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The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its median.

In the given data 5, 6, 9, 6, 12, 3, 6, 11, 6 and 7 

the frequency of 6 is maximum. 

Hence mode = 6

The given observations are 5, 3, 1,-4, 6,7,0. 

Writing the observations in ascending order, we have – 4, 0, 1, 3, 5, 6, 7. 

There are 7 observations. 

Hence the median will be ( \(\frac{7+1}{2}\) )^th observation i.e., 4^th observation. 

The 4^th observation is 3. 

Hence, the median is 3.

Medain (M) = [ l + \(\cfrac{\frac{n}{2}-c.f}{f}\) ] × h

l = lower limit of the median class, 

n = sum of the frequency 

c.f = cumulative frequency of the class preceding the median class 

f = frequency of the median class 

h = length of the class

Correct option is: C) Frequency

Number of items in a class interval is called Frequency

Correct option is: C) Frequency

B)15

Firstly arrange the ungrouped data in ascending or descending order...

14,14,14,14,15,15,15,15,15,16,17,18,19,19,20

Mode is the most repeated value

From above most repeated value=15

So mode=15....

(B) 15

We first arrange the given data in ascending order as follows 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20 From above, we see that 15 occurs most frequently i.e., 5 times.

Hence, the mode of the given data is 15.

Correct option is: B) Fisher

Fisher is known as father of statistics.

Correct option is: B) Fisher

Correct option is: C) \(\frac {Upper \, boundary + Lower \, boundary}2 \)

Class mark = \(\frac {Upper \, boundary + Lower \, boundary}2 = \frac {U+L}2\)

Correct option is: C) \(\frac{upper\,boundary\,+\,lower\,boundary}{2}\)

Correct option is: A) 6

Ascending order of observations in given data is as follows:

2, 2, 3, 5, 7, 9, 10, 11

n = Number of obs. = 8 which is even

Median = \(\frac {\frac n2 th \, obs. + (\frac n2 +1)th \, obs.}{2}\) 

= \(\frac {4^{th}\, obs. + 5^{th}\, obs.}{2}\)

= \(\frac {5+7}{2} \frac {12}2 = 6\) 

Hence, the median of given data is 6

Correct option is: A) 6

• We need to arrange them in ascending order i.e., 2,2,3,5,7,9,10,11
• since we have 8 number which is even so we need to consider 2 numbers i.e., 4th and 5th number  therefore,5,7 
• then we want to add them i.e, 5+7=12
• next step ,we need to divide them by 2 i.e.,12/2=6

Correct option is: B) n – 1

First n natural numbers are 1, 2, 3, 4,.....,n.

Minimum value = 1. Maximum value = n.

\(\therefore\) Range = Maximum value - Minimum value

= n-1

Correct option is: B) n – 1

Correct option is: C)\(\frac{7}{12}\)

Ascending order of given observations is as follows:

\(\frac 2{12}, \frac 6{12}, \frac 7{12}, \frac 8{12}, \frac 9{12}\) or \(\frac 1{6}, \frac 1{2}, \frac 7{12}, \frac 2{3}, \frac 7{12}\)

Total No of observations is n = 5 which is odd

\(\therefore\) Median = \((\frac {n+1}2)^{th}\) observation

= \(3^{rd}\) observation (\(\because\) n = 5)

= \(\frac 7{12}\)

Hence, \(\frac 7{12}\) is median of given data.

Correct option is: C) \(\frac{7}{12}\)

Correct option is: A) 5

1 to 9 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9.

Total No of obs. is n = 9 which is odd.

\(\therefore\) Median = \((\frac {n+1}2)^{th}\) obs.

= \(5^{th} \) obs. (\(\because\) n = 9 \(\Rightarrow\) \(\frac {n+1}2\) = \(\frac {9+1}2\) = 5)

= 5

Correct option is: A) 5

1. frequency
2. proportional
3. Histogram
4. grouped

(i) 330 people (110+ 220)

(ii) 150 of them (100 + 50)

(iii) No.

(iv) People should minimise the usage of cell phones.

