Maths M.C.Q

4. Is increased by 5.

Solution:

Let x₁, x₂, ...., x_n be the n observation,

Then, old mean $\bar{\text{x}}_{\text{old}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\text{x}_\text{i}}{\text{n}}\ \dots(\text{i})$

Now, adding 5 in each obsevation, the new mean becomes

$\bar{\text{x}}_{\text{new}}=\frac{(\text{x}_1+5)+(\text{x}_2+5)+\ ...\ +(\text{x}_\text{n}+5)}{\text{n}}$

$\Rightarrow\ \bar{\text{x}}_{\text{new}}=\frac{(\text{x}_1+\text{x}_2+\ ...\ +\text{x}_\text{n})+5\text{n}}{}$

$\Rightarrow\ \bar{\text{x}}_{\text{new}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\text{x}_\text{i}}{\text{n}}+5=\bar{\text{x}}_{\text{old}}+5$ [from Eq. (i)]

$\Rightarrow\ \bar{\text{x}}_\text{new}=\bar{\text{x}}_{\text{old}}+5$

Hence, the new mean is increased by 5.

3. $\frac{3}{10}$

Solution:

Total number of students = 10 + 13 + 12 + 5 = 40

Let us call event of selected students at random who have blood group B be E.

There are 12 students who has blood group B.

So, $\text{P(E)}=\frac{12}{40}=\frac{3}{10}$

2. 0

Solution:

We know that algebraic sun of deviations from mean is zero.

$\sum\limits_{\text{a}=1}^\text{b} (\text{x}_\text{t}-\bar{\text{x}})=(\text{x}_1-\bar{\text{x}})+(\text{x}_2-\bar{\text{x}})+(\text{x}_3-\bar{\text{x}})+\ ...\ +(\text{x}_\text{n}-\bar{\text{x}})$

$= (\text{x}_1+\text{x}_2+\text{x}_3+... +\text{x}_\text{n})- \text{n}\bar{\text{x}}$

$\Rightarrow\sum\limits_{\text{t}-1}^\text{b}\text{x}_\text{i}-\text{n}\bar{\text{x}}=\text{n}\bar{\text{x}}-\text{n}\bar{\text{x}}=0$ $\bigg[\because\sum\limits_{\text{i}=1}^\text{n} \text{x}_\text{i}=\text{n}\bar{\text{x}}\bigg]$

Hence, (b) is correct answer.

4. 2

Solution:

The adjusted frequency for the class 25 - 45 is

$= \frac{\text{Frequency of the class }}{\text{Class width }}\times \text{Minimum width }=\frac{8}{20} \times5=2$

4. $\frac{9}{16}$

Solution:

Total number of bulbs in a lot, n(S) = 80

Number of bulbs whose life time is less than 900h, n(E) = 10 + 12 + 23 = 45

Probability that bulbs has life time less than 900h $=\frac{\text{n(E)}}{\text{n(S)}}=\frac{45}{80}=\frac{9}{16}$

Hence, the probability that bulb has life time less than 900 is $\frac{9}{16}$

4. 0.8

Solution:

We know that empricrical probability P(E) of an event E happen, is given by

$\text{P(E)}=\frac{\text{Number of trials in which the event happed}}{\text{Total number of trials}}$

Let E be the event that a person selected at random has a high school certificate.

$\text{P(E)}=\frac{\text{Numbers of person having high school certificate}}{\text{Total number of people in a sample study}}$

$=\frac{514}{642}=0.8$

3. 54

Solution:

Arranging the data in ascending order, we get

22, 34, 39, 45, 54, 56, 78, 84

Here n = 9, which is an odd number.

$\therefore\ \text{Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}$

$\text{Value}=\Big(\frac{9+1}{2}\Big)^{\text{th}}$

$\text{value}=5^{\text{th}}\text{value}$

So, median = 54

3. 6

Solution:

First, we arrange the given numbers in ascending order is,

3, 4, 4, 5, 6, 7, 7, 7 and 12

Here, n = 9

Since, n is odd, so we use the formula for median,

Now, Median $=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{observation}$

$=\Big(\frac{9+1}{2}\Big)^{\text{th}}\text{observation}$ [Put n = 9]

$=\Big(\frac{10}{2}\Big)^{\text{th}}\text{observation}$

$=5^{\text{th}}\text{observation}=6$ 

3. 0

Solution:

Total number of bulbs = 80

Let E of the event that bulb selected at random form the lot has life time 1150hrs.

We see from the frequency table given above that none of the bulb has life time 1150hrs.

