Inter 1st Year Maths Statistics Solutions Exercise 13c

Practicing the AP Board Solutions Class 11 Maths and Chapter 13 Inter 1st Year Maths Statistics Solutions Exercise 13c Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Statistics Solutions Exercise 13c

Statistics Exercise 13c Solutions

Statistics Class 11 Exercise 13c Solutions – Statistics 13c Exercise Solutions

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, \(\bar{x}=\frac{6+7+10+12+12+13+x+y}{8}\) = 9
⇒ 60 + x + y = 72
⇒ x + y = 12 ………(1)
Variance = 9.25 = \(\frac{1}{n} \sum_{i=1}^8\left(x_i-\bar{x}\right)^2\)
⇒ 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) + 2 × 81] …. [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 x² + y² + 2xy = 144 ……..(3)
From (2) and (3), we obtain 2xy = 64 ………(4)
Subtracting (4) from (2), we obtain
x² + y² – 2xy = 80 – 64 = 16
⇒ x – y = ±4 ……..(5)
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, 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.
Mean, x = \(\bar{x}=\frac{2+4+10+12+14+x+y}{7}\) = 8
⇒ 42 + x + y = 56
⇒ x + y = 14 ……(1)
Variance = 16
⇒ \(\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^7\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2\) = 16
⇒ 16 = \(\frac {1}{7}\) [(-6)² + (-4)² + (2)² + (4)² + (6)² + x² + y² – 2 × 8(x + y) + 2 × (8)²]
⇒ 16 = \(\frac {1}{7}\) [36 + 16 + 4 + 16 + 36 + x² + y² – 16(14) + 2 × 64]
⇒ 16 = \(\frac {1}{7}\) [108 + x² + y² – 224 + 128] … [Using (1)]
⇒ x² + y² = 112 – 12 = 100
⇒ x² + y² = 100 ……….(2)
From (1), we obtain x² + y²+ 2xy = 196 ……. (3)
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 the mean is 8 and the standard deviation is 4.
Mean, \(\bar{x}=\frac{x_1+x_2+x_3+x_4+x_5}{6}\) = 8 …….(1)
If each observation is multiplied by 3 and the resulting observations are y_i, then y_i = 3x_i
⇒ x_i = \(\frac {1}{3}\)y_i, for i = 1 to 6.
∴ New mean, \(\bar{y}=\frac{y_1+y_2+y_3+y_4+y_5+y_6}{6}\)
= \(\frac{3\left(x_1+x_2+x_3+x_4+x_5\right)}{6}\)
= 3 × 8
= 24
Standard deviation, σ = \(\sqrt{\frac{1}{n} \sum_{i=1}^6\left(x_i-\bar{x}\right)^2}\)
⇒ (4)² = \(\frac{1}{6} \sum_{\mathrm{i}=1}^6\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2\)
⇒ \(\sum_{i=1}^6\left(x_i-\bar{x}\right)^2\) = 96 ……..(2)
From (1) and (2), it can be observed that \(\overline{\mathrm{y}}=3 \overline{\mathrm{x}}, \overline{\mathrm{x}}=\frac{1}{3} \overline{\mathrm{y}}\)
Substituting the values of x and \(\overline{\mathrm{x}}\) in (2), we obtain
\(\sum_{i=1}^6\left(\frac{1}{3} y_i-\frac{1}{3} \bar{y}\right)^2\) = 96
⇒ \(\sum_{i=1}^6\left(y_i-\bar{y}\right)^2\) = 864
Therefore, variance of new observations = \(\frac {1}{6}\) × 864 = 144
Hence, the standard deviation of new observations is √144 = 12.

Question 4.
Given that \(\overline{\mathrm{x}}\) is the mean and σ² is the variance of n observations x₁, x₂, …, x_n. Prove that the mean and variance of the observations ax₁, ax₂, ax₃, ………., ax_n are a\(\overline{\mathrm{x}}\) and a²σ², respectively, (a ≠ 0).
Solution:
The given n observations are x₁, x₂,…, x_n.
Mean = \(\overline{\mathrm{x}}\)
Variance (σ²) = \(\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{y}_{\mathrm{i}}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2\) …….(1)
If each observation is multiplied by a and the new observations are y_i, then y_i = ax_i
⇒ x_i = \(\frac {1}{a}\)y_i

Therefore, mean of the observations, ax₁, ax₂, …….., ax_n is a \(\overline{\mathrm{x}}\)
Substituting the values of x_i and \(\overline{\mathrm{x}}\) in (1), we obtain

Thus, the variance of the observations, ax₁, ax₂,…, ax_n, is a²σ².

Question 5.
The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
(i) If the 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

That is, the incorrect sum of observations = 200
Correct sum of observations = 200 – 8 = 192
∴ Correct mean = \(\frac{\text { Correct şum }}{19}=\frac{192}{19}\) = 10.1

(ii) When 8 is replaced by 12, Incorrect sum observations = 200
Correct sum of observations = 200 – 8 + 12 = 204
∴ Correct mean = \(\frac{\text { Correct sum }}{20}=\frac{204}{20}\) = 10.2

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 standard deviation (σ) = 3
Incorrect mean (\(\overline{\mathrm{x}}\)) = 20
⇒ 20 = \(\frac{1}{100} \sum_{\mathrm{i}=1}^{100} \mathrm{x}_{\mathrm{i}}\)
⇒ \(\sum_{i=1}^{100} x_i\) = 20 × 100 = 2000
∴ Incorrect sum observations = 2000
⇒ Correct sum observations = 2000 – 21 – 21 – 18
= 2000 – 60
= 1940