Statistics Practice Questions | Mean Correlation Regression

Module III: Statistical Techniques – I
Practice Questions with Solutions

Question 1 – Measures of Central Tendency

The marks of 10 students are:
45, 56, 62, 48, 70, 55, 60, 52, 68, 58
Find the Mean, Median and Mode.

Solution

Mean

\bar{x} = \frac{45+56+62+48+70+55+60+52+68+58}{10} = \frac{574}{10} = 57.4

Median
Arranged data: 45, 48, 52, 55, 56, 58, 60, 62, 68, 70

\text{Median} = \frac{5^{\text{th}} + 6^{\text{th}}}{2} = \frac{56+58}{2} = 57

Mode
No value repeats → Mode does not exist (or is multimodal / undefined for this set).

Question 2 – Moments, Skewness & Kurtosis

For a distribution, the first four central moments are
$\mu_2 = 16$ , $\mu_3 = -64$ , . Find the coefficients of skewness () and kurtosis (). Comment on the nature of the distribution.

Solution

\beta_1 = \frac{\mu_3^2}{\mu_2^3} = \frac{(-64)^2}{16^3} = \frac{4096}{4096} = 1

\gamma_1 = \sqrt{\beta_1} = 1 \quad\text{(positively skewed)}

\beta_2 = \frac{\mu_4}{\mu_2^2} = \frac{768}{256} = 3

\gamma_2 = \beta_2 - 3 = 0 \quad\text{(mesokurtic)}

The distribution is moderately positively skewed and mesokurtic (normal-like peakedness).

Question 3 – Moment Generating Function

Find the MGF of a random variable $X$ that takes values 0, 1, 2 with probabilities $\frac12$ , $\frac13$ , respectively. Hence find and .

Solution

M_X(t) = E(e^{tX}) = \frac12 e^{0} + \frac13 e^{t} + \frac16 e^{2t} = \frac12 + \frac13 e^{t} + \frac16 e^{2t}

E(X) = M'(0) = \frac13 + \frac26 = \frac13 + \frac13 = \frac23

E(X^2) = M''(0) = \frac13 + \frac46 = \frac13 + \frac23 = 1

\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = 1 - \left(\frac23\right)^2 = 1 - \frac49 = \frac59

Question 4 – Fitting of Straight Line

Fit a straight line $y = a + bx$ to the following data by the method of least squares:

$x$	1	2	3	4	5
$y$	2	5	3	8	7

Solution

\begin{array}{cccc} x & y & x y & x^{2} \\ 1 & 2 & 2 & 1 \\ 2 & 5 & 10 & 4 \\ 3 & 3 & 9 & 9 \\ 4 & 8 & 32 & 16 \\ 5 & 7 & 35 & 25 \\ \sum x = 15 & \sum \end{array}

Normal equations:

\sum y = na + b\sum x \implies 25 = 5a + 15b

\sum xy = a\sum x + b\sum x^2 \implies 88 = 15a + 55b

Solving:

b = \frac{5\cdot88 - 15\cdot25}{5\cdot55 - 15^2} = \frac{440-375}{275-225} = \frac{65}{50} = 1.3

a = \frac{25 - 15\cdot1.3}{5} = \frac{25-19.5}{5} = 1.1

Fitted line:

y = 1.1 + 1.3x

Question 5 – Fitting of Second Degree Parabola

Fit a parabola $y = a + bx + cx^2$ to the data:

$x$	0	1	2	3	4
$y$	1	4	10	17	27

Solution

Compute the required sums (with $n=5$ ):

\sum x=10,\ \sum y=59,\ \sum x^2=30,\ \sum x^3=100,\ \sum x^4=354, \sum xy=170,\ \sum x^2y=626

Normal equations:

\begin{aligned} 59 &= 5a + 10b + 30c \\ 170 &= 10a + 30b + 100c \\ 626 &= 30a + 100b + 354c \end{aligned}

Solving the system yields:

a = 1.0,\quad b = 2.1,\quad c = 1.1

(approximately; exact fractions may be used)

Fitted parabola:

y = 1 + 2.1x + 1.1x^2

Question 6 – Correlation Coefficient

Calculate Pearson’s correlation coefficient for the following data:

$x$	10	20	30	40	50
$y$	12	24	36	40	50

Solution

\begin{align*} \sum x &= 150,\quad \sum y = 162,\quad n=5\\ \sum x^2 &= 5500,\quad \sum y^2 = 6296,\quad \sum xy = 5800 \end{align*}

r = \frac{n\sum xy - (\sum x)(\sum y)}{\sqrt{[n\sum x^2-(\sum x)^2][n\sum y^2-(\sum y)^2]}} = \frac{5\cdot5800 - 150\cdot162}{\sqrt{[5\cdot5500-22500][5\cdot6296-26244]}} = \frac{29000-24300}{\sqrt{5000\cdot5226}} = \frac{4700}{\sqrt{26130000}} \approx 0.92

Strong positive correlation.

Question 7 – Rank Correlation

The ranks of 8 students in Mathematics and Physics are:

Maths	1	2	3	4	5	6	7	8
Physics	2	4	1	5	3	8	7	6

Find Spearman’s rank correlation coefficient.

Solution

Differences $d$ : $-1,-2,2,-1,2,-2,0,2$

\sum d^2 = 1+4+4+1+4+4+0+4 = 22

\rho = 1 - \frac{6\sum d^2}{n(n^2-1)} = 1 - \frac{6\cdot22}{8(64-1)} = 1 - \frac{132}{504} = 1 - 0.262 = 0.738

Question 8 – Regression Lines

For two variables $x$ and $y$ , the following are given:
$\bar{x}=40$ , $\bar{y}=50$ , , , .

Find the two regression lines and estimate $y$ when $x=45$ .

Solution

Regression coefficient of $y$ on $x$ :

b_{yx} = r\frac{\sigma_y}{\sigma_x} = 0.8\cdot\frac{15}{10} = 1.2

Regression coefficient of $x$ on $y$ :

b_{xy} = r\frac{\sigma_x}{\sigma_y} = 0.8\cdot\frac{10}{15} = \frac{8}{15}\approx 0.533

Regression line of $y$ on $x$

y - 50 = 1.2(x - 40) \implies y = 1.2x + 2

Regression line of $x$ on $y$

x - 40 = 0.533(y - 50)

When $x=45$ :

y = 1.2\cdot45 + 2 = 54 + 2 = 56

Practice Tips

Always arrange data in ascending order before finding median.
For least squares, form the normal equations carefully and solve the linear system.
Check that $|r| \le 1$ and $b_{yx}b_{xy}=r^2$ .