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Class 11 Applied Maths Chapter 13 Solutions
Descriptive Statistics
EXERCISE- 13.6
Q.1 Find p(x, y) if cov(x, y) = -16. 5, var(x) = 2.25 and var(y) = 144.
Ans. Using p(x, y) = Cov(x, y)/√Var(x).√Var(y)
= -16.5/√2.25 x √144
= -16.5/1.5 x 12
= -16.5/18
= -0.916
p(x, y) = -0.92 (Approx)
Strong negative relation.
Q.2 The coefficient of correlation between two variables X and Y is 0.64. Their covariance is 16. The variance of X is 9. Find the standard deviation of Y-series.
Ans. Using r(x, y) = Cov(x, y)/√Var(x).√Var(y)
0.64 = 16/√9.√σ^2(y)
0.64 = 16/3 x σ(y)
σ (y)= 16/3 x 0.64
σ (y) = (16 x 100)/(3 x 64)
σ (y) = 25/3
= 8.33
Q.3 Find the covariance and the coefficient of correlation between x and y when n =10, Σx = 60, Σy = 60, Σx^2 = 400, Σy^2 = 580 and Σxy = 305.
Ans. Using Cov(x, y) = 1/N [ΣXY – 1/N ΣXΣY]
= 1/10 [305 – 1/10 x 60 x 60]
= [305 – 360]/10
= -55/10
= 5.5
r(x, y) = [ΣXY – 1/N ΣXΣY]/[(√Σx^2 – 1/N(ΣX)^2][√ΣY^2 – 1/N(ΣY)^2])
= (305 – 1/10x60x60)/([√400-60^2/10][√580-60^2/10])
= (305 – 360)/[√400-360 x √580-360]
= -55/√44 x √220
= -55/√8800
= -55/93.5
= -0.5836 (Approx)
Moderate negative correlation.
Q.4 Find coefficient of correlation between x and y, when Σx = 375, Σy = 270, Σ(x – x̄)^2 = 136, Σ(y – ȳ)^2 = 138, Σx(x – x̄)(y – ȳ) = 122 and n= 15.
Ans. Using r(x, y) = [Σ(x – x̄)(y – ȳ)]/[√Σ(x – x̄)^2√Σ(y – ȳ)^2]
= 122/√136 x √138
= 122/√136×138
= 122/137
= 0.89 (Approx)
Strong positive correlation.
Q.5 Compute Karl Pearson’s coefficient of correlation between sales and expenditures of a firm for six months.
| Sales (in lakh) | 18 | 20 | 27 | 20 | 21 | 29 |
| Expenditure (in lakh) | 23 | 27 | 28 | 28 | 29 | 30 |
Ans. A = 23, B = 26
| Sales (in lakh) | 18 | 20 | 27 | 20 | 21 | 29 |
| Expenditure (in lakh) | 23 | 27 | 28 | 28 | 29 | 30 |
| U = X – A | -5 | -3 | 4 | -3 | -2 | 6 |
| V = Y – B | -3 | 1 | 2 | 2 | 3 | 4 |
| UV | 15 | -3 | 8 | -6 | -6 | 24 |
| U^2 | 25 | 9 | 16 | 9 | 4 | 36 |
| V^2 | 9 | 1 | 4 | 4 | 9 | 16 |
ΣU = -3, ΣV = 9, ΣUV = 32, ΣU^2 = 99, ΣV^2 = 43
Using, r(x, y) = [ΣUV – 1/N ΣU.ΣV]/√ΣU^2 – (ΣU)^2/N√ΣV^2 – (ΣV)^2/N
= [32 – 1/6 X -3 X 9]/√99^2-(-3)^2/6√43 – 9^2/6]
= [32 + 4.5]/([√99 – 9/3][√43 – 81/6]
= 36.5/[√192/2 x √59/2]
= 36.5/√(195×195)/(2×2)
= 36.5/√11505/2
= (36.5 x 2)/107.2
= 73/107.2
= 0.68
Moderate positive correlation.
