Mathematics · Quantitative Aptitude

Statistics and Dispersion

559 Questions

Statistics and dispersion involve the calculation of mean, standard deviation, variance, and coefficient of variation for data sets. These questions also cover probability distributions and cumulative frequency analysis. Such quantitative aptitude topics are heavily featured in banking and SSC examinations.

Standard deviationNormal distributionMean calculationCumulative frequencyCoefficient of variation

Statistics and Dispersion Questions

Multiple choice statistics time series moving average and variation simple moving average

For the variables $x$ and $y$, the regression equations are given as $7x-3y-18=0$ and $4x-y-11=0$. Identify the regression equation of $y$ on $x$.

  1. $4x-y-11=0$
  2. $7x-3y-18=0$
  3. $4x-y-2=0$
  4. $7x-3y-4=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let us assume that $7x-3y-18=0$ is the regression equation of $y$ on $x$.


Consider $7x-3y-18=0$

$\Rightarrow y=-6+\dfrac{7}{3}x$ 

$\therefore b_{yx}=\dfrac{7}{3}$

Now consider $4x-y-11=0$

$\Rightarrow x=\dfrac{11}{4}+\dfrac{1}{4}y$

$\therefore b_{xy}=\dfrac{1}{4}$

Now taking the product, $b_{yx} \times b_{xy}=\dfrac{7}{3} \times \dfrac{1}{4}=\dfrac{7}{12}<1$

Since the product is less than one, our assumptions are correct.

Thus $7x-3y-18=0$ is the regression equation of $y$ on $x$.

Multiple choice statistics time series moving average and variation simple moving average

The two lines of regression are $x+2y-5=0$ and $x+3y-8=0$. The coefficient of correlation between $x$ and $y$ is 

  1. $-0.72$
  2. $0.72$
  3. $-0.82$
  4. $0.82$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given two lines $x+2y-5=0, x+3y-8=0$.

Consider $x+2y-5=0$
$\Rightarrow x=-2y+5$
$\Rightarrow r_1=-2$
Consider $x+3y-8=0$
$\Rightarrow y=-\dfrac{1}{3}x+\dfrac{8}{3}$
$\Rightarrow r_2=-\dfrac{1}{3}$
We know that $r^2=r_1 \times r_2$
$\Rightarrow r^2=-2 \times -\dfrac{1}{3}$
$\Rightarrow r^2=\dfrac{2}{3}$
$\Rightarrow r=\pm \sqrt{\dfrac{2}{3}}$
We know that, If both regression coefficients are negative, $r$ would be negative.
$\Rightarrow r=-\sqrt{\dfrac{2}{3}}=-0.82$

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

Which of the following is not true regarding the normal distribution?

  1. the point of inflecting are at $X = \mu \pm \sigma$

  2. skewness is zero
  3. maximum heigth of the curve is $\dfrac{1}{\sqrt{2\pi}}$
  4. mean $=$ media $=$ mode
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\left(i\right)$Since $f\left(x\right)$ is a nonzero function we may divide both sides of the equation by this function. From this 
it is  easy to see that the inflection points occur where $X =\mu\pm\sigma$. In other words the inflection points are 
located one standard deviation above the mean and one standard deviation below the mean
$\left(ii\right)$The skewness for perfect normal distribution is $0.0$. But if sample is greater than $100$ and less 
than $200$, the acceptable absolute skewness value is $1.0$. However for large sample size $n$ greater than $200$, 
the absolute value for acceptable skewness is $1.5$.
$\left(iii\right)$The area under the normal curve is equal to $1.0$. Normal distributions are denser in the center and
 less dense in the tails. Normal distributions are defined by two parameters, the mean $\left(\mu\right)$ and the standard
deviation $\left(\sigma\right)$. $68\%$ of the area  of a normal distribution is within one standard deviation of the mean.
$\left(iv\right)$The mean, median, and mode of a normal distribution are equal. The area under the normal curve is equal to 1.0. 
Normal distributions are denser in the center and less dense in the tails. Normal distributions are defined by two 
parameters, the mean $\left(\mu\right)$ and the standard deviation $\left(\sigma\right)$.
Multiple choice statistics normal distribution distribution of measurement probability distributions introduction to normal distribution

The random variable $x$ follows normal distribution $f(x) = Ce^{\dfrac{-\dfrac{1}{2} (x - 100)^2}{25}}$. Then the value of $C$ is

  1. $\sqrt{2\pi}$
  2. $\dfrac{1}{\sqrt{2 \pi}}$
  3. $5\sqrt{2 \pi}$
  4. $\dfrac{1}{5 \sqrt{2 \pi}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since the above variable $x$ follows normal distribution, we can therefore write it as 


