[2 Mark Question Answer]

Question 12 Marks

If the mean of n observation $ax_1, ax_2, ax_3,....,ax_n$ is a$\overline{ X }$, show that $\left(a x_1-a \overline{ X }\right)+\left(a x_2-a \overline{ X }\right)+\ldots\left(a x_{ n }-a \overline{ X }\right) = 0$

Answer

We have
$a \overline{ X }=\frac{a x_1+a x_2+\ldots+a x_{ n }}{ n }$
$\Rightarrow a x_1+a x_2+\ldots+a x_{ n }= n (a \overline{ X }) \quad \ldots \text { (i) }$
$\text { Now }\left(a x_1-a \overline{ X }\right)+\left(a x_2-a \overline{ X }\right)+\ldots\left(a x_{ n }-a \overline{ X }\right)$
$ =\left(a x_1+a x_2+\ldots+a x n\right)-(a \overline{ X }+a \overline{ X }+\ldots+a \overline{ X } n \text { - times })$
$= n (a \overline{ X }- n (a \overline{ X })=0 . \quad \ldots[\text { Using (i) }]$
Hence proved.

View full question & answer→

Question 22 Marks

The following table give the marks scored by students in an examination:

Marks	0 - 5	5 - 10	10 - 15	15 - 20	20 - 25	25 - 30	30 - 35	35 - 40
No. of students	3	7	15	24	16	8	5	2

(i) Find the modal group
(ii) Which group has the least frequency?

Answer

(i) 15 – 20 is the modal group.
(ii) The group 35 – 40 has the least frequency.

View full question & answer→

Question 32 Marks

The median of the following observation 11, 12, 14, 18, (x + 4), 30, 32, 35, 41 arranged in ascending order is 24. Find x.

Answer

11, 12, 14, 18, (x + 4), 30, 32, 35, 41
No. of term are odd (9)
$\therefore$ Median $=\left(\frac{ n +1}{2}\right)^{\text {th }}$ term
$=\left(\frac{9+1}{2}\right)^{\text {th }}$ term
$=5^{\text {th }}$ term
∴ Median = x + 4
and Given Median = 24
∴ x + 4 = 24
x = 20.

View full question & answer→

Question 42 Marks

Find the mean of first five natural numbers.

Answer

First five natural numbers are $1, 2, 3, 4$ and $5$
Hence,
$\text { Mean } \overline{ X }=\frac{\sum x}{ n } $
$ =\frac{1+2+3+4+5}{5} $
$=\frac{15}{5}=3 $
$ \overline{ X }=3$

View full question & answer→

Question 52 Marks

Find the mean of $4, 7, 12, 8, 11, 9, 13, 15, 2, 7.$

Answer

Here $n = 10$
and $\sum x = 4 + 7 + 12 + 8 + 11 + 9 + 13 + 15 + 2 + 7$
$\therefore \text { Mean } \overline{ X }=\frac{\sum x}{ n } $
$ =\frac{88}{10}$
$= 8.8.$

View full question & answer→

Question 62 Marks

$A$ school has $4$ sections of Chemistry in class X having $40, 35, 45$ and $42$ students. The mean marks obtained in Chemistry test are $50, 60, 55$ and $45$ respectively for the 4 sections. Determine the overall average of marks per student.

Answer

Here $n_1 = 40, n_2 = 35, n_3 = 45, n_4 = 42,$
$\overline{ X }_1=50, \overline{ X }_2=60, \overline{ X }_3=55$ and $\overline{ X }_4=45$.
$\therefore \overline{ X }=\frac{ n _1 \overline{ X }_1+ n _2 \overline{ X }_2+ n _3 \overline{ X }_3+ n _4 \overline{ X }_4}{ n _1+ n _2+ n _3+ n _4}$
$=\frac{40 \times 50+35 \times 60+45 \times 55+42 \times 45}{40+35+45+42}$
$=\frac{2000+2100+2475+1890}{162}$
$=\frac{8465}{162}$
$= 52.25$
Hence, the overall average maeks of perstudents is $52.25$

View full question & answer→

Question 72 Marks

Find the mode from the following data:
110,120,130,120,110,140,130,120,140,120.

Answer

Arranging the data in the form of a frequency table, we have :

Value	Tally bars	Frequency
110	II	2
120	IIII	4
130	II	2
140	II	2

Since the value 120 occurs maximum number of times i.e., 4. Hence the modal value is 120.

View full question & answer→

Question 82 Marks

Find the median of the following values:
$37, 31, 42, 43, 46, 25, 39, 45, 32.$

Answer

Arraying the data in ascending order, we have
$25, 31, 32, 37, 39, 42, 45, 46.$
Here, the number of observation n =$ 9(odd)$
$\therefore$ Median $=$ Value of $\left(\frac{9+1}{2}\right)^{\text {th }}$ observation
= Value of $5^{th}$ observation
= $39$.

View full question & answer→

Question 92 Marks

Find the mean, median and mode of the following distribution:
$8,10, 7, 6,10,11, 6,13,10$

Answer

Arranging the number of in ascending order $6, 6, 7, 8, 10, 10, 11, 13.$
$\text { Mean } \overline{ X }=\frac{\sum x}{ n }=\frac{81}{9}=9 $
$\text { Median }=\left[\frac{ n +1}{2}\right]^{\text {th }} \text { term }$
$=\left(\frac{9+1}{2}\right)^{\text {th }} $
$ =5^{\text {th }} \text { term }$
Median $= 10.$
Mode $= 10$ is repeating $3$ times which is highest frequences.
So mode is $10.$

View full question & answer→

Question 102 Marks

There are $45$ students in a class, in which $15$ are girls. The average weight of $15$ girls is $45\ kg$ and $30$ boys is $52\ kg$. Find the mean weight in kg of the entire class.

Answer

Here $n_1 = 15, n_2 = 30, X_1 = 45 Kg$ and $X_2 = 52 Kg.$
$\therefore \overline{ X }=\frac{ n _1 \overline{ X }_1+ n _2 \overline{ X }_2}{ n _1+ n _2}=\frac{15 \times 45+30 \times 52}{15+30} Kg$
$=\frac{2235}{45} kg$
$= 49·67\ kg.$
Hence, the mean weight of the entire class is $49·67\ kg.$

View full question & answer→

Mathematics [2 Mark Question Answer]

Test yourself on this topic