Q.6 Calculate Karl Pearson’s coefficient of correlation between x & y for the following data:
| X | 6 | 2 | 4 | 9 | 1 | 3 | 5 | 8 |
| Y | 13 | 8 | 12 | 15 | 9 | 10 | 11 | 16 |
Ans. A = 5, B = 12
| X | 6 | 2 | 4 | 9 | 1 | 3 | 5 | 8 |
| Y | 13 | 8 | 12 | 15 | 9 | 10 | 11 | 16 |
| U = X – A | 1 | -3 | -1 | 4 | -4 | -2 | 0 | 3 |
| V = Y – B | 1 | -4 | 0 | 3 | -3 | -2 | 1 | 4 |
| UV | 1 | 12 | 0 | 12 | 12 | 4 | 0 | 12 |
| U^2 | 1 | 9 | 1 | 16 | 16 | 4 | 0 | 9 |
| V^2 | 1 | 16 | 0 | 9 | 9 | 4 | 1 | 16 |
ΣU = -2, ΣV = -2, ΣUV = 53, ΣU^2 = 56, ΣV^2 = 56
Using, r(x, y) = [ΣUV – 1/N ΣU.ΣV]/√ΣU^2 – (ΣU)^2/N√ΣV^2 – (ΣV)^2/N
= [53 – 1/8 X -2 X -2]/√56 – 4/8 √56 – 4/8]
= [53 – 4/8]/[56- 4/8]
= 52.5/55.5
= 0.9459
= 0.946
Strong positive correlation.
Q.7 Find Karl Pearson’s coefficient of correlation between x & y for the following data:
| X | 16 | 18 | 21 | 20 | 22 | 26 | 27 | 15 |
| Y | 22 | 25 | 24 | 26 | 25 | 30 | 33 | 14 |
Ans. A = 21, B = 25
| X | 16 | 18 | 21 | 20 | 22 | 26 | 27 | 15 |
| Y | 22 | 25 | 24 | 26 | 25 | 30 | 33 | 14 |
| U = X – A | -5 | -3 | 0 | -1 | 1 | 5 | 6 | -6 |
| V = Y – B | -3 | 0 | -1 | 1 | 0 | 5 | 8 | -11 |
| UV | 15 | 0 | 0 | -1 | 0 | 25 | 48 | 66 |
| U^2 | 25 | 9 | 0 | 1 | 1 | 25 | 36 | 36 |
| V^2 | 9 | 0 | 1 | 1 | 0 | 25 | 64 | 121 |
N = 8, ΣU = -3, ΣV = -1, ΣUV = 153, ΣU^2 = 133, ΣV^2 = 221
Using, r(x, y) = [ΣUV – 1/N ΣU.ΣV]/[√ΣU^2 – 1/N(ΣU)^2.√ΣV^2 – 1/N(ΣV)^2]
= [153 – 1/8 X -3 X -1]/√133 – 9/8 √221 – 1/8]
= [1221/8]/[√1055/8 x 1767/8]
= (1221/8)/[(√1055 x √1767)/8
= 1221/32.5 x 42
= 1221/1365
= 0.894
Strong positive correlation.
Q.8 The weights of sons and father (in Kilograms) are given below:
| Weight of father | 65 | 66 | 67 | 67 | 68 | 69 | 70 | 72 |
| Weight of son | 67 | 68 | 65 | 68 | 72 | 72 | 69 | 71 |
Ans. A = 68, B = 69
| Weight of father (X) | 65 | 66 | 67 | 67 | 68 | 69 | 70 | 72 |
| Weight of son (Y) | 67 | 68 | 65 | 68 | 72 | 72 | 69 | 71 |
| U = X – A | -3 | -2 | -1 | -1 | 0 | 1 | 2 | 4 |
| V = Y – B | -2 | -1 | -4 | -1 | 3 | 3 | 0 | 2 |
| UV | 6 | 2 | 4 | 1 | 0 | 3 | 0 | 8 |
| U^2 | 9 | 4 | 1 | 1 | 0 | 1 | 4 | 16 |
| V^2 | 4 | 1 | 16 | 1 | 9 | 9 | 0 | 4 |
ΣU = 0, ΣV = 0, ΣUV = 24, ΣU^2 = 36, ΣV^2 = 44
Using, r(x, y) = [ΣUV – 1/N ΣU.ΣV]/√ΣU^2 – (ΣU)^2/N√ΣV^2 – (ΣV)^2/N
= [24 – 1/8 X 0 X 0]/√36 – 0^2/8 x √44 – 0^2/8]
= 24/[√36x√44
= 24/(6 x √11×4
=4/2√11
= 2/√11
= 2/3.32
= 0.603
Strong positive correlation.
