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Correct option is: B) 50

Total number of scores is n.

\(\therefore\) 2n-1 = 99

\(\Rightarrow\) 2n = 99 +1 = 100

\(\Rightarrow\) n = \(\frac {100}2 = 50 \) which is even

Now, \(\frac n2 = \frac {50}2 = \) 25th &

\(\frac n2 +1 = 25 +1 \) = 26th

Also, 25th score = 2 \(\times\) 25 -1 = 49

26th score = 2 \(\times\) 26 -1 = 51

\(\therefore\) Median of the scores 1, 3, 5, 7......99 is the average of 25th score & 26th score (\(\because\) n = 50 is even)

\(\Rightarrow\) Median = \(\frac {25th \, score + 26th\, score}{2} = \frac {49 + 51}{2} = \frac {100}2 = 50\)  

Hence, median of the scores 1, 3, 5 ....99 is 50.

Correct option is: B) 50

Correct option is: B) 1 → q, 2 → r, 3 → p

Terms are 6, 7, x, 8, y, 14

No of terms = 6

We know that

Mean \(=\frac{Sum\,of\,all\,observations}{No.\,of\,observations}\)

\(\Rightarrow9=\frac{6+7+\text{x}+8+y+14}{6}\)

⇒ 54 = x + y + 35 

⇒ x + y = 19 

Hence, 

correct option is (B)

First data is: 

1, 3, 4, 5, 7, 4 

Given, mean = m 

And we know,

Mean \(=\frac{Sum\,of\,all\,observations}{No\,of\,observations}...[1]\)

No of observations = 6 

Sum of all observations = 1 + 3 + 4 + 5 + 7 + 4 = 24 

Hence, 

Mean \(=\frac{24}{6}=4\)

⇒ m = 4 …[2] 

Second data is : 

3, 2, 2, 4, 3, 3, p 

No of observations = 7 

Sum of observations = 3 + 2 + 2 + 4 + 3 + 3 + p = 17 + p 

Given, 

Mean = m - 1 

Using [1]

\(m-1=\frac{(17+p)}{7}\)

\(4-1=\frac{17+p}{7}\)

\(\Rightarrow3=\frac{17+p}{7}\)

⇒ 21 = 17 + p 

⇒ p = 4 …[3] 

Hence, 

series is 3, 2, 2, 4, 3, 3, 4 

For median, let us write our data in increasing order 

2, 2, 3, 3, 3, 4, 4, 

Also, as the no of terms in this data is odd

We know that if there are odd number of terms in a data, then the median of data is \((\frac{n+1}{2})^{th}\) term. Where n is no of terms 

n = 7

\(\Rightarrow\) \(\frac{n+1}{2}=4\)

median = 4^th term = 3

⇒ q = 3 …[4]

From [3] and [4]

p + q = 4 + 3 = 7

Given: 

The mode of the data: 16, 15, 17, 16, 15, x, 19, 17, 14 is 15. 

To find: 

The value of x.

Solution: 

As we know, mode of any data is the observation which occurs most.

In this case, 17 occurs two times which implies 17 is the mode.But it is given that 15 is the mode of data.In order for 15 to be mode it has to occur more than 2 times.As 15 is already occurring 2 times, the possibility of it occurring more than 2 times is that x should be 15.

⇒ x = 15

Hence, 

correct option is (B).

Correct option is (A) \(A + \frac{\sum f_i \, d_i}{\sum f_i}\)

\(A + \frac{\sum f_i \, d_i}{\sum f_i}\) is a formula to find the mean for a grouped data.

 A) A + \(\frac{∑\hat f_id_i}{\sum f_i}\)

Correct option is: B) lower limit of the median class

Formula of median for grouped data is given by 

Median = \(l+ \frac {\frac n2 - cf}{f} \times h\) 

Where l = lower limit of the median class.

Correct option is: B) lower limit of the median class

Grouped frequency distribution table

For Group A: 

Here, 

the maximum frequency is 78, the corresponding class interval 18 -20 is modal class 

l=18, h=2, f=78, f₁=50, f₂ =46

Mode = I + \(\frac{f-f_1}{2f-f_1-f_2}\times h\)

\(=18+\frac{78-50}{2\times78-50-46}\times2\)

\(=18+\frac{56}{60}\)

=18 + 0.93 =18.93 years

For Group B:

Here, 

the maximum frequency is 89, the corresponding class interval 18 -20 is modal class

l=18, h=2, f=89, f₁=54, f₂ =40

Mode = I + \(\frac{f-f_1}{2f-f_1-f_2}\times h\) 

\(=18+\frac{89-54}{2\times89-54-40}\times 2\)

\(=18+\frac{70}{84}=18+0.83=18.33\)

Hence, 

the modal age of group A is higher than that of group B.

We know that, 

Mean = \(\frac{Σ fx}{ N }\)

= \(\frac{(303 + 9p)}{(41 + p) }\)

Given, 

Mean = 7.68 

⇒ 7.68 = \(\frac{(303 + 9p)}{(41 + p)}\) 

7.68(41 + p) = 303 + 9p 

7.68p + 314.88 = 303 + 9p 

1.32p = 11.88 

∴ p = \(\frac{11.88}{1.32}\) = 9

Given, Numbers of observations are given. 

To Find: Calculate the Mean Deviation

Formula Used: Mean Deviation = \(\frac{\Sigma f|d_i|}{n}\) 

Explanation. Here we have to calculate the mean deviation from the median. 