$\therefore\ \text{P(E)}=\frac{0}{80}=0$

4. 56.5

Solution:

Given, n = 50, then mean $\bar{\text{x}}=\frac{\sum\limits^{\text{n}}_{\text{i}=1}\text{x}_\text{i}}{\text{n}}$

Then, $\bar{\text{x}}=\frac{1}{50}\times\sum\limits^{50}_{\text{i}=1}\text{x}_\text{i}$

$\Rightarrow\sum\limits^{50}_{\text{i}=1}\text{x}_\text{i}=50\bar{\text{x}}$

Now, subtract each observation from 53, we get a new mean say $\bar{\text{x}}_\text{new}$

$\therefore\ \bar{\text{x}}_\text{new}=\frac{(-\text{x}_1+53)+(-\text{x}_2+53)\ +\ .....\ +(-\text{x}_{50}+53)}{50}$

$\Rightarrow-3.5=\frac{-(\text{x}_1+\text{x}_2\ +\ .....\ +\text{x}_{50})+(53+53\ +\ ....\ +50\text{times})}{50}$

$\Rightarrow-3.5\times50=(-\text{x}_1+\text{x}_2+\ ....\ +\text{x}_{50})+53\times50$

$\therefore$ Mean of 50 observation $=\frac{1}{50}\sum\limits^{50}_{\text{i}=1}\text{x}_{\text{i}}$

$=\frac{1}{50}\times2852=56.5$ $\Bigg[\because\text{ mean}=\frac{\sum\limits^{\text{n}}_{\text{i}=1}\text{x}_\text{i}}{\text{n}}\Bigg]$

Hence, the mean of given number is 56.5

3. $11\frac{1}{3}$

Solution:

Given that, the mean of the observation x, x + 3, x + 5, x + 7 and x + 10 is 9.

$\therefore\ \frac{\text{x}+\text{x}+3+\text{x}+5+\text{x}+7+\text{x}+10}{5}=9$

$\Rightarrow {\text{5x}}+25=45$

$\Rightarrow\text{5x}=20$

$\Rightarrow \text{x}=4$

$\therefore$ Last three odservations are x + 5 = 4 + 5 = 9, x + 7 = 4 + 7 = 11

and x + 10 = 4 + 10 = 14

So, the mean of last three observations $=\frac{9+11+14}{3}=\frac{34}{3}=11\frac{1}{3}$

Hence, the mean of last three observation is $11\frac{1}{3}.$

2. 105

Solution:

$\text{Class mark }=\frac{\text{Upperlimit + Lowerlimit}}{2}$

$\Rightarrow \text{Class mark}=\frac{90+120}{2}=\frac{210}{2}=105$

4. 38

Solution:

Let x₁, x₂, x₃, x₄, and x₅ be five numbers and one of the excluded number be x₅,

Given, mean of the numbers = 30

$\Rightarrow\frac{\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4+\text{x}_5}{5}=30$

$\Rightarrow {\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4+\text{x}_5}=150$

$\Rightarrow {\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4}=150-\text{x}_5$

On dividing both sides by 4, we get

$\frac{\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4}{4}=\frac{150-\text{x}_5}{4}\ \dots(\text{i})$

Given, mean of the numbers = 28

$\therefore\ \frac{150-\text{x}_5}{4}=28$ [from Eq. (i)]

$\Rightarrow 150 -\text{x}_5=112$

$\Rightarrow \text{x}_5=150 -112$

$\Rightarrow\text{x}_5=38$

Hence, the excluded number is 38.

3. $\frac{4}{5}$

Solution:

The total number of coins tossed, n(S) = 1000

Number of outcomes in which atmost one head, n(E) = 550 + 250 = 800

$=\frac{\text{n(E)}}{\text{n(S)}}=\frac{800}{1000}=\frac{4}{5}$

Hence, the probability for atmost one head is $\frac{4}{5}$

2. $\Big(\text{a}+\frac{1}{\text{a}}\Big)\frac{\bar{\text{x}}}{2}$

Solution:

Given, mean of x₁, x₂, ......, x_n is $\bar{\text{x}}$

$\therefore\ \sum\limits^\text{n}_{\text{i}=1}\text{x}_\text{i}=\text{n}\bar{\text{x}}$

Now, let the mean of $\Big(\text{ax}_1,\text{ax}_2,\ ....\text{ ax}_\text{n},\frac{\text{x}_1}{\text{a}},\frac{\text{x}_2}{\text{a}},\ ...., \ \frac{\text{x}_\text{n}}{\text{a}}\Big)$ is $\bar{\text{z}}$

Then, $\bar{\text{z}}=\frac{(\text{ax}_1,\text{ax}_2,\ ....\text{ ax}_\text{n})+\Big(\frac{\text{x}_1}{\text{a}},\frac{\text{x}_2}{\text{a}},\ ...., \ \frac{\text{x}_\text{n}}{\text{a}}\Big)}{\text{n}+\text{n}}$

$=\frac{\text{a}(\text{x}_1+\text{x}_2+\ ....\ +\text{x}_\text{n})+\frac{1}{\text{a}}(\text{x}_1+\text{x}_2+\ ....\ +\text{x}_\text{n})}{2\text{n}}$

$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big)(\text{x}_1+\text{x}_2+\ ....\ +\text{x}_\text{n})}{2\text{n}}$

