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(B) Median

Explanation:

Since, the intersection point of less than ogive and more than ogive gives the median on the abscissa, 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

Hence, option (B) is correct

a. 5 students scored 15 marks. 

b. 20 students scored more than 15 marks. 

c. 25 students scored less than 15 marks. 

d. 6 is the lowest score of the group. 

e. 20 is the highest score of the group.

The difference between upper and lower boundary of a class is called interval or size of the class.

Some of the graphical methods of representing numerical data are

• Histogram
• Frequency polygon
• Ogive curve

The number of tally marks corresponding to a class is called the frequency of the class.

Statistics is a branch of Mathematics which deals with collection, organisation, analysis and interpretation of data.

Statistics deals mainly in the communication and analysis of facts and figures using statistical methods. Collection, classification, tabulation, representation, reasoning, testing and drawing inferences are parts of the statistical method. Graphs, tables, reasoning, estimation and prediction are the means of stastical methods.

Range = 36.8

Smallest value (S) = 13.4

Range = L – S

36.8 = L – 13.4

L = 36.8 + 13.4 = 50.2

Largest value = 50.2

(1) 0

Range = L – S = 8 – 8 = 0
 

Answer is (3) zero

Given, Numbers of observations are given.

To Find: Calculate the Mean Deviation from the mean. 

Formula Used: Mean Deviation = \(\frac{\Sigma f|d_i|}{n}\) 

Explanation. 

Here we have to calculate the mean deviation from Mean 

So, Mean = \(\frac{\Sigma f_ix_i}{f_i}\) 

Mean of the given data is  \(\frac{433}{20}\)  = 21.65

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

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

Hence, The mean Deviation is 1.25.

Given, Numbers of observations are given. 

To Find: Calculate the Mean Deviation from the mean. 

Formula Used: Mean Deviation = \(\frac{\Sigma f|d_i|}{n}\) 

Explanation. 

Here we have to calculate the mean deviation from Mean 

So, Mean = \(\frac{\Sigma f_ix_i}{f_i}\)  

Mean of given Data = \(\frac{4000}{80}\) =50

Mean =  \(\frac{4000}{80}\)  = 50

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

Mean deviation for given data = \(\frac{1280}{80}\) = 16

Hence, The mean Deviation is 16

Given, Numbers of observations are given. 

To Find: Calculate the Mean Deviation from the mean. 

Formula Used: Mean Deviation =  \(\frac{\Sigma f|d_i|}{n}\) 

Explanation. Here we have to calculate the mean deviation from Mean 

So, Mean = \(\frac{\Sigma f_ix_i}{f_i}\) 

Mean = \(\frac{350}{25}\) =14 

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

Mean deviation for given data = \(\frac{158}{25}\) = 6.32

Hence, The mean Deviation is 6.32.

Given, Numbers of observations are given. 

To Find: Calculate the Mean Deviation from the mean. 

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

Explanation. 

Here we have to calculate the mean deviation from Mean 

So, Mean = \(\frac{\Sigma f_ix_i}{f_i}\)  

Mean = \(\frac{350}{25}\) =14 

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

Mean deviation for given data = \(\frac{158}{25}\) = 6.32

Hence, The mean Deviation is 6.32.

Given, Numbers of observations are given. 

To Find: Calculate the Mean Deviation from the mean. 

Formula Used: Mean Deviation = \(\frac{\Sigma f|d_i|}{n}\) 

Explanation. 

Here we have to calculate the mean deviation from Mean 

So, Mean = \(\frac{\Sigma f_ix_i}{f_i}\) 

Mean = \(\frac{234}{26}\) = 9 

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

Mean deviation for given data = \( \frac{88}{26}\) = 3.3

Hence, The mean Deviation is 3.3.

Let the assumed mean (A) = 25

Mean = A + \(\frac{\sum f_iu_i}{N}\)

\(=25+\frac{-1170}{400}=\frac{1000-1170}{400}\)

\(=22.075\) 

Let the assumed mean (A) = 2

Average number of accidents per day workers = A + \(\frac{\sum f_iu_i}{N}\)

\(=2+(-\frac{187}{160})=\frac{320-187}{160}\)

\(=0.83\) 

To find : the average number of misprints per page.

Use the shortcut method to find the mean of given data.For that,Let the assumed mean be (A) = 2,The deviation of values xi from assumed mean be d_i = x_i – A.

Now to find the mean:First multiply the frequencies in column (ii) with the value of deviations in column (iii) as f_id_i.

Now add the sum of all entries in column (iii) to obtain \(\displaystyle\sum_{i=1}^{n} f_id_i\)

and the sum of all frequencies in the column (ii) to obtain \(\displaystyle\sum_{i=1}^{n} f_id_i=N\)

So,

Average number of misprints per day = A + \(\frac{\sum f_id_i}{N}\)

where, N = total number of observations

⇒ Mean = 2 + \(\frac{-381}{300}\)

Mean = \(\frac{600-381}{300}\)

Mean = \(\frac{219}{300}\)

Mean = 0.73

(i) Amount of monthly wages paid by firm A. 

