For Discrete Frequency Distribution MCQ Quiz - Objective Question with Answer for For Discrete Frequency Distribution - Download Free PDF

Explanation:

The mean of 10 natural numbers 

⇒ \(\overline{x} = \frac{1+2+3+4+....10}{10}\) 

= \(\frac{10\times11}{2\times10}\) = 5.5

Mean deviation = \(\frac{[1- 5.5] [2-5.5] [3- 5.5] ... [10- 5.5]}{10} = 2.5\)

∴ Option (b) is correct.

Concept:

Mean deviation = \(\frac{1}{n}|x_i-\bar{X}|\)

Where,

\(\bar{X}= \frac{∑ x}{n}\) = mean, 

∑x = sum of all observations,

n = number of observations. 

Calculation:

First 10 even natural numbers are

2, 4, 6, 8, 10, 12, 14, 16, 18, 20

Mean, 

X̅  = \(\rm \frac{2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20}{10}\) 

⇒ X̅ = 11

Therefore, the mean value is 11

Now, subtract each mean from the first 10 even natural numbers, and ignore the minus symbol if any

⇒ Mean deviation = \(\frac{1}{n}|x_i-\bar{X}|\)

⇒ Mean deviation \(=\frac{1}{10}(|2 - 11|+ |4 - 11| + |6-11|+|8-11|+|10-11|+\\|12-11|+|14-11|+|16-11|+|18-11|+|20-11|\)

⇒  \(\rm\frac{9 + 7 + 5 + 3 + 1 + 1 + 3 + 5 + 7 + 9}{10}\) = \(\frac{50}{10}\)

∴ The mean deviation of the first 10 even natural numbers is 5.

Concept:

Mean deviation = \(\frac{1}{n}|x_i-\bar{X}|\)

Where,

\(\bar{X}= \frac{∑ x}{n}\) = mean, 

∑x = sum of all observations,

n = number of observations. 

Calculation:

First 10 even natural numbers are

2, 4, 6, 8, 10, 12, 14, 16, 18, 20

Mean, 

X̅  = \(\rm \frac{2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20}{10}\) 

⇒ X̅ = 11

Therefore, the mean value is 11

Now, subtract each mean from the first 10 even natural numbers, and ignore the minus symbol if any

⇒ Mean deviation = \(\frac{1}{n}|x_i-\bar{X}|\)

⇒ Mean deviation \(=\frac{1}{10}(|2 - 11|+ |4 - 11| + |6-11|+|8-11|+|10-11|+\\|12-11|+|14-11|+|16-11|+|18-11|+|20-11|\)

⇒  \(\rm\frac{9 + 7 + 5 + 3 + 1 + 1 + 3 + 5 + 7 + 9}{10}\) = \(\frac{50}{10}\)

∴ The mean deviation of the first 10 even natural numbers is 5.

Formula

Median = (n + 1)/2

Calculation

Median = value of (n + 1)/2^th item

⇒ (48 + 1)/2 = 24.5^th item

24.5 lies in cf of 36

∴ Median = 12

Mean deviation about median = (1/48)(36)

⇒ 0.75

∴ Mean deviation about median is 0.75

Formula

Median = (n + 1)/2

Calculation

Median = value of (n + 1)/2^th item

⇒ (48 + 1)/2 = 24.5^th item

24.5 lies in cf of 36

∴ Median = 12

Mean deviation about median = (1/48)(36)

⇒ 0.75

∴ Mean deviation about median is 0.75

Concept:

Mean deviation = \(\frac{1}{n}|x_i-\bar{X}|\)

Where,

\(\bar{X}= \frac{∑ x}{n}\) = mean, 

∑x = sum of all observations,

n = number of observations. 

