MATHS M.C.Q

Arranging the numbers in ascending order, we have:
$31, 35, 36, 38, 40, 44, 45, 52, 55, 60$
Here, $n$ is $10,$ which is an even number. Thus, we have:
Median $=$ Mean of $\Big(\frac{\text{n}}{2}\Big)\text{th}$ obervation & $\Big(\frac{\text{n}}{2}+1\Big)\text{th}$ obervation Median Weight $=$ mean of the weights of $\Big(\frac{10}{2}\Big)\text{th}$ student & $\Big(\frac{10}{2}+1\Big)\text{th}$ student $=$ mean of the weights of $5th$ student & $6th$ student$=\frac{1}{2}(40+44)=42$
Hence, the median weight is $42kg.$

Required mean $=\frac{(\text{ax}_1+\text{ax}_2+_{\dots}+\text{ax}_\text{n})+(\frac{\text{x}_1}{\text{a}},\frac{\text{x}_2}{\text{a}},\ \dots,\frac{\text{x}_\text{n}}{\text{a}})}{2\text{n}}$
$=\frac{1}{2}\bigg[\frac{\text{a}(\text{x}_1+\text{x}_2+{\dots}+\text{x}_\text{n})}{\text{n}}+\frac{\frac{1}{2}(\text{x}_1+\text{x}_2+{\dots}+\text{x}_\text{n})}{\text{n}}\bigg]$
$=\frac{1}{2}\Big[\text{a}\bar{\text{x}}+\frac{1}{\text{a}}\bar{\text{x}}\Big]$
$=\Big[\text{a}+\frac{1}{\text{a}}\Big]\frac{\bar{\text{x}}}{2}$

$\text{Mean}=\frac{7+5+13+\text{x}+9}{5}=10$
$\Rightarrow34+\text{x}=50$
$\Rightarrow\text{x}=16$

The empirical relationship between the three measures of central tendencies is $3$ Median $=$ Mode $+\ 2$ Mean.
The relationship is as per observation. A distribution in which the values of mean, median and mode coincide $($i.e., mean $=$ median $=$ mode$)$ is called symmetrical
Distribution. conversely, when the values of mean, median, mode are not equal, the distribution is called asymmetrical or skewed.
Knowing any two values, the third can be computed by this formula
$3$ median $= 2$ mean $+$ mode
$2$ mean $= 3$ median $–$ mode
$\text{Mean}=\frac{1}{2}(3 \text{ median - mode})$

Difference between the maximum and minimum value of the observation is called as range.
So, $99 - 47 = 52$

For $1st$ class mark
Class mark $=\frac{\text{(upper limit +lower limit)}}{2}$
$1st$ class mark $= 212$
Add the class size to get the class mark of the next class
$2nd$ class mark $= 212 + 25$
$2nd$ class mark $= 237$
Hence, the class mark of first two intervals are $212$ and $237.$

The difference between the maximum and minimum value is called range.
Thus, $30 - 15 = 15$

Data: $0, 2, 2, 2, -3, 5, -1, 5, 5, -3, 6, 6, 5, 6$
Rearranging data in increasing order, we have
$-3, -3, -1, 0, 2, 2, 5, 5, 5, 5, 6, 6, 6$
Number of observations $= n = 14$ (even)
Now,
$\text{Median}=\frac{\Big(\frac{\text{n}}{2}\Big)^{\text{th}}\text{observation}+\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{observation}}{2}$
$=\frac{7^{\text{th}}\text{observation}+8^{\text{th}}\text{observation}}{2}$
$=\frac{2+5}{2}$
$\Rightarrow\text{Median}=3.5$

The mean is equal to the sum of all the values in the data set divided by the number of values in the data set.
The first $4$ prime numbers are $2, 3, 5, 7$
So, mean is $\frac{17}{4}=4.25$

A cumulative frequency distribution is the sum of the class and all classes below it in a frequency distribution.
Less than $30$ has the class interval 20-30. Frequency of this class interval will be corresponding to

In frequency polygon $x$-axis repesents data value whereas $y$-axis is used represent the frequencies of the data.
We include one class below lowest value and one class above highest value with zero frequencies.
The graph touches the $x$-axis at these points.
So, one sides of the frequency polygon is $x$-axis.

The classes are $10-15, 15-20, 20-25, 25-30, 30-35$ so that upper limit of the highest class is $35$.