Q.9 Calculate Karl Pearson’s coefficient of correlation from the following data and interpret the result:
| Serial number of student | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Marks in mathematics | 15 | 18 | 21 | 24 | 27 | 30 | 36 | 39 | 42 | 48 |
| Marks in statistics | 25 | 25 | 27 | 27 | 31 | 33 | 35 | 41 | 41 | 45 |
Ans. A = 68, B = 69
| Marks in mathematics (X) | 15 | 18 | 21 | 24 | 27 | 30 | 36 | 39 | 42 | 48 |
| Marks in statistics (Y) | 25 | 25 | 27 | 27 | 31 | 33 | 35 | 41 | 41 | 45 |
| U = X – A | -16 | -13 | -10 | -7 | -4 | -1 | 5 | 8 | 11 | 17 |
| V = Y – B | -8 | -8 | -6 | -6 | -2 | 0 | 2 | 8 | 8 | 2 |
| UV | 128 | 104 | 60 | 42 | 8 | 0 | 10 | 64 | 88 | 204 |
| U^2 | 256 | 169 | 100 | 49 | 16 | 1 | 25 | 64 | 121 | 289 |
| V^2 | 64 | 64 | 36 | 36 | 4 | 0 | 4 | 64 | 64 | 144 |
ΣU = -9, ΣV = 0, ΣUV = 708, ΣU^2 = 1090, ΣV^2 = 480, N = 10
Using, r(x, y) = [ΣUV – 1/N ΣU.ΣV]/√ΣU^2 – (ΣU)^2/N√ΣV^2 – (ΣV)^2/N
= [708 – 1/10 X -9 X 0]/√1090 – 81/10 x √480 – 0^2/10]
= 708/[√1090-8x√480
= 708/√1089×4√30
= 117/33x√30
= 59/11x√30
= 59/11×5.5
= 59/60.5
= 0.98
Strong positive correlation.
Q.10 From the following table, calculate the Karl Pearson’s coefficient of correlation.
| X | 6 | 2 | 10 | 4 | 8 |
| Y | 9 | 11 | ? | 8 | 7 |
The arithmetic means of the X and Y series are 6 and 8 respectively.
Ans. x̄ = 6, ȳ = 8
ȳ = Sum/no.
8 = (35 + x)/5
8 x 5 = 35 + x
40 – 35 = x
x = 5
| x | 6 | 2 | 10 | 4 | 8 |
| y | 9 | 11 | 5 | 8 | 7 |
| x – x̄ | 0 | -4 | 4 | -2 | 2 |
| y – ȳ | 1 | 3 | -3 | 0 | -1 |
| (x – x̄)^2 | 0 | 16 | 16 | 4 | 4 |
| (y – ȳ)^2 | 1 | 9 | 9 | 0 | 1 |
| (x – x̄)(y – ȳ) | 0 | -12 | -12 | 0 | -2 |
Σ(x – x̄)(y – ȳ) = -26, Σ(x – x̄)^2 = 40, Σ(y – ȳ)^2 = 20
r(x, y) = [Σ(x – x̄)(y – ȳ)]/[√(x – x̄)^2.√(y – ȳ)^2]
= -26/√40.√20
= -26/[√2x√20x√20]
= -26/20√2
= -26/(20×1.414
= -26/28.28
= -13/14.14
= -0.92
Strong negative correlation.
Q.11 In calculating the coefficient the coefficient of correlation between two variables x and y for 12 pairs of observations, the following data was obtained:
Σx = 30, Σy = 5, Σx^2 = 670, Σy^2 = 285 and Σxy = 334.
On rechecking, it was found that one pair (11, 4) was wrongly copied while the correct pair being (10, 14). Find the correct value of coefficient of correlation.
Ans. Correct Σx = 30 -11 + 10 = 29
Correct Σy = 5 – 4 + 14 = 15
Correct Σx^2 = 670 – 11^2 + 10^2 = 649
Correct Σy^2 = 285 – 4^2 + 14^2 = 465
Correct Σxy = 334 – 44 + 140 = 430
r(x, y) = [ΣUV – 1/N ΣU.ΣV]/√ΣU^2 – (ΣU)^2/N√ΣV^2 – (ΣV)^2/N
= [430 – 1/12 X 29 X 15]/√649 – 29^2/12 x √465 – 15^2/12]
= [430 – 145/4]/[√(649×12-841)/12.√(465×12-225)/12
= (430 – 36.25/√6947/12.√5355/12
= 393.75/√37201185/12
= 393.75×12/6100
= 4725/6100
= 0.77
Moderate positive correlation.
FAQ’s related to Class 11 Applied Maths Chapter 13 on Descriptive Statistics:
Q.1 What is Descriptive Statistics?
Ans. Descriptive Statistics involves methods for summarizing and organizing the information in a data set. This includes measures of central tendency (mean, median, mode) and measures of variability (range, variance, standard deviation).
Q.2 What are Measures of Central Tendency?
Ans. Measures of central tendency are statistical metrics used to determine the center or typical value of a data set. They include:
- Mean: The average of all data points.
- Median: The middle value when data points are ordered.
- Mode: The most frequently occurring value(s) in the data set.
These are a few Frequently Asked Questions relating to Class 11 Applied Maths Chapter 13
In Class 11 Applied Maths chapter 13, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Class 11 Applied Maths Chapter 13 promises an enriching learning experience that will set you on the path to success. Class 11 Applied Maths Chapter 13, we delve deep into advanced mathematical concepts that are crucial for understanding.