= 586 x mean wages 

= 586 x 5253 = Rs.3078258 

And amount of monthly wages paid by form B 

= 648 x mean wages = 648 x 5253 = Rs. 3403944. 

Clearly, firm B pays more wages. 

Alternatively, 

The variance of wages in A is 100. 

∴ S.D. of distribution of wage in A = 10 

Also the variance of distribution of wages in firm B is 121. 

∴ S.D. of distribution of wages in B = 11. 

since the average monthly wages in both firms is same i.e., Rs. 5253, therefore the firm with greater S.D. will have more variability. 

∴ Firm B pays more wages. 

(ii) Firm B with greater S.D. shows greater variability in individual wages.

Correct answer is

(A) 2,16,000

Measure of the central angle

\(=\frac{\text{Expenditure of cement}}{\text{Total expenditure}} \times 360^\circ\)

\(\therefore\) Total expenditure = \(\frac{45000 \times 360^\circ}{75^\circ}\) = Rs. 2,16,000

Correct option is (C) 10

The width of the class interval 40–50

= 50 - 40 = 10

Correct option is   C) 10

We know, 

M.D = \(\frac{\Sigma fd}{\Sigma f}\) 

Variance = \(\Big(\frac{\Sigma fd^2}{\Sigma f}-\Big(\frac{\Sigma fd}{\Sigma f}\Big)^2\Big)\) 

SD \(\sigma = \sqrt{Variance}\)  

Hence,   \(\sigma = \sqrt{\Big(\frac{\Sigma fd^2}{\Sigma f}- \Big(\frac{\Sigma fd}{\Sigma f}\Big)^2}\Big)\) 

Correct option is (A) Assumed mean

In formula \(\overline X = A + \frac{\sum f_i d_i}{\sum f_i} ,\) letter A represents for assumed mean.

A) Assumed mean

Correct option is (A) 6

Mean \(=\frac{\text{Sum of first five even natural numbers}}5\)

\(=\frac{2+4+6+8+10}5\)

\(=\frac{30}5=6\)

Correct option is  A) 6

Correct option is (B) 17

Ascending order of given observations is 7, 10, 15, X, 27, 30.

Median = 16   (given)

\(\therefore\) \(\frac{15+X}2=16\)

\(\Rightarrow\) 15 + X = 32

\(\Rightarrow\) X = 32 - 15 = 17

Correct option is  B) 17

Correct option is (D) 83

Average pulse \(=\frac{72+78+90+102+73}5\)

\(=\frac{415}5\) = 83

Correct option is  D) 83

Statistical methods are commonly used for analyzing experiments results, and testing their significance in biology, physics, chemistry, mathematics, meteorology, research, chambers of commerce, sociology, business, public administration, communications and information technology, etc.

The statistical problems in real life consist of sampling, inferential statistics, probability, estimating, enabling a team to develop effective projects in a problem-solving frame. For instance, car manufacturers looking to paint the cars might include a wide range of people that include supervisors, painters, paint representatives, or the same professionals to collect the data, which is necessary for the whole process and make it successful.

Correct option is: C) continuous

To find median, classes should be continuous.

Correct option is: C) continuous

Correct option is: B) \((\frac{n+1}{2})^{th}\)

For a distribution with odd numbers (n) of observations, the median is \((\frac {n+1}2)^{th}\) observation  in ascending or descending order of observations.

Correct option is: B) \((\frac{n+1}{2})\)^th

(i) 4, 7, 8, 9, 10, 12, 13, 17

We know that,

MD = 1/n ∑ⁿ_i=1|d_i|

Where, |d_i| = |x_i – x|

So, let ‘x’ be the mean of the given observation.

x = [4 + 7 + 8 + 9 + 10 + 12 + 13 + 17]/8

= 80/8

= 10

Number of observations, ‘n’ = 8

MD = 1/n ∑ⁿ_i=1|d_i|

= 1/8 × 24

= 3

∴ The Mean Deviation is 3.

(ii) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

We know that,

MD = 1/n ∑ⁿ_i=1|d_i|

Where, |d_i| = |x_i – x|

So, let ‘x’ be the mean of the given observation.

x = [13 + 17 + 16 + 14 + 11 + 13 + 10 + 16 + 11 + 18 + 12 + 17]/12

= 168/12

= 14

Number of observations, ‘n’ = 12

MD = 1/n ∑ⁿ_i=1|d_i|

= 1/12 × 28

= 2.33

∴ The Mean Deviation is 2.33.

(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

We know that,

MD = 1/n ∑ⁿ_i=1|d_i|

Where, |d_i| = |x_i – x|

So, let ‘x’ be the mean of the given observation.

x = [38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44]/10

= 500/10

= 50

Number of observations, ‘n’ = 10

MD = 1/n ∑ⁿ_i=1|d_i|

= 1/10 × 84

= 8.4

∴ The Mean Deviation is 8.4.