$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big)\sum\limits^\text{n}_{\text{i}=1}\text{x}_\text{i}}{2\text{n}}$

$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big)\cdot\text{n}\bar{\text{x}}}{2\text{n}}$

$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big){\bar{\text{x}}}}{2}$

2. $\frac{\bar{\text{x}}+\bar{\text{y}}}{2}$

Solution:

We have $\bar{\text{x}}$ is the mean of x₁, x₂, ......, x_n, and $\bar{\text{y}}$ is the mean of y₁ , y₂ , ... , y_n

So, $\bar{\text{x}}=\frac{(\text{x}_1+\text{x}_2+\text{x}_3+\ ....\ +\text{x}_\text{n})}{\text{n}}$

$\Rightarrow\text{x}_1+\text{x}_2+\text{x}_3+\ .....\ +\text{x}_\text{n}=\text{n}\bar{\text{x}}$

And $\bar{\text{y}}=\frac{(\text{y}_1+\text{y}_2+\text{y}_3+\ ....\ +\text{y}_\text{n})}{\text{n}}$

$\Rightarrow\text{y}_1+\text{y}_2+\text{y}_3+\ ....\ +\text{y}_\text{n}=\text{n}\bar{\text{y}}$

If $\bar{\text{z}}$ is the mean of x₁, x₂, ...., y₁ , y₂ , ... , y_n then,

_{$\bar{\text{z}}=\frac{\text{n}\bar{\text{x}}+\text{n}\bar{\text{y}}}{\text{n}+\text{n}}=\frac{\text{n}(\bar{\text{x}}+\bar{\text{y}})}{2\text{n}}=\frac{\bar{\text{x}}+\bar{\text{y}}}{2}$}

2. 2m - l

Solution:

Let x and y be the lower and upper class limit of a continuous frequency distribution.

Now, mid-point of a class $=\frac{(\text{x}+\text{y})}{2}=\text{m}$ [given]

⇒ x + y = 2m = x + l = 2m

$\big[\therefore$ y = l = upper class limit (given)$\big]$

⇒ x = 2m - l

Hence, the lower class limit of the class is 2m - l.

3. 0.75

Solution:

Total number of survey children’s age from 19 - 36 months, n(S) = 364 In those of them 91 out of them liked to eat potato chips.

$\therefore$ Number of children who do not like to eat potato chips, n(E) = 364 - 91 = 273

$\therefore$ Probability that he/she does not like to eat potato chips $=\frac{\text{n(E)}}{\text{n(s)}}=\frac{273}{364}=0.75$

Hence, the probability that he/ she does not like to eat potato chips is 0.75.

2. 36

Solution:

Mean of first 13 observation = 32

$\therefore$ Sum of all first 13 observation = (32 × 13) = 416

Mean of last 13 observation = 40

$\therefore$ Sum of all last 13 observation = (40 × 13) = 520

Mean of 25 observation = 36

$\therefore$ Sum of all first 25 observation = (36 × 25) = 900

Hence, 13^th observation = 416 + 520 - 900 = 36

2. 7

Solution:

Let x and y be the uppel and lower and lower class limit in a frequency distribution.

Now, mid value of a class $\frac{(\text{x}+\text{y})}{2}=10$ [given]

⇒ x + y = 20 ...(i)

Also, given that, width of class x - y = 6 ...(ii)

On adding Eqs. (i) and (ii), we get

2x = 20 + 6

⇒ 2x = 26

⇒ x = 13

On putting x = 13 in Eq. (i), we get

13 + y = 20

⇒ y = 7

Hence, the lower limit of the class is 7.

3. $\frac{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{\text{n}_2}$

Solution:

If $\bar{\text{x}}_1, \bar{\text{x}}_2, \bar{\text{x}}_3, ....., \bar{\text{x}}_\text{n}$ are the mean of n group with n₁, n₂, ...., n_n number of observation respectively, then the mean $\bar{\text{x}}$

$\bar{\text{x}}=\frac{\text{n}_1​​​​\bar{\text{x}}_1+\text{n}_2\bar{\text{x}}_2+\text{n}_3\bar{\text{x}}_3+\ ....\ +\text{n}_\text{n}\bar{\text{x}}_\text{n}}{\text{n}_1+\text{n}_2+\text{n}_3+\ ....\ +\text{n}_\text{n}}$

$=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}}$

Hence, the mean $\bar{\text{x}}$ of all group taken together is gien up $=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}}$

2. 51

Solution:

We have, $\bar{\text{x}}=\frac{\sum\text{x}_\text{i}}{\text{n}}$

$\Rightarrow50=\frac{\sum\text{x}_\text{i}}{100}$

$\Rightarrow\sum\text{x}_\text{i}=50\times100=5,000$

If one of the observation which was 50 is resplaced by 150, then

$\sum\text{x}_\text{i}=5,000-50+150=5100$

Then, the resulting mean $=\frac{5100}{100}=51$