Calculation:

First 10 even natural numbers are

2, 4, 6, 8, 10, 12, 14, 16, 18, 20

Mean, 

X̅  = \(\rm \frac{2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20}{10}\) 

⇒ X̅ = 11

Therefore, the mean value is 11

Now, subtract each mean from the first 10 even natural numbers, and ignore the minus symbol if any

⇒ Mean deviation = \(\frac{1}{n}|x_i-\bar{X}|\)

⇒ Mean deviation \(=\frac{1}{10}(|2 - 11|+ |4 - 11| + |6-11|+|8-11|+|10-11|+\\|12-11|+|14-11|+|16-11|+|18-11|+|20-11|\)

⇒  \(\rm\frac{9 + 7 + 5 + 3 + 1 + 1 + 3 + 5 + 7 + 9}{10}\) = \(\frac{50}{10}\)

∴ The mean deviation of the first 10 even natural numbers is 5.

Explanation:

The mean of 10 natural numbers 

⇒ \(\overline{x} = \frac{1+2+3+4+....10}{10}\) 

= \(\frac{10\times11}{2\times10}\) = 5.5

Mean deviation = \(\frac{[1- 5.5] [2-5.5] [3- 5.5] ... [10- 5.5]}{10} = 2.5\)

∴ Option (b) is correct.

Formula

Median = (n + 1)/2

Calculation

Median = value of (n + 1)/2^th item

⇒ (48 + 1)/2 = 24.5^th item

24.5 lies in cf of 36

∴ Median = 12

Mean deviation about median = (1/48)(36)

⇒ 0.75

∴ Mean deviation about median is 0.75

Concept:

Mean deviation = \(\frac{1}{n}|x_i-\bar{X}|\)

Where,

\(\bar{X}= \frac{∑ x}{n}\) = mean, 

∑x = sum of all observations,

n = number of observations. 

Calculation:

First 10 even natural numbers are

2, 4, 6, 8, 10, 12, 14, 16, 18, 20

Mean, 

X̅  = \(\rm \frac{2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20}{10}\) 

⇒ X̅ = 11

Therefore, the mean value is 11

Now, subtract each mean from the first 10 even natural numbers, and ignore the minus symbol if any

⇒ Mean deviation = \(\frac{1}{n}|x_i-\bar{X}|\)

⇒ Mean deviation \(=\frac{1}{10}(|2 - 11|+ |4 - 11| + |6-11|+|8-11|+|10-11|+\\|12-11|+|14-11|+|16-11|+|18-11|+|20-11|\)

⇒  \(\rm\frac{9 + 7 + 5 + 3 + 1 + 1 + 3 + 5 + 7 + 9}{10}\) = \(\frac{50}{10}\)

∴ The mean deviation of the first 10 even natural numbers is 5.

Explanation:

The mean of 10 natural numbers 

⇒ \(\overline{x} = \frac{1+2+3+4+....10}{10}\) 

= \(\frac{10\times11}{2\times10}\) = 5.5

Mean deviation = \(\frac{[1- 5.5] [2-5.5] [3- 5.5] ... [10- 5.5]}{10} = 2.5\)

∴ Option (b) is correct.

Concept:

Mean deviation = \(\frac{1}{n}|x_i-\bar{X}|\)

Where,

\(\bar{X}= \frac{∑ x}{n}\) = mean, 

∑x = sum of all observations,

n = number of observations. 

Calculation:

First 10 even natural numbers are

2, 4, 6, 8, 10, 12, 14, 16, 18, 20

Mean, 

X̅  = \(\rm \frac{2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20}{10}\) 

⇒ X̅ = 11

Therefore, the mean value is 11

Now, subtract each mean from the first 10 even natural numbers, and ignore the minus symbol if any

⇒ Mean deviation = \(\frac{1}{n}|x_i-\bar{X}|\)

⇒ Mean deviation \(=\frac{1}{10}(|2 - 11|+ |4 - 11| + |6-11|+|8-11|+|10-11|+\\|12-11|+|14-11|+|16-11|+|18-11|+|20-11|\)

⇒  \(\rm\frac{9 + 7 + 5 + 3 + 1 + 1 + 3 + 5 + 7 + 9}{10}\) = \(\frac{50}{10}\)

∴ The mean deviation of the first 10 even natural numbers is 5.