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16 is not the mode of the data. The mode of a given data is the observation with highest frequency and not the observation with highest value.

No, the above representation is not correct.

Reason:

The classes 0 – 20, 20 – 40, 40 – 60 and 60 – 100 are not of uniform width but of varying widths.

Consider 10 positive integer 

Let Assume, 1,2,3,4,5,6,7,8,9,10 

Now, If we multiply by -1 in each number we get, 

-1,-2,-3,-4,-5,-6,-7,-8,-9,-10 

And then we add 1 in each number 

0,-1,-2,-3,-4,-5,-6,-7,-8,-9 

Now,

Standard deviation Variance = \(\Big(\frac{\Sigma x^2_i}{N} - \Big(\frac{\Sigma x_i}{N}\Big)^2\Big)\)  

Variance = \(\Big(\frac{285}{10} - \Big(\frac{-45}{10}\Big)^2\Big)\) 

Variance = (28.5 - 20.25)

Var = 8.25

Hence, variance is 8.25

Median will be a good representative of the data given in the question, because

a) each value occurs once.

b) the data is influenced by extreme values.

Since the child says that the median of 3, 14, 18, 20, 5 is 18, it is clear that the child doesn’t understand the fact that the given data should be arranged in ascending or descending order before finding the middle term, i.e., median.

Once the child is familiar with the concept,

He/she will understand that,

Arranging the given data in ascending order, we get,

3, 5, 14, 18, 20

And hence,

The median = 14

In finding the median, the observations must be in order. The data is to be arranged either in ascending/descending order to divide the data into two equal groups.

The standard deviation of the data = 4.5

Each data is decreased by 5

The new standard deviation = 4.5

If the standard deviation of a data is 3.6, and each of the data is divided by 3 then the new standard deviation is also divided by 3.

∴ The new standard deviation = \(\frac{3.6}{3}\)

= 1.2

The new variance = (standard deviation)²

= σ² = 1.2² = 1.44

The average of the rain fall per day during that month = \(\frac{Total\,rain\,fall}{Number\,of\,days}\)

= \(\frac{1545}{30}\) = 51.5mm

Mean age = \(\frac{Total\,age}{No\,of\,persons}\)

= \(\frac{2775}{70}\) = 39.64

Average rainfall for 5 years \(=\frac{\text{sum of annual rainfall in five years}}{\text{number of years}}\)

\(=\frac{900+650+450+733+400}{5}\)

\(=\frac{3133}{5}\)

= 626.6

∴ The average rainfall in Vidarbha for 5 years was 626.6 mm.

Mean height = \(\frac{Total\,height}{No\,of\,children}\)

= \(\frac{7892}{50}\) = 157.84 cm

(Average weight of the sacks ) = (sum of weight of each sack)/(number of sacks).

= (49.8+49.7+49.5+49.3+50+48.9+49.2+48.8)/8

= 395.2/8

= 3952/80

= 49.4 

∴ The average weight of the sacks is 49.4 kg.

(Average rainfall during the week) = Sum of rainfall for each day of the week)/(number of days)

= (9 + 11 + 8 + 20 + 10 +10 +16 +12)/7

= 86/7

= 12.285 ≈ 12.29 

∴ The average rainfall during the week is 12.29 mm.

Average \(=\frac{\text{Sum of the number of jumps ons even days}}{\text{Total number of days}}\)

\(=\frac{[60]+[62]+[61]+[60]+[59]+[63]+[58]}{7}\)

\(=\frac{423}{7}\)

= 60.42

∴ Average number of jumps per minute = 60.4

(Average hourly sales) = (sum of sales every hour)/( number of hours)

= (960 + 830 + 945 + 800 + 847 + 970)/6

= 5352/6

= Rs. 892 

∴ The average of the hourly sales was Rs. 892.

Average = (Sum of the number of jumps ons even days)/(Total number of days)

= ([60]+[62]+[61]+[60]+[59]+[63]+[58])/7

= 423/7

= 60.42

∴ Average number of jumps per minute = 60.42

We Know that, 

If any constant is added in each observation, than standard deviation remains same.

∴ The standard deviation of 

a+k,b+k,c+k,d+k and e+k is 'S'.

New variance after multiplying each observation by constant k is k²σ²

∴ New standard deviation of 

ka,kb,kc,kd and ke is

\(=\sqrt{k^2σ^2}\)

\(=kσ\)

\(=ks.\)

(A) is correct answer. If each observation is increased by a constant k, then standard deviation is unchanged

(C) is the correct answer. When each observation is multiplied by 2, then variance is also multiplied by 2.

We have, 

N = 250

\(\frac{N}{2}=\frac{250}{2}=125\)

The cumulative frequency is just greater than \(\frac{N}{2}\) is 127 then median class is 50-60 such that:

l = 50, f = 31, F = 96, h = 60-50 = 10

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

\(=50+\frac{125-96}{31}\times 10\)

\(=50+\frac{29\times10}{31}\)

\(=\frac{445}{31}=59.35\)

From Direct method:

Mean = \(\frac{\sum f_ix_i}{N}\)

\(=\frac{3620}{140}=25.857\)

Assumed mean method: 

let assumed mean (A) = 25

Mean = A + \(\frac{\sum f_iu_i}{N}\)

Mean = A + \(\frac{\sum f_ix_i}{N}\)

\(=25+\frac{120}{140}=25+0.857\)

\(=25.857\)

Step deviation method: Let the assumed mean (A) = 25

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

\(=25+0.857=25.857\)

\(=25+\frac{12}{140}\times10\)

\(=25+0.857=25.857\)

Let the assumed mean (A) = 25

Average number of marks = A + \(\frac{\sum f_iu_i}{N}\)

\(=25+\frac{110}{102}=\frac{2550+110}{102}\)

\(=\frac{2660}{102}=26.08\) (approx)

(D) 26

Explanation:

According to the question,

The minimum and maximum values of given data are 6 and 32 respectively.

Hence,

Range of the data = 32 – 6

= 26

Hence, option (D) is the correct answer.

(B) 10

Explanation:

According to the question,

The given frequency varies from 14 to 112.

So the class intervals are as follows:

13-22, 23-32, 33-42, 43-52, 53-62, 63-72, 73-82, 83-92, 93-102, 103-112.

Number of class interval = 10.

Hence, option (B) is the correct answer.