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Correct option is (B) Median

Cumulative frequency is used in finding median of the data.

Correct option is  B) Median

Correct option is  A) 7

(B) Centered at the class marks of the classes

Explanation:

In computing the mean of grouped data, the frequencies are centered at the class marks of the classes.

Hence, the option (B) is correct

Correct answer is

(B) Median

Let a, b, c, d, e be the observation and mean is m

m = \(\frac{a+b+c+d+e}{5}\) 

Let suppose new mean be m₁

m₁ = \(\frac{a+k+b+k+c+k+d+k+e+k}{5}\) 

m₁ =  \(\frac{5k}{5}+\frac{a+b+c+d+e}{5}\) 

M₁ = m + k

Now, The standard deviation

S = \(\sqrt{\frac{(a-m)^2+(b-m)^2+(c-m)^2+(e-m)^2}{5}}\) 

So, The standard deviation for new observation

S₁ = 

\(\sqrt{\frac{(a+k-n)^2+(b+k-n)^2+(c+k-n)^2+(e+k-n)^2+(d+k-n)^2}{5}}\) 

Now, we can compare both observation 

a+k-n=a+k-(m+k) 

a+k–n = a+k–m- k 

a+k-n=a-m

Similary 

b+k-n=b-m 

c+k-n=c-m 

d+k-n=d-m 

e+k-n=e-m 

when we substitute the values, we get, 

S₁ = S 

Hence, The Sd is S

First 10 natural numbers are 1,2,3,4,5,6,7,8,9,10 

So, N =10

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

\(\sigma =\sqrt{\Big(\frac{385}{10} - \Big(\frac{55}{10}\Big)^2\Big)}\) 

\(\sigma = \sqrt{38.5-30.25}\) 

\(\sigma = \sqrt{8.25}\) 

Hence, SD is 2.87

It is correct. Since the 2nd data is obtained by multiplying each observation of 1st data by 2, therefore, the mean will be 2 times the mean of the 1st data.

Consider numbers are 1,2,3,4,5,6,7,8,9,10 

If one is added to each number then, numbers will be 

Let say x_i = 2,3,4,5,6,7,8,9,10,11

 So, N= 10

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

Variance = \(\Big(\frac{505}{10} - \Big(\frac{65}{10}\Big)^2\Big)\)  

Var = 8.25

Hence, the variance is 8.25

No. It is true only when the class sizes are the same.

Given, The data is given in table

To Find: Find the standard deviation 

The formula used: SD = \(\sqrt{Var(X)}\) 

Explanation:

Now, N = 60, Σx_if_i = 277, Σx_i²f_i = 1393

Mean \(\bar{X}\) = \(\Big(\frac{\Sigma x_if_i}{N}\Big)\) 

  \(\bar{X}\) = \(\frac{277}{60}\) 

  \(\bar{X}\) = 4.62

Var (X) = \(\frac{1393}{60}\) - (4.62)²

Variance = 1.88 

Standard Deviation σ = \(\sqrt{1.88}\) 

SD = 1.37

Hence, The standard deviation is 1.37

Correct option is (D) 5 and 3

Since 3 and 5 occur most often.

\(\therefore\) 3 and 5 are mode of given data.

Hence, it is a bi-modal data.

Correct option is  D) 5 and 3

B) secondary data

Correct option is (B) 42

Smallest observation = 3 and

highest observation = 45

\(\therefore\) Range = 45 - 3 = 42

Correct option is  B) 42

Correct option is (C) 20.5

Classes are 0–10, 11–20, 21–30, 31–40

Difference between lower limit of next class and upper limit of class = 1.

\(\therefore\) Classes converts to 0.5-10.5, 10.5-20.5, 20.5-30.5, 30.5-40.5

\(\therefore\) Real lower limit of the class 21-30 is 20.5.

Correct option is  C) 20.5