Referring to the AP Inter 1st Year Maths Study Material Chapter 13 Statistics Exercise 13c Solutions makes it easier to understand complex problems.
AP Inter 1st Year Maths Statistics Solutions Exercise 13c
Question 1.
The mean and variance of eight observations are 9 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.
Solution:
Let the remaining two observations be x and y.
Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y
Mean x̄ = \(\frac{6+7+10+12+12+13+x+y}{8}\) = 9
⇒ 60 + x + y = 72
⇒ x + y = 12 ………(1)
Variance = 9.25 = \(\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^8\)(xi – x̄)2
9.25 = \(\frac{1}{8}\)[(-3)2 + (-2)2 + (1)2 + (3)2 + (3)2 + (4)2 + x2 + y2 – 2 × 9(x + y) + 2(9)2]
9.25 = \(\frac{1}{8}\)[9 + 4 + 1 + 9 + 9 + 16 + x2 + y2 – 18(12) + 162] ………..[Using (1)]
9.25= \(\frac{1}{8}\)[48 + x2 + y2 – 216 + 162]
9.25 = \(\frac{1}{8}\)[x2 + y2 – 6]
⇒ x2 + y2 = 80 …. (2)
From (1), we obtain (x + y)2 = (12)2
⇒ x2 + y2 + 2xy = 144
From (2) and (3), we obtain 2xy = 64
Subtracting (4) from (2), we obtain
x2 + y2 – 2xy = 80 – 64 = 16
⇒ x – y = ±4
Therefore, from (1) and (5), we obtain x = 8 and y = 4, when x – y = 4 x = 4 and y = 8, when x – y = – 4 Thus, the remaining observations are 4 and 8.
Question 2.
The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12 and 14. Find the remaining two observations.
Solution:
Let the remaining two observations be x and y.
The observations are 2, 4, 10, 12, 14, x, y.
i.e Mean, x̄ = \(\frac{2+4+10+12+14+x+y}{7}\)
⇒ 56 = 42 + x + y
⇒ x + y = 14 ………..(1)
Variance = 16 = \(\frac{1}{n} \sum_{i=1}^7\left(x_i-\bar{x}\right)^2\)
16 = \(\frac{1}{7}\)[36 + 16 + 4 +16 + 36 + x2 + y2 – 16(x + y) + 2(64)]
16 = \(\frac{1}{7}\) [36 + 16 + 4 + 16 + 36 + x2 + y2 – 16(14) + 2(64)]
16 = \(\frac{1}{7}\)[108 + x2 + y2 – 224 + 128]
16 = \(\frac{1}{7}\)[12 + x2 + y2]
⇒ x2 + y2 = 112 – 12
⇒ x2 + y2 = 100
From (1), we obtain x2 + y2 + 2xy = 196
From (2) and (3), we obtain 2xy = 196 – 100
⇒ 2xy = 96 …. (4)
Subtracting (4) from (2), we obtain x2 + y2 – 2xy = 100 – 96
⇒ (x – y)2 = 4 ⇒ x – y = ±2 ….(5)
Therefore, from (1) and (5), we obtain
x = 8 and y = 6 when x – y = 2; x = 6 and y = 8 when x – y = – 2
Thus, the remaining observations are 6 and 8.
Question 3.
The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.
Solution:
Let the observations be x1, x2, x3, x4, x5 and x6
It is given that mean is 8 and standard deviation is 4.
Mean x̄ = \(\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}\) …. (1)
If each observation is multiplied by 3 and the resulting observation be y., then
yi = 3xi ⇒ xi = \(\frac{1}{3}\)yi, for i = 1 to 6
New Mean ȳ = \(\frac{y_1+y_2+y_3+y_4+y_5+y_6}{6}=\frac{3\left(x_1+x_2+x_3+x_4+x_5+x_6\right)}{6}\)
= 3 × 8 = 24 ……….. [using (1)]
Standard deviation σ = \(\sqrt{\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^6\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2}\)
⇒ (4)2 = \(\frac{1}{6} \sum_{i=1}^6\)(xi – x̄)2
From (1) and (2), it can be observed that, ȳ = 3x̄ ⇒ x̄ = \(\frac{1}{3}\) ȳ
Substituting the values of xi and x̄ in (2), we obtain
\(\sum_{i=1}^6\left(\frac{1}{3} y_i-\frac{1}{3} \bar{y}\right)^2\) = 96 ⇒ \(\sum_{\mathrm{l}=1}^6\)(yi – ȳ)2 = 864
Therefore, variance of new observations = (\(\frac{1}{6}\) × 864) = 144
Hence, the standard deviation of new observations is \(\sqrt{144}\) = 12.
Question 4.
Given that x̄ is the mean and σ2 is the variance of n observations x1, x2,…. xn. Prove that the mean and variance of the observations ax1, ax2, ax3 ………….. axn are ax̄ and a2σ2, respectively, (a * 0)
Solution:
The given n observations x1, x1 ….xn.
Mean = x̄; Variance = σ2.