So, We know, Median in the even terms \(\frac{3+5}{2}\)   , 

Therefore, Median = 4 

Let x_i =Number of calls 

And, f_i = Frequency

N = 245 

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

Mean deviation for given data = \(\frac{336}{245}\) = 1.49

Hence, The mean deviation is 1.49

Let the assumed mean(A) = 3

Mean number of calls = \(A + \frac{Σ f_ix_i }{N }\)

= 3 + \(\frac{135}{250 }\)

= \(\frac{(750 + 135)}{250}\) 

= \(\frac{885}{250}\) 

= 3.54

Let the assumed mean (A) = 3

Mean number of calls = A + \(\frac{\sum f_iu_i}{N}\)

= 3 + \(\frac{135}{250}=\frac{885}{250}\)

= 3.54

Correct option is: A) 17

Arithmetic mean = \(\frac {12+15+13+20+25}5 = \frac {85}5 = 17\)

Correct option is: A) 17

(c) Scaling

The representation of ‘one picture too many objects’ in a Pictograph is called Scaling.

(c) Scaling

The representation of‘one picture to many objects’ in a pictograph is called Scaling.

1. c) 20 units

2. a) 19.64 units

3. c) 20-30 units

4. b) 43.6 units

5. c) 38.75 units

1. c) 36.82

2. d) 45

3. d) 35.4

4. a) 41.4

5. b) 40.2

Median can be find graphically by drawing any of the ogive or both ogives. 

And Mode can be find graphically by drawing histogram of the given data. 

But mean can't be determined graphically.

Correct option is: C) Mode

Mean depends over the sum of the data, so it will change when we add any number to observation.

Median is the mid one data when arranged in ascending or descending order. So, if we add if we add more entry to our data certainly median will change.

Mode is frequently occurring data.

So, if we add more No. of entries to our data, then these may be a wide chances that most frequent occurring remains same.

So, Mode may be more consistent

Correct option is: C) Mode

Given series is,

30, 34, 35, 36, 37, 38, 39, 40

We know that, if even no of terms or observations are given, then the median of data is mean of the values of \((\frac{n}{2})^{th}\) term and \((\frac{n}{2}+1)^{th}\) term. Where n is no of terms.

In this case,no of terms,n = 8

\(\frac{n}{2}=4\) and \(\frac{n}{2}+1=5\)

i.e. median of above data is mean of 4^th and 5^th term

In this case, 

4^th term = 36 

5^th term = 37

⇒ median \(=\frac{36+37}{2}=\frac{73}{2}\)

⇒ median = 37.5

If 35 is removed, the series will be

30, 34, 36, 37, 38, 39, 40

No of terms = 7

We know that if there are odd number of terms in a data, then the median of data is \((\frac{n+1}{2})^{th}\) term. Where n is no of terms

n = 7

\(\Rightarrow \frac{n+1}{2}=4\)

median = 4th term = 37

Difference in both medians = 37.5 - 37 = 0.5

Hence, 

median increases by 0.5.

Since, \(\frac x2\) occurs most frequently in given data.

\(\therefore\) Mode = \(\frac x2\)

But mode of the given data is 9.

\(\therefore\) \(\frac x2\) = 9 \(\Rightarrow\) x = 9 \(\times\) 2 = 18

Hence, the value of x is 18.

Correct option is: D) 18

Correct option is: D) 4.5

Total No of observations is n = 6.

\(\therefore\) \(\frac n2 = \frac 62 = 3^{rd}\) and \(\frac n2 +1 \) = 3 + 1 = 4th

\(\therefore\) Median = \(\frac {3^{rd} \,observation + 4th \, observation}{2}\) = \(\frac {4+5}2 = \frac92 = 4.5\) 

Hence, the median of 2, 3, 4, 5, 6, 7 is 4.5.

Correct option is: D) 4.5

Correct option is: B) Mode

To elect the leader of the class from 3 contestants, we have to considered mode of votes which 3 contestants will get.

Correct option is: B) Mode

Correct option is: B) Median

Correct option is: A) median

The abscissa of the point of intersection of less than type and more than type ogive curves gives median of the data.

Correct option is: A) median

Correct option is: A) 3 

 Since, 3 occurs most frequently in the given data.

\(\therefore\) Mode of the given data is 3.

Correct option is: A) 3

Correct option is: B) 7 

Since, 7 occurs most frequently in the given data.

\(\therefore\) 7 is made of the given data.

Correct option is: B) 7

The correct option is: 

The correct option is:

(C) 64 kg

The correct option is: (A) 64 degree

Explanation:

Let the temperatures on first, second, third, fourth and fifth day be x₁, x₂, x₃, x₄ and x₅, respectively. 

Sum of temperatures of first four days = 58 x 4 

i.e., x₁ + x₂ + x₃ + x₄ = 232       ...(i)

and sum of temperatures on second, third, fourth and fifth day = 60 x 4

i.e., x₂ + x₃ + x₄ + x₅ = 240            ...(ii) 

Subtracting (i) from (ii), we get x₅ - x₁ = 8        ...(iii) 

Also, temperature of first and fifth day were in the ratio 7 : 8. 

Let the temperatures be 7 k and 8 k respectively. 

From (iii), we have 8k - 7 k = 8 => k = 8 

Temperature of fifth day - 8 x 8 = 64 degree

We may calculate class marks (x_i) for each interval by using the relation

x_i \(=\frac{Upper\,class\,limit+lower\,class\,limits}{2}\)

Class size = 50

Now taking 225 as assumed mean we can calculate as follows:

Mean (x̅) = A + \(\frac{\sum f_iu_i}{N}\times h\) 

\(=225+\frac{-7}{25}\times50\)

\(=225-14=211\)

So mean expenditure on food is 211

(60 + 50 + 54 + 46 + 50)/5 = 260/5 = 52

∴ Average number of pages read daily is 52