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 maths average arithmetic mean of ap introduction to averages means

The A.M. of 'n' observations is M. If the sum of $(n - 4)$ observation is 'a', what is the mean of remaining $4$ observations?

  1. $nM + a$
  2. $\dfrac {nM - a}{2}$
  3. $\dfrac {nM + a}{2}$
  4. $\dfrac {nM - a}{4}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given $\cfrac { { a } _{ 1 }+{ a } _{ 2 }+{ a } _{ 3 }+{ a } _{ 4 }+....+{ a } _{ n } }{ n } =M\quad \quad \quad (1)\\ \left( { a } _{ 1 }+{ a } _{ 2 }+{ a } _{ 3 }+{ a } _{ 4 }+....+{ a } _{ n-4 } \right) =a\quad \quad \quad (2)$

From $(1)$ and $(2)$
$a+{ a } _{ n-3 }+{ a } _{ n-2 }+{ a } _{ n-1 }+{ a } _{ n }=nM\\ { a } _{ n-3 }+{ a } _{ n-2 }+{ a } _{ n-1 }+{ a } _{ n }=(nM-a)\\ Mean=\cfrac { { a } _{ n-3 }+{ a } _{ n-2 }+{ a } _{ n-1 }+{ a } _{ n } }{ 4 } =\cfrac { nM-a }{ 4 } \\ Mean=\cfrac { (nM-a) }{ 4 } $
Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

The mean of a data set consisting of $20$ observations is $40$. If one observation $53$ was wrongly recorded as $33$, then the correct mean will be:

  1. $41$
  2. $49$
  3. $40.5$
  4. $42.5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Mean=\dfrac{S}{n}$


where S=sum of all observations
            n=number of observations

hence, $40=\dfrac{S}{20}\Rightarrow S=800$

but Since, 53 is recorded as 33 so we need to add $(53-33=20)$ to get the correct mean which is 
$Mean _{correct}=\dfrac{800+20}{20}=41$

Multiple choice

If the expected value of a random variable $X$ is 5 and its variance is 4, what is the standard deviation of $X$?

  1. 2

  2. 4

  3. 6

  4. 8
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The standard deviation of $X$ is the square root of the variance of $X$, which is 4. Therefore, the standard deviation of $X$ is 2.

Multiple choice

The arithmetic mean for the following frequency distribution is: | | | | | || |---|---|---|---|---|---| | Class | 0 - 4 | 4 - 8 | 8 - 12 | 12 - 16 | 16 - 20 | | Frequency | 6 | 11 | 20 | 7 | 4 |

  1. 8

  2. 9.2

  3. 9.33

  4. 9.66
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The mean is calculated by finding the sum of (midpoint * frequency) divided by the total frequency. The midpoints are 2, 6, 10, 14, 18, so the sum is (2*6 + 6*11 + 10*20 + 14*7 + 18*4) = (12 + 66 + 200 + 98 + 72) = 448. Dividing 448 by the total frequency of 48 gives 9.33.

Multiple choice

A test was conducted for the students of Class X in a school for 14 students which resulted in the following frequency table for the marks obtained by the 14 students: | Marks Obtained | 0-10 | 10-20 | 20-30 | | --- | --- | --- | --- | | Number of students | 4 | 8 | 2 | Find the mode of the given data.

  1. 10

  2. 14

  3. 18

  4. 24
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The modal class is 10-20 (highest frequency 8). Mode = L + ((f1 - f0) / (2*f1 - f0 - f2)) * h. L=10, f1=8, f0=4, f2=2, h=10. Mode = 10 + ((8-4) / (16-4-2)) * 10 = 10 + (4/10) * 10 = 14.

Multiple choice

The median of a grouped data is 19. The lower limit of the median class is 16. The frequency of the median class and the cumulative frequency of its preceding class are 4 and 20 respectively. If the number of observations is 48, then the class size of the median class will be _____.

  1. 2

  2. 3

  3. 4

  4. 5
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Median formula: M = L + ((N/2 - cf) / f) * h. Given M=19, L=16, N=48 (so N/2=24), cf=20, f=4. 19 = 16 + ((24 - 20) / 4) * h. 3 = (4 / 4) * h. h = 3.