∴ σ2 = \(\frac{1}{n} \sum_{i=1}^n\) yn …………(1)
If each observation is multiplied by a and the new observations be yi, then
yi = axi ⇒ xi = \(\frac{1}{a}\)yi
∴ ȳ = \(\frac{1}{n} \sum_{i=1}^n\)y1
= \(\frac{1}{n} \sum_{i=1}^n\) axi
= \(\frac{a}{n} \sum_{i=1}^n\) xi
= ax
(x̄ = \(\frac{1}{n} \sum_{i=1}^n\)xi)
Therefore, mean of the observations ax1, ax2, ……… axn, is ax̄
Substituting the values of xi and x̄ in (!), we obtain
σ2 = \(\frac{1}{n} \sum_{i=1}^n\left(\frac{1}{a} y_i-\frac{1}{a} \bar{y}\right)^2\)
⇒ a2σ2 = \(\frac{1}{n} \sum_{i=1}^n\)(yi – ȳ)2
Thus, the variance of the observations, ax1, ax2 ……….. axn is a2a2.
Question 5.
The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases.
i) If wrong item is omitted
ii) If it is replaced by 12.
Solution:
i) Number of observations (n) = 20
Incorrect mean = 10
Incorrect standard deviation = 2
x̄ = \(\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{20}\)xi
⇒ 10 = \(\frac{1}{20} \sum_{i=1}^{20}\)xi
⇒ \(\sum_{i=1}^{20}\)xi = 200
That is, incorrect sum of observations = 200
Correct sum of observations = 200 – 8 = 192
∴ Correct mean = \(\frac{\text { correct sum }}{19}=\frac{192}{19}\) = 10.105
Standard deviation σ = \(\sqrt{\frac{1}{n} \sum_{i=1}^n x_i^2-\frac{1}{n^2}\left(\sum_{i=1}^n x_i\right)^2}=\sqrt{\frac{1}{n} \sum_{i=1}^n x_i^2-(\bar{x})^2}\)
⇒ 2 = \(\sqrt{\frac{1}{20} \text { Incorrect } \sum_{i=1}^n x_i^2-(10)^2}\)
⇒ 4 = \(\frac{1}{20}\) Incorrect \(\sum_{i=1}^{\mathrm{n}}\) xi2 – 100
⇒ Incorrect \(\sum_{i=1}^n\) xi = 2080
∴ Correct = \(\sum_{i=1}^n\)xi2 = Incorrect \(\sum_{i=1}^n\)x – (8)2 + (12)2
= 2080 – 64 + 144
= 2016
Corect standard deviation = \(\sqrt{\frac{\text { Correct } \sum x_1^2}{n}-(\text { Correct mean })^2}\)
= \(\sqrt{\frac{2016}{19}-(10.105)^2}\)
= \(=\sqrt{106.105-102.111}\)
= \(\sqrt{3.984}\) = 1.995
ii) When 8 is replaced by 12.
Incorrect sum of observations = 200
∴ Correct sum of observation = 200 – 8 + 12 = 204
Correct mean = \(\frac{\text { Correct sum }}{20}=\frac{204}{20}\) = 10.2
Standard deviation σ = \(\sqrt{\frac{1}{n} \sum_{i=1}^n x_i^2-\frac{1}{n^2}\left(\sum_{i=1}^n x_i\right)^2}=\sqrt{\frac{1}{n} \sum_{i=1}^n x_i^2-(\bar{x})^2}\)
2 = \(\sqrt{\frac{1}{20} \text { Incorrect } \sum_{i=1}^n x_i^2-(10)^2}\)
⇒ 4 = \(\frac{1}{20}\) Incorrect \(\sum_{l=1}^{\mathrm{n}}\) xi2 – 100
Incorrect \(\sum_{l=1}^{\mathrm{n}}\)xi = 2080
Correct Σxi2 = Incorrect \(\sum_{i=1}^{\mathrm{n}}\) i – (8)2 + (12)2
= 2080 – 64 + 144
= 2160
Correct standard deviation
= \(\sqrt{\frac{\text { Correct } \sum x_i^2}{n}-(\text { Correct mean })^2}=\sqrt{\frac{2160}{20}-(10.2)^2}\)
= \(\sqrt{108-104.04}=\sqrt{3.96}\) = 1.98
Question 6.
The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.
Solution:
Number of observations (n) = 100
Incorrect mean (x) = 20 ; Incorrect standard deviation (σ) = 3
20 = \(\frac{1}{100} \sum_{i=1}^{100}\)xi ⇒ \(\sum_{i=1}^{100}\)xi = 20 × 100 = 2,000
Incorrect sum of observations = 2,000
Correct sum of observations = 2000 – 21 – 21 – 18 = 1940
Now \(\frac{1}{2}\)xi2 – (x̄)2
⇒ \(\frac{1}{100}\) Σxi2 – (20)2 = 9
⇒ Σxi2 = 40, 900
Correct mean = \(\frac{1940}{97}\) = 20
Also Σxi2 = incorrect Σxi2 – (21)2 – (21)2 – (18)2
= 40,900 – 441 – 441 – 324
= 39,694
Correct variance = \(\frac{1}{97}\) correct Σxi2 – (correct mean)2
= \(\frac{1}{97}\) × 39,694 – (20)2
= 409.22 – 400 = 9.22
⇒ Correct SD = \(\sqrt{9.22}\) = 3.036