dfrac{10}{3}$
  4. $\dfrac{20}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mean = (41+45+54+57+43+x)/6 = 48. 240+x = 288, so x = 48. Data: 41, 45, 54, 57, 43, 48. Deviations from 48: -7, -3, 6, 9, -5, 0. Squares: 49, 9, 36, 81, 25, 0. Sum = 200. Variance = 200/6 = 100/3. Standard deviation = sqrt(100/3) = 10/sqrt(3).

Multiple choice

A student scores the following marks in five test: $45, 54, 41, 57, 43$. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six test is
  1. $\dfrac{10}{\sqrt{3}}$
  2. $\dfrac{100}{\sqrt{3}}$
  3. $\dfrac{130}{3}$
  4. $\dfrac{10}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Sum of 5 = 240. Mean of 6 = 48, so sum of 6 = 288. The 6th mark is 288 - 240 = 48. The marks are 45, 54, 41, 57, 43, 48. Mean = 48. Variance = [(45-48)^2 + (54-48)^2 + (41-48)^2 + (57-48)^2 + (43-48)^2 + (48-48)^2] / 6 = [9 + 36 + 49 + 81 + 25 + 0] / 6 = 200 / 6 = 100 / 3. Standard deviation = sqrt(100/3) = 10/sqrt(3).