$Ce^{-\dfrac{(x-100)^{2}}{50}}=\dfrac{1}{\sqrt{2\sigma^{2}\pi}}e^{-\dfrac{(x-\mu)^{2}}{2\sigma^{2}}}$

Comparing LHS with RHS gives 

$2\sigma^{2}=50$...(i)

And $C=\dfrac{1}{\sqrt{2\sigma^{2}\pi}}$...(ii)

Now from equation we know that $2\sigma^{2}=50$. Substituting in eq (i) gives

$C=\dfrac{1}{\sqrt{2\sigma^{2}\pi}}=\dfrac{1}{\sqrt{50\pi}}$

$=\dfrac{1}{5\sqrt{2\pi}}$

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

Which of the following are correct regarding normal distribution curve?
(i) Symmetrical about the line $X=\mu $ (Mean)
(ii) Mean $=$ Median $=$ Mode
(iii) Unimodal
(iv) Points of inflexion are at $X=\mu \pm \sigma $

  1. (i), (ii)

  2. (ii), (iv)

  3. (i), (ii), (iii)

  4. All of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a normal distribution , mean, median and mode are equal. The curve is bell type curve which is symmetric about 

$x=$ mean (mode or median).
It is unimodal as it has its first derivative $0$ at $x=\mu$ (mean) and the derivative is less than $0$, for $x>\mu$ and greater than $0$ for $x<\mu$
It has its point of inflection (second derivative $0$) at $x=\mu+\sigma$ and $x=\mu-\sigma$ where $\sigma$ is the standard deviation.

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

$X$ is a Normally distributed variable with mean $ = 30$ and standard deviation $ = 4$. Find $P(30 < x<35)$

  1. $0.3698$
  2. $0.3956$
  3. $0.2134$
  4. $0.3944$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Here we need to determine the value of $ Z=\dfrac{x-\mu}{\sigma} $
Here, $\mu=\text{Mean}=30$ and $\sigma=\text{Stand Deviation}=4$
For, $x=30\Rightarrow Z=\dfrac{30-30}{4}=0$
For, $x=35\Rightarrow Z=\dfrac{35-30}{4}=1.25$
$\Rightarrow P(30<x<35)=P(0<Z<1.25) $
$\Rightarrow P(0<Z<1.25)=P(Z<1.25)-P(Z<0) $                         $(\because P(a<Z<b)=P(Z<b)-P(Z<a))$
From the Normal Distribution table, $P(Z<1.25)=0.8944$ and $P(Z<0)=0.5$
$\Rightarrow P(30<x<35)=0.8944-0.5=0.3944$

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 length of life of an instrument produced by a machine has a normal distribution with a mean of $12$ months and standard deviation of $2$ months. Find the probability that an instrument produced by this machine will last less than $7$ months. 

  1. $0.2316$
  2. $0.0062$
  3. $0.0072$
  4. $0.2136$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Life of intsrument follows normal distribution.
Given mean i.e. $ \mu $ $= 12 $months
Given standard deviation i.e. $ \sigma $ $= 2$ months

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 $X =22$
$Z= (7-12)/2 = -2.5$

$P( X < 7) = P(Z < -2.5)$
              $ = 0.0062$
Multiple choice data handling analysis analysis of frequency distributions frequency table evs analysing data collection of data and presentation of data

Consider the following frequency distribution:

Class $0-10$ $0-20$ $0-30$ $0-40$ $0-50$
Frequency $3$ $8$ $14$ $20$ $25$

What is the above frequency distribution known as?

  1. Cumulative distribution in more than type

  2. Cumulative distribution in less than type

  3. Continuous frequency distribution

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The given table lists the upper limits of the classes alongside the accumulated frequencies up to that point, which defines a less than type cumulative frequency distribution. More than type distributions use lower limits with frequencies accumulated downward.

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

Find a 9596 confidence interval for the population mean from the following data:

Sample size n = 65, Mean = 6300, Standard deviation = 9.5, N=1000.

  1. (6298 - 6302)

  2. (6301 - 6308)

  3. (6290 - 4310)

  4. (6288 - 6315)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The confidence interval is calculated as mean +/- Z * (standard error). For a 95% confidence interval, Z is approximately 1.96. Standard error = 9.5 / sqrt(65) is approx 1.18. 1.96 * 1.18 is approx 2.3. 6300 +/- 2.3 gives 6297.7 to 6302.3, which rounds to 6298-6302.

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

In a truly normal frequency distribution _________.

  1. the mean always is the same as the standard deviation

  2. the mean is never the same as the mode

  3. the mode is never the same as the median

  4. the mean always is the same as the median
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a perfectly normal distribution, the curve is symmetrical, meaning the mean, median, and mode are all located at the same central point. Options A, B, and C are incorrect because they contradict the fundamental symmetry of the normal distribution.