Difference between the maximum and minimum value of the observations is called as range.
Let, minimum value be $'x'$
$75 - x = 20$
So, $x = 55$

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

Prime numbers between $30$ and $40$ are $31$ and $37$.
Mean $=\frac{31+37}{2}$
Mean $=34$

Arrange the given data in ascending order.
$7, 7, 7, 8, 9, 10, 13$
For odd number $(n)$ of observation median $=\Big[\frac{(\text{n+1)}}{2}\Big]\text{th}$ term,
Here $n = 7$ so median $=\Big[\frac{{(7+1)}}{2}\Big]\text{th}$ term $= 4th$ term that is $8$,
Hence median $= 8$

Since, Mean – Mode + Mean + $2$ Mode = $3$ Median
$\Rightarrow $ Mean – Mode = $3$ Median – Mean – $2$ Mode
$= 3$ Median – Mean – $2 (3$ Median – $2$ Mean$)$
$= 3$ Median – Mean – $6$ Median + $4$ Mean
$= 3$ Mean – $3$ Median
$= 3$ (Mean – Median)

The observations in ascending order can be written as:
$3, 4, 4, 5, 6, 7, 7, 7, 12$
Median $=\Big(\frac{9+1}{2}\Big)\text{th}$ term $= 5th$ term $= 6$

Range of observations = Highest observation - Lowest observation
$= 95 - 62 = 33$

We first arrange the given data in ascending order as follows $14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20$.
From above, we see that $15$ occurs most frequently i.e., $5$ times.
Hence, the mode of the given data is $15$.

Since $x^{-}$and $y^{-}$are two numbers, though being means, their arithmetic mean is given by:
$\text{z}=\frac{\text{x and y}}{2}$

$3$ Median $= 2 \times $ Mean + Mode
$\Rightarrow 3 \times 20 = 2 \times $ Mean + $18$
$\Rightarrow 2 \times$ Mean $= 60 - 18 = 42$
$\Rightarrow $ Mean $= 21$

A cumulative frequency distribution is the sum of the class and all classes below it in a frequency distribution.
Subtract cumulative frequency of class more than $70$ from the next cummulative Frequency of class more than $69$.
$30 − 18 = 12$

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

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 $n = 9$]
$=\Big(\frac{10}{2}\Big)^{\text{th}}\text{observation}$
$=5^{\text{th}}\text{observation}=6$

Let the numbers be $x_1, x_2 \ldots ., x_n$
Now the new numbers after decrasing every number by $8:\left(x_1-8\right),\left(x_2-8\right) \ldots\left(x_n-8\right)$
New mean $=\frac{(\text{x}_1 - 8)+(\text{x}_2 - 8)+...+(\text{x}_\text{n} - 8)}{\text{n}}$
$=\frac{\text{x}_1+\text{x}_2+...+\text{x}_\text{n}-\text{8n}}{\text{n}}$
$=\frac{\text{x}_1+\text{x}_2+...+\text{x}_\text{n}}{\text{n}}-8$
New mean $=$ mean $-8$
Hence, mean is decreased by $8.$

Tallies are usually marked in a bunch of $4$.
Hence, the correct option is $(b)$.

Let $l$ and $m$ respectively be the lower and upper limits of the class.
Then the mid-value of the class is $\frac{\text{l+m}}{2}$ and the class-size is $(m - l)$.
Therefore, we have two equations
$\frac{\text{l+m}}{2}=15$
$\Rightarrow l + m = 30,$
$m - l = 4$
Subtracting the second equation from the first equation, we have
$(l + m) - (m - l) = 30 - 4$
$\Rightarrow l + m - m + l = 26$
$\Rightarrow 2l = 26$
$\Rightarrow l = 13$
Hence, the lower limit of the class is $13$. Thus, the correct choice is $(c)$.

upper limit - lower limit = class width
upper limit - lower limit $= 10$
$\frac{\text{(upper limit + lower limit)}}{2}=\text{mid}-\text{value}$
upper limit + lower limit $= 2 \times 60.5$
upper limit + lower limit $= 121$
By solving the above two equations, we get
upper limit $= 65.5$
Lower limit $= 55.5$

We have:
Maximum value $= 32$
Minimum value $= 6$
We know:
Range = Maximum value - minimum value
$= 32 - 6$
$= 26$