TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a)

Students must practice this TS Intermediate Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 8 Measures of Dispersion Ex 8(a)

I.
Question 1.
Find the mean deviation about the mean for the following data.
i) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
ii) 3, 6, 10, 4, 9, 10
Solution:
i) Mean of the given data is \(\bar{x}=\frac{38+70+48+40+42+55+63+46+54+44}{10}\)
= \(\frac{500}{10}\) = 50
The absolute values of the deviations are \(\left|x_i-\bar{x}\right|\) =12, 20, 2, 10, 8, 5, 13, 4, 4, 6
∴ Mean Deviation about the Mean = \(\frac{\sum_{\mathrm{i}=1}^{10}\left|\mathbf{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{n}\)
= \(\frac{12+20+2+10+8+5+13+4+4+6}{10}\)
= \(\frac{84}{10}\) = 8.4.

ii) Mean of the given data (\(\bar{x}\)) = \(\frac{\sum_{\mathbf{i}=1}^6 x_i}{n}\)
∴ \(\bar{x}\) = \(\frac{3+6+10+4+9+10}{6}=\frac{42}{6}\) = 7
The absolute values of the deviations are |x_i – \(\bar{x}\)| = 4, 1, 3, 3, 2, 3
Mean Deviation about the Mean = \(\frac{\sum_{i=1}^6\left|x_i-\bar{x}\right|}{n}\)
= \(\frac{4+1+3+3+2+3}{6}=\frac{16}{6}\) = 2.67.

Question 2.
Find the mean deviation about the median for the following data.
i) 13, 17, 16, 11, 13, 10, 16, 11, 18, 12, 17
ii) 4, 6, 9, 3, 10, 13, 2
Solution:
i) Expressing the given data in the ascending order, we get 10, 11, 11,12,13, 13, 16, 16, 17, 17, 18
Median (M) of these 11 observations is 13.
The absolute values of deviations are |x_i – M| = \(\frac{3+2+2+1+0+0+3+3+4+4+5}{11}\)
= \(\frac{27}{11}\) = 2.45.

ii) Expressing the given data in the ascending order, we get 2, 3, 4, 6, 9, 10, 13.
Median (M) of given data = 6
The absolute values of deviations are |x_i – M | = 4, 3, 2, 0, 3, 4, 7
∴ Mean Deviation about the Median = \(\frac{\sum_{\mathrm{i}=1}^7\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\mathrm{n}}=\frac{4+3+2+0+3+4+7}{7}\)
= \(\frac{23}{7}\) = 3.29.

Question 3.
Find the mean deviation about the mean for the following distribution.
i)

ii)
Solution:
i)

∴ Mean (\(\bar{x}\)) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\mathrm{N}}=\frac{534}{45}\) = 11.87

∴ Mean Deviation about the Mean = \(\frac{\sum_{i=1}^4 f_i\left|x_i-\bar{x}\right|}{N}=\frac{31.95}{45}\) = 0.71.

ii)

∴ Mean (\(\bar{x}\)) = \(\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\mathrm{N}}=\frac{4000}{80}\) = 50

∴ Mean Deviation about the Mean = \(\frac{\sum_{i=1}^5 f_i\left|x_i-\bar{x}\right|}{N}=\frac{1280}{80}\) = 16.

Question 4.
Find the mean deviation about the median for the following frequency distribution.

Solution:

Hence N = 26 and \(\frac{N}{2}\) = 13
Median (M) = 7
Median Deviation about Median = \(\frac{\sum_{\mathrm{i}=1}^6 \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\mathrm{N}}=\frac{84}{26}\) = 3023.

Note:
We shall identify the observation whose cumulative frequency is equal to or just greater than N/2. This is the median of the data. Here median is “7”.

II.
Question 1.
Find the mean deviation about the median for the following continuous distribution.
i)

ii)
Solution:

i)

Hence L = 20, \(\frac{N}{2}\) = 25, f₁ = 14, f = 14, h = 10
Median (M) = L + \(\left[\frac{\left[\frac{\mathrm{N}}{2}-\mathrm{f}_1\right]}{\mathrm{f}}\right]\) h
= 20 + \(\frac{25-14}{14}\) × 10
= 20 + \(\frac{110}{14}\)
= 20 + 7.86 = 27.86.
∴ Mean Deviation about Median = \(\frac{\sum_{\mathrm{i}=1}^6 \mathrm{t}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\mathrm{N}}\)
= \(\frac{517.16}{50}\) = 10.34.

ii)

Here N = 100, \(\frac{N}{2}\) = 50, L = 40, f₁ = 32, f = 28, h = 10
Median (M) = L + \(\left\{\frac{\left[\frac{\mathrm{N}}{2}-\mathrm{f}_1\right]}{\mathrm{f}}\right\}\) h
= 40 + \(\frac{50-32}{28}\) × 10
= 40 + \(\frac{180}{28}\)
= 40 + 6.43 = 46.43.
∴ Mean Deviation about Median = \(\frac{\sum_{i=1}^8 f_i\left|x_i-M\right|}{N}=\frac{1428.6}{100}\) = 14.29.

