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By Definition of mode, mode is most frequent value.

Given, 

Median = 24

Then median class = 20 - 30

l = 20, h = 10, f = x and F = 30

Median = l + \(\frac{\frac{N}{2}-F}{f}\)

\(24 = 20+\frac{55+\frac{x}{2}30}{x}\times10\)

4x = 275 + 5x – 300

4x – 5x = -25

-x = -25

x = 25

Therefore, 

missing frequency = 25

Given, 

N = 200

= 46 + x + y + 25 + 10 + 5 = 200

= x + y = 200 – 46 – 25 – 10 – 5

= x + y = 114 (i)

And Mean = 1.46

\(\frac{\sum fx}{N}=1.46\)

\(=\frac{x+2y+140}{200}=1.46\)

= x + 2y + 140 = 292

= x + 2y = 292 – 140

= x + 2y = 152 (ii)

Subtract (i) from (ii), we get

X + 2y – x – y = 152 – 114

y = 38 

Put the value of y in (i), we get 

x = 114 – 38 = 76

We have, 

N = 200

\(\frac{N}{2}=\frac{200}{2}=100\)

The cumulative frequency just more than \(\frac{N}{2}\) is 122 so the median is 1

Given, 

Mean = 18

Let missing frequency be V

Mean = \(\frac{\sum f_ix_i}{N}\)

 \(18 = \frac{752+20V}{44+V}\)

792 + 18V = 752 + 20V 

792 – 752 = 20V – 18V 

40 = 2V 

V = 20

Correct option is: C) pictograph

i. Given data: 5, 4, 0, 2, 2, 4, 4, 3, 3. 

Total number of observations = 9

\(Mean = \cfrac{The\,sum \,of\, all \,observatiojns \,in \,the \,data }{Total \,number \,of \,observations}\)

= \(\frac{5+4+0+2+2+4+4+3+3}{9}\)

= \(\frac{27}{9}\)

∴ The mean of the given data is 3. 

ii. Given data in ascending order: 

0,2, 2, 3, 3, 4, 4, 4,5 

∴ Number of observations(n) = 9 (i.e., odd) 

∴ Median is the middle most observation 

Here, the 5 number is at the middle position, which is 3. 

∴ The median of the given data is 3.

iii. Given data in ascending order:

 0,2, 2, 3, 3, 4, 4, 4,5 

Here, the observation repeated maximum number of times = 4 

∴ The mode of the given data is 4.

We know that, 

For finding median from a less than ogive or more than ogive curve, we follow below steps.

1. we find the sum of all frequencies or the last cumulative frequency in our given data, let that be N

2. Then we calculate \(\frac{N}{2}\)and locate the point corresponding to \(\frac{N}{2}\)th on the curve.

3. The X coordinate of the point located i.e. the class corresponding to \(\frac{N}{2}\)th cumulative frequency is the median of data. 

From the graph, we locate last cumulative frequency as 40 i.e. sum of all the frequencies is 40.

i.e. N = 40 and \(\frac{N}{2}\) = 20

Median is the marks corresponding to \(\frac{N}{2}\)th student.

In order to find the median, we first locate the point corresponding to 20^th student on Y axis. 

And from graph, that point is (50, 20) 

So, 

marks corresponding to 20^th student is 50. 

So, 

the median of above data is 50

So, 

the median of above data is 50

Correct option is  B) 40 – 50

Correct option is: B) 10

Given that the mean of 10 numbers is 7.

\(\therefore\) Sum of these 10 observations = \(n_1\overline x_2\)= 10 \(\times\)7 = 70

Also given that the mean of 15 numbers is 12.

\(\therefore\) Sum of these 15 observations =  \(n_2\overline x_2\)= 15 \(\times\)12 = 180.

There are total 10 + 15 or 25 observations whose sum are \(n_1\overline x_2\) + \(n_2\overline x_2\) = 70 + 180 = 250

\(\therefore\) Mean of all observations = \(\frac {Sum \, of\, all\, obs.}{Total\, No\, of \,obs}\) = \(\frac {n_1\overline x_2 + n_2\overline x_2}{n + n_2}\)

=  = \(\frac {70 + 180}{10 + 15} = \frac {250}{25} = 10\)

Hence, mean of all observations is 10.

Correct option is: B) 10

Correct option is: D) no mode

Since all the observations in the given data occurs equal times (1 times).

\(\therefore\) Mode of the observations in the given data does not exists.

Hence, given data has no mode.

Correct option is: D) no mode

Correct option is: D) 50

AM (Algebraic mean) of 10 numbers is 20.

\(\therefore\) Sum of 10 numbers = \(n_1\overline x_1 = 10 \times 20 = 200\)

AM (Algebraic mean) of 30 numbers is 60.

\(\therefore\) Sum of 30 numbers = \(n_2\overline x_2 = 30 \times 60 = 1800\)

\(\therefore\) AM of combined data is \(\overline x\) = \(\frac {n_1\overline x_1+ n_2\overline x_2}{n_1+n_2}\) 

 =\( \frac {200 + 1800}{10 + 30} = \frac {2000}{40} = 50\)

Hence, algebraic mean (AM) of combined data is 50.

Correct option is: D) 50

Correct option is: A) 15

Mid - value of class 10-20 = \(\frac {Upper \, limit + Lower \, limit}{2}\) 

= \(\frac {20+10}{2} = \frac {30}{2} = 15\) 

Hence, mid-value of class 10-20 is 15.

Correct option is: A) 15

Correct option is: A) 28

Empirical relation between mean, median and mode is 

Mode = 3 Median - 2 Mean

\(\Rightarrow\) Median = \(\frac {Mode + 2 \, Mean}{3} = \frac {24.5 + 2\times 29.75}{3}\) 

= \(= \frac {24.5 + 59.5}{3}\)= \(\frac {84}3 = 28\) 

Correct option is: A) 28

D. 51

In a deck of 52 cards, there are 13 cards of heart and 1 is ace of heart.

Hence, the number of outcomes favorable to E = 52 - 1 = 51.

We know that

a = 15

b = 35 – 15 = 20

c = 35 + 12 = 47

d = 50 – 47 = 3

e = 55 – 50 = 5

f = 55 + 5 = 60

g = 60

Frequency Distribution Table

Explanation: Mean \(\bar X = \frac{3+10+10+4+7+10+5}{7}\) 

\(\bar X= \frac{49}{7}\) 

\(\bar X = 7\) 

Mean Deviation = \(\frac{\Sigma d_i}{N}\) 

Mean Deviation = \(\frac{18}{7}\) 

Hence, The MD is 2.57

Mean(y) = a.mean(x) + b 

Therefore, 

55 = a.48 + b 

We can see that only first option satisfies this equation. Therefore, a is the correct answer.

Let x₁,x₂,…x_n be n observation 

And X is the arithemetic mean then, 

We know, Standard deviation 
\(\sigma = \sqrt{\Big(\frac{\Sigma x^2_i}{N}-\Big(\frac{\Sigma x_i}{N}\Big)^2\Big)}\)  

So \(\sigma = \frac{1}{n}\sqrt{(\Sigma x^2_i-\bar X)^2}\) 

(A) The answer is 15 

(D) 66

5^th number = Sum of five numbers – Sum of four numbers

= (5 x 50) – (4 x 46) 

= 250 – 184

= 66

New mean =  \(\cfrac{4000-30+70}{100}\)

= 40.4 

Correct option is (B) 40.4