(2)      80

(2)

Since, $N=66,$ then $\frac{N}{2}=33$

and cumulative frequency greater than or equal to 33 lies in class 10 – 15
So, median class is 10 – 15

$\therefore$ Lower limit of median class is 10

and highest frequency is 20 lie in class 15 – 20
So, modal class is 15 – 20.

$\therefore$ Lower limit of modal class is 15.

Hence, sum of lower limits of the median and modal class is 10 + 15 = 25.

(2)

mode – median = 24 (given)

$\therefore$ mode = 24 + median

Since, mode = 3 median – 2 mean [By empirical relation]

$\therefore$ 24 + median = 3 median – 2 mean

⇒ 2 median – 2 mean = 24

⇒ median – mean = 12

(2)

The middle most observation, after arranging all observations in ascending or descending order is called the median.

(4)     0

(2)

Mode = The Most Common or (Maximum). Number that appears in your set of data.

(1)

When value of each observation in data is increased by 2.

So, median of data is Increases by 2

(4)

$$\begin{gathered} \sum _{i=1}^n{f}_i(x_i-x)=f1(x1-x)+f2(x2-x)+\cdots +fn(xn-x) \\ =(f_1{x}_1+f_2{x}_2+\dots +f_n{x}_n)-(f_{1}x+f2x+\cdots +fnx) \\ =\sum _{i=1}^n{f}_i{x}_i-xi=1nfi \\ =\sum _{i=1}^n{f}_i{x}_i-\frac{{\sum }_{i=1}^n{f}_i{x}_i}{{\sum }_{i=1}^n{f}_i}·\sum _{i=1}^n{f}_i=0 \\ \overline{x}=\frac{\sum _{i=1}^n{f}_i{x}_i}{\sum _{i=1}^n{f}_i} \end{gathered}$$

(1)

Let n terms of data be

$$\begin{gathered} x_1,x_2,x_3,\dots ,x_n. \\ \therefore \text{Mean }\left(\overline{x}\right)=\frac{{\sum }_{i=1}^n{x}_i}{n} \\ Now, when each term is decreased by 2. \\ \therefore Data will be x_1-2,x_2-2,x_3-2,\dots ,x_n-2 \\ \therefore New mean==\frac{{\displaystyle {\sum }_{i=1}^n(x_i-2)}}{{\displaystyle n}}=\frac{{\displaystyle {\sum }_{i=1}^n{x}_i}}{{\displaystyle n}}-\frac{{\displaystyle 2n}}{{\displaystyle n}}=\overline{x}-2 \\ \Rightarrow " Mean is also decreased by 2." \end{gathered}$$

(2)

We have, mean of first n natural numbers

= $$\begin{gathered} \frac{n(n+1)}{2n} \\ \Rightarrow 15=\frac{(n+1)}{2} \\ \Rightarrow n+1=30 \\ \Rightarrow n=30-1=29 \\ \therefore n=29 \end{gathered}$$

(2)

The middle most observation of every data arranged in order is called median.

(4)

An ogive is a plot of the cumulative frequencies of a frequency distribution.
When the cumulative frequencies of class-intervals are plotted against upper or lower limits of class intervals, the graph obtained by joining the points by free hand forming smooth curve is called the cumulative frequency curve or ogive.
Ogive graphs are used to find the median of the given set of data.

(3)

Given, median of the data 16, 18, 20, 24 − x, 22 + 2x, 28, 30, 32 is 24.

$$\begin{gathered} Here, n=8\left(Even\right) \\ \therefore \text{Median}=\frac{{\left(\frac{n}{2}\right)}^{\text{th}}\text{ observation }+{\left(\frac{n}{2}+1\right)}^{\text{th}}\text{ observation}}{2} \\ 24=\frac{{\displaystyle \text{4th observation}+\text{5th observation}}}{{\displaystyle 2}} \\ \Rightarrow 48=(24-x)+(22+2x) \\ \Rightarrow 48=46+x\Rightarrow x=2 \\ Hence, the value of x is 2. \end{gathered}$$

(4)

$$\begin{gathered} Since a < b < c \\ \text{Median}=b\Rightarrow b=35 \\ \text{Mean}=\frac{a+b+c}{3}=50 \\ \Rightarrow a+b+c=150 \\ \Rightarrow a+c=150-35=115 \\ \therefore \text{(ii) is correct.} \\ Given, a + c = 115 c - a = 55 \Rightarrow a=30 \text{and} c=85 \\ \Rightarrow b-a=35-30=5 \\ \therefore \text{Option (d) is correct.} \end{gathered}$$

(1)

We know that the empirical relation is given by,

Mode = 3 Median − 2 Mean

(1)

Given, mean = 8.32 and median = 8.05

∴ Mode = 3 × Median − 2 × Mean

= 3 × 8.05 − 2 × 8.32
= 24.15 − 16.64
= 7.51

(3)

We have,

From the above cumulative frequency table, 82 athletes completed the race in or equal to 14.6 seconds.

(2)

In the above dataset, value 6 and 7 have occurred most number of times, i.e., 3 times.

Also, mode = K.

So, either 6 or 7 would be the mode.

If K = 6,
$K² - 12K + 42 = 6² - 12\left(6\right) + 42 = 6$

If K = 7,
$K² - 12K + 42 = 7² - 12\left(7\right) + 42 = 7$

∴ K = 6 and K = 7

So, (i) and (iii) are correct.