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(B) 40.4

New mean = (4000 − 30 + 70)/100

= 40.4

proof:

Given that, n = 2m + 1

∴ The sequence of the terms of scores can be 1, 2, 3, ...., 2m + 1

Here,  (n +1)/2 = (2m + 1 + 1)/2 

= (2m + 2)/2 = 2(m + 1)/2 

= m + 1

The sequence of the terms of scores is 

1,2, 3, …….., m, m + 1, m + 2, …, 2m + 1 

Thus, we have to prove that m + 1 is the middle term if the number of scores is 2m + 1 

i.e. to prove 

number of terms from 1 to m = number of terms from m + 2 to 2m + 1 …(i) 

Consider the L.H.S. of equation (i) 

The sequence is an A.P. with a = 1,d = 1,

t_n1 = m 

t_n1 = a + (n₁ – 1) d 

∴ m = 1 + (n₁ – 1)1 

∴ m = 1 + n₁ – 1 

∴ m = n₁ 

Consider the R.H.S. of equation (ii)

The sequence is an A.P. with a = m + 2, d = 1, t_n2 = 2m + 1 

t_n2 = a + (n₂ – 1)d 

∴ 2m + 1 = m + 2 + (n₂ – 1)1 

∴ 2m + 1 = m + n₂ + 1 

∴ m = n₂ 

∴ number of terms from 1 to m = number of terms from m + 2 to 2m + 1 = m = (n-1)/2

∴ m + 1 is the middle term if the number of scores is 2m + 1.

Given data in ascending order: 

50, 60, 65, 70, 70, 80 85, 90, 95, 95 

∴ Number of observations (n) = 10 (i.e., even) 

∴ Median is the average of middle two observations 

Here, 5th and 6th numbers are in the middle position 

∴ Median = (70 + 80)/2

∴ Median = 150/2

∴ The median of the weights of tomatoes is 75 grams.

The correct answer is : (C) 16 – 18 

Correct option is (B) 16 Kgs

Average weight of 15 packets \(=\frac{5\times24+10\times12}{15}\)

\(=\frac{120+120}{15}\)

\(=\frac{240}{15}\) = 16 kg

Correct option is  B) 16 Kgs

 (C) (5,8)

Class mark = 5 

Frequency = 8 

∴ Co-ordinates of the point = (5, 8)

i. The class 60 – 70 has the maximum number of students. 

ii. The classes 20 – 30 and 90 – 100 have frequency zero. 

iii. The class mark of the class having 50 students is 55. 

iv. The lower and upper class limits of the class having class mark 85 are 80 and 90 respectively

v. There are 15 students in the class 80 – 90.

i. Primary data 

ii. Primary data

By observing the number of girls per 1000 boys and literacy percentages in the given respective states, we can conclude that the literacy rate of girls is least in Bihar and is highest in Kerala.

Lower class limit = 20 

Upper class limit = 25

i. Primary data 

ii. Secondary data 

iii. Primary data 

iv. Primary data

i. Primary data 

ii. Secondary data

We may find class mark of each interval (xi) by using the relation.

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

Class size of this data = 3

Now taking 75.5 as assumed mean (A), we may calculate d_i, u_i, f_iu_i as following:

Now we may observe from table that ∑f_i = 30, ∑f_iu_i = 4

Mean (X̅) = di + \(\frac{\sum f_ iu_i }{\sum f_i}\times h\)

\(=75.5\,+\frac{4}{30}\times3\)

\(=75.5 \,+0.4 = 75.9\) 

The demand for accounting information by investors, lenders, creditors, etc., creates fundamental qualitative characteristics that are desirable in accounting information. There are six qualitative characteristics of accounting information. Two of the six qualitative characteristics are fundamental (must have), while the remaining four qualitative characteristics are enhancing.

The qualitative characteristics of accounting information are as follows:-

1. Reliability-The first qualitative characteristic of accounting information is reliability. Reliability means the users must be able to depend on the information. It is believed that reliable information should be free from error and bias and faithfully represents what it is meant to represent.
2. Relevance-The second qualitative characteristic of accounting information is relevance. It is believed that relevant, information must be available in time, must help in prediction and feedback and must influence the decisions of users in a positive manner.
3. Understandability- Understandability is the third most important qualitative characteristic of accounting information. Understandability means decision-makers must interpret accounting information in the same sense as it is prepared and conveyed to them.
4. Comparability-The last qualitative characteristic of accounting information is comparability. It is believed that it is not sufficient that the financial information is relevant and reliable at a particular time, in a particular circumstance or for a particular reporting entity. 

Mean = \(\frac{\sum yx}{N}\)

\(=\frac{698}{40}=17.45\)

Therefore, 

mean age is 17.45 years.

Correct option is (D) 40°

Average temperature \(=\frac{\text{Sum of temperatures in a week}}7\)

\(=\frac{42^\circ+35^\circ+44^\circ+33^\circ+39^\circ+45^\circ+42^\circ}7\)

\(=\frac{280^\circ}7=40^\circ\)

Correct option is  D) 40°

We know, 

class marks of a class interval is

\(=\frac{1}{2}(lower\,limit+upper\,limit)\)

For 10 - 25 

Lower limit = 10 

Upper limit = 25

Class mark = \(\frac{1}{2}(10+25)=17.5\)

For 35 - 55 

Lower limit = 35 

Upper limit = 55

Class mark = \(\frac{1}{2}(35+55)=45\)

Correct option is (B) 144.1 cm

The mean height of 10 boys \(=\frac{3\times142+7\times145}{10}\)

\(=\frac{426+1015}{10}\)

\(=\frac{1441}{10}\) = 144.1 cm

Correct option is  B) 144.1 cm

Correct option is: C) 7^th

\(\because\) n = Number of observations = 13 which is odd

\(\therefore\) \(\frac {n+1}{2} = \frac {13+1}{2} = \frac {14}2 = 7^{th}\) observation is median of the given data.

Hence, \(7^{th}\) observation arranged in ascending or descending order is median of the data.

Correct option is: C) 7th

The correct option is: (C) 14 

Explanation:

The incorrect sum of observations = 12.5 x 20 = 250

Now, correct sum = 250 - (-15) + 15 = 280

.'. Correct mean = 280/20 = 14