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Answer is (3) 2 and 1

(i) 16

\(\frac{x+y}{2}\) = 20

x + y = 40 

24 + y = 40 

y = 40 – 24 

= 16

(iv) No mode

Answer is (i) blue

The total number started = 22 members

Number of dropped = 2

Number completed the tour = 20

The amount withdrawn by 2 members = 500

So these five hundred rupees is to be beared by 20 members.

Additional share to each = 500/200 = Rs. 25/-

(3) 15

\(=\frac{1}{9}\times135=15\)

(3) \(\frac{23}{26}\)

\(=1-\frac{3}{26}=\frac{23}{26}\)

We know that, in English alphabets, there are (5 vowels + 21 consonants) = 26 letters. So, total number of outcomes in English alphabets is,n(S) = 26

Let E = Event of choosing an English alphabet, which is a consonant

= {b, c, d, f, g, h, j, k, l, m, n, p, q, r, s t, v, w, x, y, z}

∴ n(E) = 21

Hence, required probability = n(E)/n(S) = 21/26

We know that the minimum observation is 1 and the maximum observation is 24. So the classes of equal size which covers the given data are 0-5, 5-10, 10-15, 15-20 and 20-25

Frequency distribution table

(i) Our class intervals will be 0 − 5, 5 − 10, 10 −15…..

The grouped frequency distribution table can be constructed as follows.

(ii) The number of children who watched TV for 15 or more hours a week is 2 (i.e., the number of children in class interval 15 − 20).

We know that the minimum observation is 0 and the maximum observation is 6. So the classes of equal size which covers the given data are 0-2, 2-4, 4-6 and 6-8.

Frequency distribution table

Here, the marks 8 has the maximum frequency ‘16’.

∴ Mode = 8

D. 13/50

Total numbers of outcomes = 100

So, the prime numbers between 1 to 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 56, 61, 67, 71, 73, 79, 83, 89 and 97.

∴ Total number of possible outcomes = 25

∴ Required probability = 13/50

We know that mean or average of observations, is the sum of the values of all the observations divided by the total number of observations.

and, we have given series

1, 2, 3, …, n

Clearly the above series is an AP(Arithmetic progression) with

first term, a = 1 and

common difference, d = 1

And no of terms is clearly n.

And last term is also n.

We know, sum of terms of an AP if first and last terms are known is:

\(S_n=\frac{n}{2}(a\,+a_n)\)

Putting the values in above equation we have sum of series i.e.

\(1+2+3+...+n=\frac{n}{2}(1+n)=\frac{n(n+1)}{2}...[1]\)

As,

Mean = \(\frac{sum\,of\,all\,terms}{no\,of\,terms}\)

\(=\frac{1+2+3+...+n}{n}\)

⇒ Mean \(=\frac{\frac{n(n+1)}{2}}{n}=\frac{n+1}{2}\)

Terms are x, x + 3, x + 6, x + 9, x + 12

No of terms = 5

We know that

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

\(\Rightarrow=\frac{x+x+3+x+6+x+9+x+12}{5}\)

⇒ 50 = 5x + 30 

⇒ 5x = 20 

⇒ x = 4

Correct option is: C) 1

Given that mean of the data is 4.

\(\therefore\) \(\frac {8+6+4+x+3+6+0}7 = 4\)

\(\Rightarrow\) 27 + x = 7 \(\times\) 4 = 28

= x = 28-27 = 1

Hence, the value of x is 1.

Correct option is: C) 1

i. The crop production of wheat has increased consistently in 3 years. 

ii. The production ofjowar has reduced by 3 quintals in 2012 as compared to 2011. 

iii. The difference between the production of wheat in 2010 and 2012 = 48 – 30 = 18 quintals

iv.

i. Percentage bar diagram 

ii. Sub-divided bar diagram 

iii. Sub-divided bar diagram 

iv. Sub-divided bar diagram 

v. Sub-divided bar diagram 

vi. Sub-divided bar diagram

Answer is (B) 8

Answer is (B) 4

We find that the observation 14 occurs frequently maximum number of times i.e. 4 times. 

Hence, mode is 14.

