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 general knowledge

The exam scores of all 500 students were recorded and it was determined that these scores were normally distributed. If Jane's score is 0.8 standard deviation above the mean, then how many, to the nearest unit, students scored above Jane?

  1. 394

  2. 250

  3. 400

  4. 106
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a normal distribution, 0.8 standard deviations above the mean corresponds to approximately 78.81% of the population below that point. This means about 21.19% scored above Jane. For 500 students: 500 × 0.2119 ≈ 106 students scored higher. The other options (394, 250, 400) don't match this calculation.

Multiple choice general knowledge science & technology

The mean of a data set is equal to 10 and its standard deviation is equal to 1. If we add 5 to each data value, then the mean and standard deviation become

  1. mean = 15 , standard deviation = 6

  2. mean = 10 , standard deviation = 6

  3. mean = 15 , standard deviation = 1

  4. mean = 10 , standard deviation = 1
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Adding a constant to every data value increases the mean by that constant but does not change the spread. New mean = 10 + 5 = 15. Standard deviation remains 1 because adding a constant shifts all values equally without changing their dispersion.

Multiple choice general knowledge

When the number of observations is odd, the median is

  1. The value of ((n+1)/2) th observation

  2. The value of (n/2) th observation

  3. The mean of (n/2) and ((n/2) + 1) th observations

  4. The value of (n+2) th observation
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For an odd number of observations (n), the median is the middle value when arranged in order, which is at position (n+1)/2. For example, with 7 observations, the median is the 4th value since (7+1)/2 = 4. Option C describes the formula for even-numbered observations.

Multiple choice

Which of the following assertions are CORRECT? P: Adding 7 to each entry in a list adds 7 to the mean of the list. Q: Adding 7 to each entry in a list adds 7 to the standard deviation of the list. R: Doubling each entry in a list doubles the mean of the list. S: Doubling each entry in a list leaves the standard deviation of the list unchanged.

  1. P, Q

  2. Q, R

  3. P, R

  4. R, S
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

P and R always hold true. Otherwise, consider a sample set {1, 2, 3, 4} and check accordingly.

Multiple choice statistics normal distribution distribution of measurement probability distributions introduction to normal distribution

A large group of students took a test in Physics and the final grades have a mean of $70$ and a standard deviation of $10$. If we can approximate the distribution of these grades by a normal distribution, what percent of the students should fail the test (grades$<60$)?

  1. $15.21$
  2. $23.21$
  3. $15.87$%
  4. $16.23$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Final grades follows normal distribution.
Given mean i.e. $ \mu $ $= 70$
Given standard deviation i.e. $ \sigma $ $= 10$

The normal random variable of a standard normal distribution is called a standard score or a z-score. Every normal random variable X can be transformed into a z score via the following equation:

$z = (X - μ) / σ$

where $X$ is a normal random variable, $μ$ is the mean of $X$, and $σ$ is the standard deviation of $X$.


For $X =60$
$Z= (60-70)/10 = -1$

$P(X < 60) = P(Z < -1)$ 
                $= 0.1587$
Hence percent of students failed in test is $15.87\%$

Multiple choice statistics normal distribution distribution of measurement probability distributions introduction to normal distribution

The marks secured by $400$ students in a Mathematics test were normally distributed with mean $65$. If $120$ students got marks above $85$, the number of students securing marks between $45$ and $65$ is

  1. $120$
  2. $20$
  3. $80$
  4. $160$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $X$ denote the marks secured.

Given, $\mu =65$
Thus, $X\sim N(65,\rho)$
$\Rightarrow z=\dfrac {X-\mu}{\rho}=\dfrac {X-65}{\rho}$
$\Rightarrow P(X>85)=\dfrac {120}{400}$
$\Rightarrow P\left (z>\dfrac {85-65}{\rho}\right)=\dfrac {3}{10}$
$\Rightarrow P\left (z>\dfrac {20}{\rho}\right)=\dfrac {3}{10}$ ....(1)
$\Rightarrow P(45<x<65)$ $=P\left (\dfrac {45-65}{\rho}<z<\dfrac {65-65}{\rho}\right)$
$=P\left (\dfrac {-20}{\rho}<z<0\right)$
$=P\left (0<z<\dfrac {20}{\rho}\right)$
$=0.5-P\left (z>\dfrac {20}{\rho}\right)$
$=\dfrac {1}{2}-\dfrac {3}{10}$
$=\dfrac {1}{5}$
Number of students secured marks between $45$ and $65$ $=\dfrac {1}{5}\times 400=80$.
Hence, the correct answer is option .

