Maths M.C.Q

The mean of $50$ observations is $39.$
So sum of these $50$ observations is $50 × 39 = 1950$
After replacing the observation value $23$ by $43,$
Sum becomes $1970$
So the mean is $\frac{1970}{50}=39.4$

Since, $3$ Median $=$ Mode $+\ 2$ Mean
$\Rightarrow\text{Median}=\frac{\text{Mode}}{3}+\frac{2}{3}\text{Mean}$
$=\frac{\text{Mode}}{3}+\frac{2}{3}\text{Mean}-\frac{2}{3}\text{Mode}+\frac{2}{3}\text{Mode}$
$\text{Median}=\text{Mode}+\frac{2}{3}(\text{Mean - Mode})$

First, we arrange the given observations in ascending order as follows
$22, 34, 39, 45, 54, 54, 56, 68, 78$ and $84$
Here, total number of observation, $n = 10$
Since, $n$ is even, so we use the formula for median,
Median $=\frac{\big(\frac{\text{n}}{2}\big)^{-1}\text{observation +}\big(\frac{\text{n}}{2}+1\big)\text{th observation}}{2}$
$=\frac{\big(\frac{10}{2}\big)\text{ th observation +}\big(\frac{10}{2}+1\big)\text{th observation}}{2} [$put $n = 10]$
$=\frac{5\text{th observation $+ 6th$ obsevation}}{2}=\frac{54+54}{2}=\frac{108}{2}=54$
Hence, the median of given data is $54.$

The adjusted frequency for the class $25 - 45$ is
$= \frac{\text{Frequency of the class }}{\text{Class width }}\times \text{Minimum width }=\frac{8}{20} \times5=2$

Median $= 4$
Mode $= 2$
$\text{Mean}=\frac{2+2+4+5+12}{5}=\frac{25}{5}=5$
Hence, $($Mean $= 5) > ($Mode $= 2)$

The given data is $5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10$ and $8$
Make the frequency table.

The given data is $7, 5, 13, x$ and $9.$ They are $5$ in numbers.
The mean is $\frac{7+5+13+\text{x}+9}{5}=\frac{34+\text{x}}{5}$
But, it is given that the mean is $10$. Hence, we have
$\frac{34+\text{x}}{5}=10$
$⇒ 34 + x = 50$
$⇒ x = 50 - 34$
$⇒ x = 16$

Add the values corresponding to the height of the bar before $40.$
$6 + 3 + 4 + 2 = 15$

Difference between the maximum & minimum value of the observation is called as range.
So, $32 - 6 = 26$

Sol. Width of each of the five continuous classes in a frequency distribution is $5.$
Lower class limit of the lowest class $= 10$
Upper class limit of the lowest class is $10 + 5 = 15$
So, the five continuous classes are
$10 - 15, 15 - 20, 20 - 25, 25 - 30, 30 - 35$
Hence, the upper-class limit of the height class is $35.$

The class width is the difference between the upper- or lower-class limits of consecutive classes. In this case, class width equals to the difference between the lower limits of the first two classes.
$w = 10 - 0$
So, the class width is $10$

The number $20$ is included in $20–30.$
Hence, $(b)$ is the correct answer.

As the class size $= 5,$ Class interval is got by subtracting and adding $2.5$ to $43.$

Let the lower limit $= k$
Mid-point $= m$
Upper limit $= l$
Mid-point $=\frac{\text{(upper limit + lower limit)}}{2}$
$\text{m}=\frac{(\text{k+l})}{2}$
$2\text{m}=\text{k}+\text{l}$
$\text{k}=2\text{m}-\text{l}$
Therefore, lower limit $= 2m - l$

The mean of the six numbers is $23.$
So the sum of six numbers is $23 × 6 = 138$
After excluding one number, the mean of the remaining numbers is $20.$
So the sum of five numbers is $20 × 5 = 100$
The difference between them is$138 - 100 = 38$

The median is the middle score for a set of data that has been arranged in ascending or descending order of magnitude.
for even number of observations, median is calculated as average of two middle number.
for the given example $30$ and $x$ are in the middle and median is $35.$
So,
$35=\frac{30+\text{x}}{2}$
$70=30+\text{x}$
$\text{x}=40$

The mean of five observations is $15$
So the sum of these five observations is $15 × 5 = 75$
The mean of first three observations is $14$
So the sum of the first three observations is $14 × 3 = 42$
So the sum of the last two numbers is $75 - 42 = 33$
The mean of the last three observations is $17.$
So sum of last three observations is $17 × 3 = 51$
So the middle number is $51 - 33 = 18.$

In statistics, the mode in a list of numbers refers to the integers that occurs most number of times.
for the set of numbers, $9$ occurs three times i.e more than any other number in the list.