Correct option is: B) 7

Ascending order of observations in the given data is as follows:

1, 3, 6, 8, 12, 16

Total number of observations is n = 6

\(\therefore\) Median = \(\frac {\frac n2^{th} \, obs. +(\frac n2 +1)^{th}\, obs. }{2}\)

= \(\frac {3^{rd}\, obs. + 4^{th} \, obs.}2\)

= \(\frac {6+8}2 = \frac {14}2 = 7\)

Correct option is: B) 7

Correct option is: D) doesn’t exist

There is no mode when all observed values appear the same number of time in a data set.

Here, 1 & 5 are only observations in the data and both appears 4 times.

So, mode of data 1, 1, 1,1, 5, 5, 5, 5 does not exists.

Correct option is: D) doesn’t exist

Clearly, the value 225 occurs mavimum number of times So, the modal wage is  ₹225

As we know that, the x-coordinate of the point of intersection of the more than ogive and less than ogive give us median of the data.

So, 

median can be determined graphically

We know that,

Mode = 3 Median – 2 Mean

(i) Mode = 3 

(ii) Mode = 22

1. The given graph is a percentage bar graph. 

2. Percent of tur production to the total production in Ajita’s farm is 60%. 

3. Production of Gram in the farm of Yash = 50% 

Production of Gram in the farm of Ravi = 30% 

∴ Difference in the production = 50% – 30% = 20% 

∴ Yash’s production of Gram is more and by 20%. 

4. Sudha’s percentage production of Tur is the least. 

5. Production percentages of Tur and Gram in Sudha’s farm are 40% and 60% respectively.

i. The given graph is a subdivided bar graph. 

ii. Vaishali’s savings in the month of April is Rs 600. 

iii. Total savings of Saroj in the months of March and April is Rs 800. 

iv. Savita’s total saving = Rs 1000, Megha’s total saving 

= Rs 500.

∴ difference in their savings = 1000 – 500 = Rs 500.

Savita’s saving is Rs 500 more than Megha. 

v. Megha’s savings in the month of April is the least.

Correct option is: D) Range

Arithmetic mean, median & mode are measures of the central tendency.

So, extreme values does not much affects on these measures.

But Range = Highest observation - Lowest observation.

Hence, range depends on the extreme values.

Thus, the extreme values in the data affects range the most.

Correct option is: D) Range

Correct option is: A) median

Constructing of a cumulative frequency table is useful in finding median.

Correct option is: A) median

Median

 As we know that, the x-coordinate of the point of intersection of the more than ogive and less than ogive give us median of the data.

Correct option is: A) Mean

Mean is affected by extreme values because mean affects by changing in any observation of the data.

Correct option is: A) Mean

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

So, We know, Median is the Middle term, 

Therefore, Median = 61 

Let x_i =Heights in inches 

And, f_i = Number of students

N=197 

Mean deviation= \(\frac{336}{197}\) = 1.70

Hence, The mean deviation is 1.70.

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, Here, N =50 

Then, \(\frac{N}{2} = \frac{50}{2}\) = 25

So, The median Corresponding to 25 is 13

N = 50 

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

Mean deviation for given data = \(\frac{126}{50}\) = 2.72  

Hence, The mean Deviation is 2.72.

(A) 2,16,000

Measure of the central angle = (Expenditure of cement) / (Total expenditure ) x 360^∘

∴ Total expenditure = (45000 x 360^∘)/ 75^∘ = Rs. 2,16,000

(D) 144°

Measure of the central angle = 40/100 × 360° = 144°

1. The given graph is a percentage bar graph. 

2. Percent of tur production to the total production in Ajita’s farm is 60%. 

3. Production of Gram in the farm of Yash = 50%

Production of Gram in the farm of Ravi = 30% 

∴ Difference in the production = 50% – 30% =20%

∴ Yash’s production of Gram is more and by 20%.

4. Sudha’s percentage production of Tur is the least. 

5. Production percentages of Tur and Gram in Sudha’s farm are 40% and 60% respectively.

Lives in hours of 15 pieces are = 715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719 Arrange the above in ascending order:

694, 696, 699, 705, 710, 712, 716, 719, 724, 725, 728, 729, 734, 745

N = 15 (Odd)

Median = \((\frac{N+1}{2})\) Term

\(=(\frac{15+1}{2})\) Term

= 16th Term = 716

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

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

Class size (h) for this data = 10

Now taking 70 as assumed mean (A) we can calculate as follows:

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

\(=70+\frac{-2}{35}\times10\)

\(=70-0.57\)

= \(69.43\)

So, 

mean literacy rate is 69.437