Statistics IIT JEE Previous Year Questions ST1.7

Описание: \subsection{Problem 1 [2013]}
All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks were given?

(1) mean \quad (2) median \quad (3) mode \quad (4) variance

\textbf{Solution:}

Variance is independent of origin.

\textbf{Answer: (4) variance}

\subsection{Problem 2 [2014]}
The variance of first 50 even natural numbers is

(1) $\frac{833}{4}$ \quad (2) $833$ \quad (3) $437$ \quad (4) $\frac{437}{4}$

\textbf{Solution:}

$$\sigma^2 = \left(\frac{\sum x_i^2}{n}\right) - \overline{x}^2$$

$$\sum_{i=1}^{50} 2r = 50 \times 51$$

$$\overline{x} = \frac{50 \times 51}{50} = 51$$

$$\sum_{i=1}^{50} 4r^2 = \frac{4 \times 50 \times 51 \times 101}{6}$$

$$\sigma^2 = \frac{4 \times 50 \times 51 \times 101}{6 \times 50} - (51)^2 = 833$$

\textbf{Answer: (2) $833$}

\subsection{Problem 3 [2016]}
If the standard deviation of the numbers $2, 3, a$ and $11$ is $3.5$, then which of the following is true?

(1) $3a^2 - 32a + 84 = 0$ \quad (2) $3a^2 - 34a + 91 = 0$

(3) $3a^2 - 23a + 44 = 0$ \quad (4) $3a^2 - 26a + 55 = 0$

\textbf{Solution:}

Standard deviation of numbers $2, 3, a$ and $11$ is $3.5$

$$\therefore (3.5)^2 = \frac{\sum x_i^2}{4} - (\overline{x})^2$$

$$\Rightarrow (3.4)^2 = \frac{4 + 9 + a^2 + 121}{4} - \left(\frac{2+3+a+11}{4}\right)^2$$

On solving, we get
$$3a^2 - 32a + 84 = 0$$

\textbf{Answer: (1) $3a^2 - 32a + 84 = 0$}

\subsection{Problem 4 [2018]}
If $\sum_{i=1}^{9}(x_i - 5) = 9$ and $\sum_{i=1}^{9}(x_i - 5)^2 = 45$, then the standard deviation of the 9 items $x_1, x_2, ..., x_9$ is

(1) $3$ \quad (2) $9$ \quad (3) $4$ \quad (4) $2$

\textbf{Solution:}

Let $x_i - 5 = y_i$

$$\therefore \sum_{i=1}^{9} y_i = 9 \text{ and } \sum_{i=1}^{9} y_i^2 = 45$$

So, required standard deviation is

$$\sigma = \sqrt{\frac{\sum_{i=1}^{9} y_i^2}{9} - \left(\frac{\sum_{i=1}^{9} y_i}{9}\right)^2} = \sqrt{\frac{45}{9} - \left(\frac{9}{9}\right)^2} = 2$$

\textbf{Answer: (4) $2$}

\subsection{Problem 5 [2019]}
Five students of a class have an average height 150 cm and variance 18 cm. A new student, whose height is 156 cm, joined them. The variance (in cm$^2$) of the height of these six students is

(1) $22$ \quad (2) $20$ \quad (3) $16$ \quad (4) $18$

\textbf{Solution:}

Given $\overline{x} = \frac{\sum_{i=1}^{5} x_i}{5} = 150$

$$\Rightarrow \sum_{i=1}^{5} x_i = 750 \quad \ldots(1)$$

Now, $\frac{\sum x_i^2}{5} - (\overline{x})^2 = 18$

$$\sum_{i=1}^{5} x_i^2 = 5 \times 18 + (150)^2 = 112590 \quad \ldots(2)$$

Given that height of new student is $x_6 = 156$.

$$\therefore \overline{x}_1 = \frac{\sum_{i=1}^{6} x_i}{6} = \frac{750 + 156}{6} = 151$$

Also, new variance $= \frac{\sum_{i=1}^{6} x_i^2}{6} - (\overline{x}_1)^2$

$$= \frac{112590 + (156)^2}{6} - (151)^2$$

$$= 22821 - 22801 = 20$$

\textbf{Answer: (2) $20$}



\subsection{Problem 6 [2019]}
A data consists of $n$ observations $x_1, x_2, \ldots, x_n$. If
$$\sum_{i=1}^{n}(x_i + 1)^2 = 9n \quad \text{and} \quad \sum_{i=1}^{n}(x_i - 1)^2 = 5n$$
then the standard deviation of this data is

(1) $5$ \quad (2) $\sqrt{5}$ \quad (3) $\sqrt{7}$ \quad (4) $2$

\textbf{Solution:}

Given:
$$\sum_{i=1}^{n}(x_i + 1)^2 = 9n \quad \ldots (1)$$
$$\sum_{i=1}^{n}(x_i - 1)^2 = 5n \quad \ldots (2)$$

Adding (1) and (2):
$$\sum_{i=1}^{n}[(x_i + 1)^2 + (x_i - 1)^2] = 14n$$
$$\sum_{i=1}^{n}[x_i^2 + 2x_i + 1 + x_i^2 - 2x_i + 1] = 14n$$
$$\sum_{i=1}^{n}(2x_i^2 + 2) = 14n$$
$$2\sum_{i=1}^{n}x_i^2 + 2n = 14n$$
$$\sum_{i=1}^{n}x_i^2 = 6n \quad \ldots (3)$$

Subtracting (2) from (1):
$$\sum_{i=1}^{n}[(x_i + 1)^2 - (x_i - 1)^2] = 4n$$
$$\sum_{i=1}^{n}[4x_i] = 4n$$
$$\sum_{i=1}^{n}x_i = n$$

Therefore, $\bar{x} = \frac{\sum x_i}{n} = 1$

Variance: $\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2 = \frac{6n}{n} - 1^2 = 6 - 1 = 5$

Standard deviation: $\sigma = \sqrt{5}$

\textbf{Answer: (2) $\sqrt{5}$}