Arranging the data in an ascending order,

14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28

It can be observed that 14 has the highest frequency, i.e. 4, in the given data.

Therefore, mode of the given data is 14.

When any data has a few observations such that these are very far from the other observations in it, it is better to calculate the median than the mean of the data as median gives a better estimate of average in this case.

(i) Consider the following example − the following data represents the heights of the members of a family.

154.9 cm, 162.8 cm, 170.6 cm, 158.8 cm, 163.3 cm, 166.8 cm, 160.2 cm

In this case, it can be observed that the observations in the given data are close to each other. Therefore, mean will be calculated as an appropriate measure of central tendency.

(ii) The following data represents the marks obtained by 12 students in a test. 48, 59, 46, 52, 54, 46, 97, 42, 49, 58, 60, 99

In this case, it can be observed that there are some observations which are very far from other observations. Therefore, here, median will be calculated as an appropriate measure of central tendency.

(i) Each score is in lesser difference, it is easy to calculate average than median. 

Ex : monthly salary of 5 persons, 10000, 10100, 10200, 10300, 10400 Mean of the scores is 10200. 

(ii) If scores have more difference, it is easy to calculate median than mean. 

Ex.: Marks obtained by 7 students in Mathematics : 2, 10, 20, 15, 4, 23, 3 

Ascending Order : 

2, 3, 4, 10, 15, 20, 23 

Median =10 but mean is 11.

Total sum = 60 × 6 = 360

i. 20, 30, 40, 50, 100, 120

ii. 5, 15, 70, 80, 90, 100

Other ways are also possible.

i. Total weight of 6 players = 60 × 6 = 360 kg 

The team contains players having weights 60, less than 60 or greater than 60. If the weights of all the players are less than 60, the average will also be less than 60. This is not possible. Therefore there will be at least one player having weight greater than 60.

ii. If the weight of all the players are greater than 60, the average will also be greater than 60. Therefore there will be at least one player having weight less than 60.

Average weight of the sacks \(=\frac{\text{sum of weight of each sack}}{\text{number of sacks}}\)

\(=\frac{49.8+49.7+49.5+49.3+50+48.9+49.2+48.8}{8}\)

\(=\frac{395.2}{8}\)

\(=\frac{3952}{80}\)

= 49.4

∴ The average weight of the sacks is 49.4 kg.

i. The mean of 6 numbers is 20.

ie; sum = 6 x 20 = 120 (Write 3 pairs with sum 40)

i.e., 15, 25, 18, 22, 19, 21

ii. The mean of 6 numbers is 15 i.e; sum = 6 x 15 = 90

(Write 30 pairs with sum 3)

12, 18, 13, 17, 14, 16

iii. Mean is 25

sum = 25 x 6 = 150

(Write 50 pairs with sum 3)

22, 28, 23, 27, 24, 26

Arranging the data in ascending order: 1, 1, 2, 2, 2, 3, 3, 3, 5, 5, 5 

Here 2, 3 and 5 occurs 3 times each. 

Which is the maximum number of times. 

∴ Mode is 2, 3 and 5.

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 mean. So, 

Mean = \(\Sigma \frac{f_ix_i}{n}\)  

Here, Mean = \(\frac{13630}{106}\) = 128.6

Therefore, Mean = 49

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

Mean deviation for given data = \(\frac{1314}{106}\) = 12.39

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 mean. So, 

Mean = \(\Sigma \frac{f_ix_i}{n}\) 

Here, Mean =  \(\frac{17900}{50}\) 

Therefore, Mean = 358

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

Mean deviation for given data \(\frac{7896}{50}\) = 157.92 

Hence, The Mean Deviation is 157.92

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,

N = 20 

\(\frac{N}{2}\) = 10

So, The median Corresponding to 10 is 12

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

Mean deviation for given data \(\frac{25}{20}\) = 1.25

Hence, The Mean Deviation is 1.25.

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, Median is the middle term of the X_i, 

Here, The middle term is 25 

Therefore, Median = 25

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

Mean deviation for given data \(\frac{450}{50}\) = 9

Hence, The Mean Deviation is 9