Question 2.
Find the mean deviation about the mean for the following continuous distribution.

Solution:

Mean (\(\bar{x}\)) = A + \(\frac{\Sigma f_i \mathrm{~d}_{\mathrm{i}}}{\mathrm{N}}\) . h
= 130 + \(\left(\frac{-47}{100}\right)\) . 10
= 130 – 1.7 = 125.3.

∴ Mean Deviation about Mean = \(\frac{\sum_{i=1}^6 f_i\left|x_i-\bar{x}\right|}{N}\)
= \(\frac{1128.8}{100}\) = 11.29.

Question 3.
Find the variance for the discrete data given below.
i) 6, 7, 10, 12, 13, 4, 8, 12
ii) 350, 361, 370, 373, 376, 379, 385, 387, 394, 395.
Solution:
i) Mean (\(\bar{x}\)) = \(\frac{6+7+10+12+13+4+8+12}{8}=\frac{72}{8}\) = 9

Variance (σ²) = \(\frac{\sum_{\mathrm{i}=1}^8\left(x_i-\bar{x}\right)^2}{n}=\frac{74}{8}\) = 9.25.

ii)

Mean (\(\bar{x}\)) = \(\frac{350+361+370+373+376+379+385+387+394+395}{10}\)
= \(\frac{3770}{10}\) = 377.
Variance (σ²) = \(\frac{\sum_{i=1}^{10}\left(x_i-\bar{x}\right)^2}{n}=\frac{1832}{10}\) = 183.2.

Question 4.
Find the variance and standard deviation of the following frequency distribution.

Solution:

Mean (x) = \(\frac{760}{40}\) = 19
Variance (σ²) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2}{\mathrm{~N}}=\frac{1736}{40}\) = 43.4
Standard Deviation (σ) = \(\sqrt{43.4}\) = 6.59.

III.
Question 1.
Find the mean and variance using the step deviation method, of the following tabular data, giving the age distribution of 542 members.

Solution:

Question 2.
The coefficent of variation of two distributions are 60 and 70 and their standard deviations are 21 and 16 respectively. Find their arithmetic means.
Solution:
C.V = \(\frac{\sigma}{\overline{\bar{X}}}\) × 100

i) 60 = \(\frac{21}{\overline{\mathrm{X}}}\) × 100
\(\overline{\mathrm{X}}\) = \(\frac{21 \times 100}{60}\) = 35

ii) 70 = \(\frac{16}{\overline{\mathrm{Y}}}\) × 100
\(\overline{\mathrm{Y}}\) = \(\frac{16 \times 100}{70}\) = 22.857.

Question 3.
From the prices of shares X and Y given below, for 10 days of trading, find out which share is more stable?

Solution:
Variance is independent ol change of origin.

V(X) = \(\frac{\Sigma x_i^2}{n}-(\bar{x})^2\)
= \(\frac{360}{10}-\left(\frac{10}{10}\right)^2\)
= 36 – 1 = 35.

V(Y) = \(\frac{\Sigma Y_i^2}{n}-(\bar{Y})^2\)
= \(\frac{290}{10}-\left(\frac{50}{10}\right)^2\)
= 29 – 25 = 4.

Question 4.
The mean of 5 observations is 4.4. Their variance is 8.24. If three of the observations are 1, 2 and 6. Find the other
two observations.
Solution:

S.D = \(\sqrt{\frac{\Sigma \mathrm{m}^2}{\mathrm{n}}-(\overline{\mathrm{x}})^2}\)
\(\bar{x}\) = 4.4
⇒ 4.4 = \(\frac{1+2+6+x+y}{5}\)
⇒ 9 + x + y = 22
⇒ x + y = 13 …………..(1)
S.D² = \(\frac{1+4+36+x^2+y^2}{5}\) – (4.4)²
= \(\frac{41+x^2+y^2}{5}\) – 19.36
S.D² = Variance
Variance = \(\frac{41+x^2+y^2}{5}\) – 19.36
8.24 + 19.36 = \(\frac{41+x^2+y^2}{5}\)
41 + x² + y² = 5×27.6
x² + y² = 138 – 41
x² + y² = 97 …………..(2)
From (1) and (2),
x² + (13 – x)² = 97
x² + 169 + x² – 26x = 97
2x² – 26x + 72 = 0
x² – 13x + 36 = 0
x² – 9x – 4x + 36 = 0
x (x – 9) – 4 (x – 9) = 0
(x – 9) (x – 4) = 0
x = 4, 9
Put x = 4 in (1)
y = 13 – 4= 9
Put x = 9 in (1)
y = 13 – 9 = 4
∴ If x = 4, then y = 9.
If x = 9, then y = 4.

Question 5.
The arithmetic mean and standard deviation of a set of 9 items are 43 and 5 respectively. If an item of value 63 is added to that set, find the new mean and standard deviation of 10 items set given.
Solution: