Statistics Class 11 Notes AP Inter 1st Year Maths Chapter 13

Students can go through AP Inter 1st Year Maths Notes 13th Lesson Statistics will help students in revising the entire concepts quickly.

Statistics Class 11 Notes AP Inter 1st Year Maths 13th Lesson

→ Measures of dispersion: Range, Quartile deviation, mean deviation, variance, and standard deviation are measures of dispersion.

→ Range = Maximum Value – Minimum Value

→ Mean deviation for ungrouped data

→ Mean deviation for grouped data

→ Variance and standard deviation for ungrouped data

→ Variance and standard deviation of a discrete frequency distribution
σ² = \(\frac{1}{\mathrm{~N}} \sum f_{\mathrm{i}}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2\)
σ = \(\sqrt{\frac{1}{\mathrm{~N}} \sum f_{\mathrm{i}}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2}\)

→ Variance and standard deviation of a continuous frequency distribution

→ Shortcut method to find variance and standard deviation.

Example Problems

Question 1.
Find the mean deviation about the mean for the following data:
6, 7, 10, 12, 13, 4, 8, 12
Solution:
We proceed step-wise and get the following:
Step 1: The Mean of the given data is
\(\bar{x}=\frac{6+7+10+12+13+4+8+12}{8}=\frac{72}{8}\) = 9
Step 2: The deviations of the respective observations from the mean \(\overline{\mathbf{x}},\),
i.e., x_i – \(\overline{\mathbf{x}},\) are
6 – 9, 7 – 9, 10 – 9, 12 – 9, 13 – 9, 4 – 9, 8 – 9, 12 – 9
or -3, -2, 1, 3, 4, -5, -1, 3
Step 3: The absolute values of the deviations, i.e., \(\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|\) are 3, 2, 1, 3, 4, 5, 1, 3
Step 4: The required mean deviation about the mean is

Question 2.
Find the mean deviation about the mean for the following data:
12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5
Solution:
We have to first find the mean (\(\overline{\mathrm{x}}\)) of the given data
\(\bar{x}=\frac{1}{20} \sum_{i=1}^{20} x_i=\frac{200}{20}\) = 10
The respective absolute values of the deviations from mean,
i.e., \(\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|\) are 2, 7, 8, 7, 6, 1, 7, 9, 10, 5, 2, 7, 8, 7, 6, 1, 7, 9, 10, 5
Therefore \(\sum_{i=1}^{20}\left|x_i-\bar{x}\right|\) = 124
and M.D. (\(\overline{\mathrm{x}}\)) = \(\frac {124}{20}\) = 6.2

Question 3.
Find the mean deviation about the median for the following data:
3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21
Solution:
Here, the number of observations is 11, which is an odd number.
Arranging the data into ascending order, we have 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21
Now Median = \(\left(\frac{11+1}{2}\right)^{\mathrm{th}}\) or 6th observation = 9
The absolute values of the respective deviations from the median, i.e., |x_i – M|, are 6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12
Therefore \(\sum_{i=1}^{n 1}\left|x_i-M\right|\) = 58
and M.D.(M) = \(\frac{1}{11} \sum_{i=1}^{11}\left|x_i-M\right|\)
= \(\frac {1}{11}\) × 58
= 5.27

Question 4.
Find the mean deviation about the mean for the following data:

Solution:
Let us make a table of the given data and append other columns after calculations.

Question 5.
Find the mean deviation about the median for the following data:

Solution:
The given observations are already in ascending order.
Adding a row corresponding to cumulative frequencies to the given data, we get the table

Now, N = 30, which is even.
Median is the mean of the 15th and 16th observations.
Both of these observations lie in the cumulative frequency 18, for which the corresponding observation is 13.
Therefore, Median (M) = \(\frac{15^{\text {th }} \text { observation }+16^{\text {th }} \text { observation }}{2}=\frac{13+13}{2}\) = 13
Now, absolute values of the deviations from the median, i.e., |x_i – M|, are shown in the Table.

Question 6.
Find the mean deviation about the mean for the following data:

Solution:
We make the following Table from the given data:

Question 7.
Calculate the mean deviation about the median for the following data:

Solution:
Form the following table from the given data:

The class interval containing \(\frac{\mathrm{N}^{\text {th }}}{2}\) or 25th item is 20-30.
Therefore, 20-30 is the median class.
We know that Median = \(l+\frac{\frac{\mathrm{N}}{2}-\mathrm{C}}{f} \times \mathrm{h}\)
Here l = 20, C = 13, f = 15, h = 10 and N = 50
Therefore, Median = 20 + \(\frac{25-13}{15}\) × 10
= 20 + 8
= 28
Thus, the Mean deviation about the median is given by
M.D (M) = \(\frac{1}{\mathrm{~N}} \sum_{\mathrm{i}=1}^6 f_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|\)
= \(\frac {1}{50}\) × 508
= 10.16