Class mark $=\frac{\text{( upper limit + Lower limit ) }}{2}$
$=\frac{50+60}{2}$
$=\frac{110}{2}$
So class mark is $55.$

The median is the middle score for a set of data that has been arranged in ascending or descending order of magnitude.
Since $2x + 1$ is in the middle of the arranged numbers, so it is median
Hence, $2x + 1 = 7$
Now since $7$ occurs more number of times then other numbers so mode of the list is $7.$

$1\ cm = 30\ km$
So for $75\ km$
$\frac{75}{30}=2.5\text{cm}$

Average is equal to the sum of all the values in the data set divided by the number of values in the data set.
Average $=\frac{\text{x}+\text{x}+\text{y}+\text{y}+\text{y}}{5}$
Average $=\frac{2\text{x}+3\text{y}}{5}$

Given that the mean of $7$ numbers is $81$. Let us denote the numbers by $X_1, \ldots . X_7$
If $\overline{\text{X}}$ be the mean of the n observations $X_1, \ldots X_n$, then we have
$\overline{\text{X}}=\frac{1}{2}\sum_{\text{i}=1}^{\text{n}}\text{x}_\text{i}$
$\Rightarrow\sum_{\text{i}=1}^{\text{n}}\text{x}_\text{i}=\text{n}\overline{\text{X}}$
Hence the sum of $7$ numbers is
$\sum_{\text{i}=1}^{7}\text{x}_\text{i}=7\times81=567$
If one number is discarded then the mean becomes $78$ and the total numbers become $6.$
Let the number discarded is $x.$
After discarding one number the sum becomes $567 - x $ and then the mean is
$\frac{567-\text{x}}{6}$
But it is given that after discarding one number the mean becomes $78.$
Hence we have
$\frac{567-\text{x}}{6}=78$
$\Rightarrow 567 - x = 468$
$\Rightarrow 567 = x + 468$
$\Rightarrow x = 468 = 567$
$\Rightarrow x = 567 - 468$
$\Rightarrow x = 99$
Thus the excluded number is $99.$

Minimum value $= 14$
Maximum value $= 112$
The classes are
$13 - 22, 23 - 32, 33 - 42, 43 - 52, 53 - 62, 63 - 72, 73 - 82, 83 - 92, 93 - 102$ and $103 - 112.$
The number of classes in the distribution will be $10.$

The mean of $50$ observations is $39.$
So sum of these $50$ observations is $50 × 39 = 1950$
After replacing the observation value $23$ by $43,$
Sum becomes $1970$
So the mean is $\frac{1970}{50}=39.4$

$\text{Mean}=\frac{\text{a}+\text{b}+\text{c}+\text{d}+\text{e}}{5}=28$
$⇒ a + b + c + d + e = 140 ...(1)$
Also, $\text{Mean}=\frac{\text{a}+\text{c}+\text{e}}{3}=24$
$⇒ a+ c + e = 72 ...(2)$
Subtracting equation $(2)$ from $(1),$ we have
$b + d = 68$
$\text{Mean}=\frac{\text{b}+\text{d}}{2}=\frac{68}{2}=34$

Mean of first $13$ observation $= 32$
$\therefore$ Sum of all first $13$ observation $= (32 \times 13) = 416$
Mean of last $13$ observation $= 40$
$\therefore$ Sum of all last $13$ observation $= (40 \times 13) = 520$
Mean of $25$ observation $= 36$
$\therefore$ Sum of all first $25$ observation $= (36 \times 25) = 900$
Hence, $13^{th}$ observation $= 416 + 520 - 900 = 36$

Since mean is equal to the sum of all the values in the data set divided by the number of values in the data set.
So increase in the value of each observation will also increase the mean (average) by $3.$

The median is the middle score for a set of data that has been arranged in ascending or descending order of magnitude.
So for numbers $1, 2, 3, ... 9$
$5$ is the median.