Multiple choice data handling analysis analysis of frequency distributions frequency table evs analysing data collection of data and presentation of data

The frequency distribution of marks obtained by $60$ students of a class is given.

X $30-34$ $40-44$ $45-49$ $50-54$ $55-59$ $60-64$
f $3$ $5$ $12$ $18$ $14$ $62$


Find mode of the distribution.

  1. $42.50$
  2. $52.50$
  3. $52.05$
  4. $42.05$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given class intervals should be converted into class boundaries. Since the distribution is regular, the modal class, by inspection, is $49.5-54.5$
Further, $L _m=49.5, f _m=18, f _1=12, f _2=12, h=5$
Mode, $Z=49.5+\displaystyle\frac{18-12}{(2\times 18)-12-14}\times 5=52.5$

Multiple choice maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

The mean weight of 9 students is 24 kg. If one more student is joined in the group the mean is unaltered, then the weight of the 10th student is

  1. 25 kg

  2. 24 kg

  3. 26 kg

  4. 23 kg
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The sum of weights of the 9 students = $25\times 9=225$ kg
If one more student is joined in the group, then total number of students is 10, and the mean isn 25.
The sum of weights of the 10 students is $25\times 10=250$ kg
The weight of the tenth student is $250-225=25$ kg
Multiple choice maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

Find the arithmetic mean for the following grouped frequency distribution:

Class-intervals$ $6-10$ $10-14$ $14-18$ $18-22$  $22-26$  $26-30$
Frequency  $ 4$   $ 6$   $9$   $ 12$    $ 7 $    $2$


  1. $15.8$
  2. $16.8$
  3. $17.8$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Consider the following table, to calculate mean:
|  $ci$ |  $f _i$ | $x _i$  |  $f _ix _i$ | | --- | --- | --- | --- | |  $6-10$ |  $4$ | $ 8$ | $ 32$ | | $ 10-14$ |  $6$ |  $12$ |  $72$ | |  $14-18$ |  $9$ |  $16$ | $ 144$ | |  $18-22$ | $12$ | $ 20$ |  $240$ | | $ 22-26$ |  $7$ | $ 24$ |  $168$ | |  $26-30$ | $ 2$ |  $28$ |  $56$ |
$N=\Sigma f _i=40$          
$\Sigma f _ix _i=712$

Mean $\overline x=\dfrac {\Sigma f _ix _i}{N}$
$\therefore \overline x=\dfrac{712}{40}=17.8$
Hence, option $C$ is correct.
Multiple choice maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

The frequency distribution of marks in English are given in the table:

Marks 50-60 60-70 70-80 80-90
Number of students 12 24 14 10

Find the mean by step deviation method.

  1. $58$
  2. $48$
  3. $69$
  4. $71$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

 X Mid point(x)  Frequency (F)  i=class interval width  A=65 Assumed mean  d'=$\dfrac{x-A}{i} $  fd'
 50-60 55  12  10  65  -1  -12 
 60-70 65=A  24  10  65  0  0 
70 -80  75  14 10  65  1  14 
80-90  85  10  10  65  2  20 
    $\Sigma f=60$        $Sigma fd'=22$ 


The formula used for arithmetic mean of grouped data by step deviation method is, $\overline {X} =A + \dfrac{\sum fd'}{\sum f} \times i$ 
$A =$ Assumed mean of the given data
$\sum$ = Summation of the frequencies given in the grouped data
$\sum fd'$ = Summation of the frequencies and deviation of a given mean data
$d' = \dfrac{(x - A)}{i}$
$i =$ Class interval width
$\overline {X}$ = arithmetic mean
$\overline {X} =65 +\dfrac{22}{60}\times 10$
$= 65 + 3.666$
$= 68.666$ $\approx$ $69$ marks