Question 8.
Find the variance of the following data:
6, 8, 10, 12, 14, 16, 18, 20, 22, 24
Solution:
From the given data, we can form the following Table.
The mean is calculated by the step deviation method, taking 14 as the assumed mean.
The number of observations is n = 10

Therefore Mean(\(\bar{x}\)) = Assumed mean + \(\frac{\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{~d}_{\mathrm{i}}}{\mathrm{n}} \times \mathrm{h}\)
= 14 + \(\frac {5}{10}\) × 2
= 15
and Variance (σ²) = \(\frac{1}{n} \sum_{i=1}^{10}\left(x_i-\bar{x}\right)^2\)
= \(\frac {1}{10}\) × 330
= 33
The Standard deviation (σ) = √33 = 5.74

Question 9.
Find the variance and standard deviation for the following data:

Solution:
Presenting the data in tabular form (Table), we get

Question 10.
Calculate the mean, variance, and standard deviation for the following distribution:

Solution:
From the given data, we construct the following Table.

Question 11.
Find the standard deviation for the following data:

Solution:
Let us form the following Table:

Now, by formula (3), we have

Therefore, Standard deviation (σ) = 6.12

Question 12.
Calculate mean, variance, and standard deviation for the following distribution.

Solution:
Let the assumed mean A = 65.
Here h = 10
We obtain the following Table from the given data:

Question 13.
The variance of 20 observations is 5. If each observation is multiplied by 2, find the new variance of the resulting observations.
Solution:
Let the observations be x₁, x₂, …., x₂₀ and \(\bar{x}\) be their mean.
Given that variance = 5 and n = 20.
We know that

If each observation is multiplied by 2, and the new resulting observations are y_i, then

Thus the variance of new observations = \(\frac {1}{20}\) × 400 = 20 = 22 × 5

Question 14.
The mean of 5 observations is 4.4, and their variance is 8.24. If three of the observations are 1, 2, and 6, find the other two observations.
Solution:
Let the other two observations be x and y.
Therefore, the series is 1, 2, 6, x, y
Now Mean (\(\bar{x}\)) = 4.4 = \(\frac{1+2+6+x+y}{5}\)
Or 22 = 9 + x + y
Therefore x + y = 13 ……….(1)
Also variance = 8.24 = \(\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^5\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2\)
i.e., 8.24 = \(\frac {1}{5}\)[(3.4)² + (2.4)² + (1.6)² + x² + y² – 2 × 4.4 (x + y) + 2 × (4.4)²]
or 41.20 = 11.56 + 5.76 + 2.56 + x² + y² – 8.8 × 13 + 38.72
Therefore x²+ y² = 97 ……(2)
But from (1), we have
x² + y² + 2xy = 169 ……(3)
From (2) and (3), we have
2xy = 72 ……(4)
Subtracting (4) from (2), we get
x² + y² – 2xy = 97 – 72
i.e., (x – y)² = 25
or x – y = ±5 …….(5)
So, from (1) and (5), we get
x = 9, y = 4 when x – y = 5
or x = 4, y = 9 when x – y = -5
Thus, the remaining observations are 4 and 9.

Question 15.
If each of the observations x₁, x₂,…, x_n is increased by a, where a is a negative or positive number, show that the variance remains unchanged.
Solution:
Let \(\bar{x}\) be the mean of x₁, x₂,…, x_n
Then the variance is given by
\(\sigma_1^2=\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{n}}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2\)
If a is added to each observation, the new observations will be
y_i = x_i + a ……(1)
Let the mean of the new observations be \(\bar{y}\). Then

Thus, the variance of the new observations is the same as that of the original observations.

Question 16.
The mean and standard deviation of 100 observations were calculated as 40 and 5.1, respectively, by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation?
Solution:
Given that the number of observations (n) = 100
Incorrect mean (\(\bar{x}\)) = 40
Incorrect standard deviation (σ) = 5.1
We know that \(\overline{\mathrm{x}}=\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}\)
i.e., 40 = \(\frac{1}{100} \sum_{\mathrm{i}=1}^{100} \mathrm{x}_{\mathrm{i}}\)
or \(\sum_{i=1}^{100} x_i\) = 4000
i.e., Incorrect sum of observations = 4000
Thus the correct sum of observations = Incorrect sum – 50 + 40
= 4000 – 50 + 40
= 3990
Hence Correct mean = \(\frac{\text { Correct sum }}{100}=\frac{3990}{100}\) = 39.9
Also Standard deviation (σ) = \(\sqrt{\frac{1}{n} \sum_{i=1}^n x_i^2-\frac{1}{n^2}\left(\sum_{i=1}^n x_i\right)^2}=\sqrt{\frac{1}{n} \sum_{i=1}^n x_i^2-(\bar{x})^2}\)
i.e., 5.1 = \(\sqrt{\frac{1}{100} \times \text { incorrect } \sum_{i=1}^n x_i^2-(40)^2}\)
or 26.01 = \(\frac {1}{100}\) × Incorrect \(\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}^2\) – 1600
Therefore Incorrect \(\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}^2\) = 100(26.01 + 1600) = 162601
Now Correct \(\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}^2\) = Incorrect \(\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}^2\) – (50)² + (40)²
= 162601 – 2500 + 1600
= 161701
Therefore Correct Standard deviation = \([latex]\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}^2\)[/latex]
= \(\sqrt{\frac{161701}{100}-(39.9)^2}\)
= \(\sqrt{1617.01-1592.01}\)
= √25
= 5

Multiple Choice Questions

Question 1.
The range of the following data (marks scored by 10 students in mathematics) is 15, 20, 31, 62, 13, 6, 41, 86, 21, 74 is
(1) 68
(2) 56
(3) 80
(4) 71
Answer:
(3) 80

Question 2.
In a test match, two batsmen, A and B, scored 116, 2 and 76, 138 in two innings respectively. Whose score is more scattered?
(1) A
(2) B
(3) can’t say
(4) equal
Answer:
(1) A

Question 3.
The formula \(\frac{1}{N} \sum_{i=1}^n f_i\left|x_i-\bar{x}\right|\) is used to calculate which of the following dispersion?
(1) Mean deviation
(2) Standard deviation
(3) Range
(4) Q.D
Answer:
(1) Mean deviation

Question 4.
Which of the following formulas is used to measure mean deviation for grouped data about the median?
(1) \(\sum_{\mathrm{i}=1}^{\mathrm{n}} f_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|\)
(2) \(\frac{1}{\mathrm{~N}} \sum_{\mathrm{i}=1}^{\mathrm{n}} f_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|\)
(3) \(\frac{1}{\mathrm{~N}} \sum_{\mathrm{i}=1}^{\mathrm{n}} f_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|\)
(4) \(\frac{1}{N} \sum_{i=1}^n\left|x_i-M\right|\)
Answer:
(3) \(\frac{1}{\mathrm{~N}} \sum_{\mathrm{i}=1}^{\mathrm{n}} f_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|\)

Question 5.
The standard deviation of a discrete frequency distribution is measured by using
(1) \(\frac{1}{\mathrm{~N}} \sqrt{\sum_{\mathrm{i}=1}^{\mathrm{n}} f_{\mathrm{i}}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2}\)
(2) \(\sqrt{\sum_{\mathrm{i}=1}^{\mathrm{n}} f_{\mathrm{i}}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2}\)
(3) \(\sqrt{\frac{1}{\mathrm{~N}} \sum_{\mathrm{i}=1}^{\mathrm{n}} f_{\mathrm{i}}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2}\)
(4) \(\sqrt{\frac{1}{\mathrm{~N}} \sum_{\mathrm{i}=1}^{\mathrm{n}} f_{\mathrm{i}}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}\)
Answer:
(3) \(\sqrt{\frac{1}{\mathrm{~N}} \sum_{\mathrm{i}=1}^{\mathrm{n}} f_{\mathrm{i}}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2}\)

Question 6.
If each observation is multipied by a constant k(≠ 0), the variance of the resulting observations becomes ____ times the original variance.
(1) k (remains unchanged)
(2) √k
(3) k³
(4) k²
Answer:
(4) k²

Question 7.
The variance of 20 observations is 5. If each observation is added by 3, then the new variance of the resulting observations
(1) 5
(2) 8
(3) 15
(4) 9
Answer:
(1) 5

Question 8.
Mean of the squares of the deviations from the mean is known as
(1) Mean deviation
(2) Standard deviation
(3) Quartile deviation
(4) Variance
Answer:
(4) Variance

Question 9.
The relation between the Mean deviation and the Standard deviation is
(1) σ ≤ MD
(2) M.D ≤ σ
(3) σ ≤ \(\sqrt{\mathrm{MD}}\)
(4) σ² ≤ \(\sqrt{\mathrm{MD}}\)
Answer:
(2) M.D ≤ σ

Question 10.
The sum of 10 items is 12, and the sum of their squares is 18. The standard deviation is
(1) \(\frac {4}{5}\)
(2) \(\frac {3}{5}\)
(3) \(\frac {2}{5}\)
(4) \(\frac {1}{5}\)
Answer:
(2) \(\frac {3}{5}\)