Welcome to Class 11 Applied Maths Chapter 13, where we embark on an exciting journey into the world of advanced mathematical concepts tailored for Class 11 students.” Unlock the power of applied mathematics with expert solutions crafted by professionals at AppliedMath.com. Designed to propel students towards academic success, our meticulously curated ML Aggarwal Solutions for Applied Mathematics cater to Class 11 and class 12 students seeking mastery in their examinations. Every query from the CBSE ML Aggarwal Books finds a comprehensive answer on our platform, complete with detailed explanations and step-by-step solutions presented in an easily understandable language.
Dive into the world of applied mathematics and discover how our resources can elevate your understanding and performance. Keep reading to explore the wealth of ML Aggarwal Solutions for Class 11 and Class 12 Applied Mathematics.
Here we provide you with Class 11 Applied Maths Chapter 13, to help you gain a comprehensive understanding of the chapter and its concepts. https://appliedmathsolution.com/wp-admin/post.php?post=6&action=edit
Class 11 Applied Maths Chapter 13 Solutions
Descriptive Statistics
EXERCISE- 13.3
Q.1 Find the first four central moments for the following set of observations:
(i) 2, 3, 5, 7, 8 (ii) 4, 6, 7, 10 ,13
Ans. (i) 2, 3, 5, 7, 8
x̄ = Sum/No.
= 25/5 = 5
| (xi – x̄) | -3 | -2 | 0 | 2 | 3 | Σ(xi – x̄) = 0 |
| (xi – x̄)^2 | 9 | 4 | 0 | 4 | 9 | Σ(xi – x̄)^2 = 26 |
| (xi – x̄)^3 | -27 | -8 | 0 | 8 | 27 | Σ(xi – x̄)^3 = 0 |
| (xi – x̄)^4 | 81 | 16 | 0 | 16 | 81 | Σ(xi – x̄)^4 = 194 |
Using, μn = Σ(xi – x̄)^n/No. of observations
μ1 = 0/5 = 0
μ2 = 26/5 = 5.2
μ3 = 0/5 = 0
μ4 = 194/5 = 38.8
(ii) 4, 6, 7, 10 ,13
x̄ = Sum/No.
= 40/5 = 8
| (xi – x̄) | -4 | -2 | -1 | 2 | 5 | Σ(xi – x̄) = 0 |
| (xi – x̄)^2 | 16 | 4 | 1 | 4 | 25 | Σ(xi – x̄)^2 = 50 |
| (xi – x̄)^3 | -64 | -8 | -1 | 8 | 125 | Σ(xi – x̄)^3 = 60 |
| (xi – x̄)^4 | 256 | 16 | 1 | 16 | 625 | Σ(xi – x̄)^4 = 914 |
Using, μp = Σ(xi – x̄)^p/No. of observations
μ1 = 0/5 = 0
μ2 = 50/5 = 10
μ3 = 60/5 = 12
μ4 = 914/5 = 182.8
Q.2 Find the first four central moments for the following frequency distributions:
| xi | 3 | 5 | 7 | 9 | 11 | 12 |
| fi | 4 | 3 | 5 | 4 | 6 | 3 |
Ans.
| xi | fi | fi xi | xi – x̄ [I] | (xi – x̄)^2 [II] | (xi – x̄)^3 [III] | (xi – x̄)^4 [IV] | fi [I] | fi [II] | fi [III] | fi [IV] |
| 3 | 4 | 12 | -5 | 25 | -125 | 625 | -20 | 100 | -500 | 2500 |
| 5 | 3 | 15 | -3 | 9 | -27 | 81 | -9 | 27 | -81 | 243 |
| 7 | 5 | 35 | -1 | 1 | -1 | 1 | -5 | 5 | -5 | 5 |
| 9 | 4 | 36 | 1 | 1 | 1 | 1 | 4 | 4 | 4 | 4 |
| 11 | 6 | 66 | 3 | 9 | 27 | 81 | 18 | 54 | 162 | 486 |
| 12 | 3 | 36 | 4 | 16 | 64 | 256 | 12 | 48 | 192 | 768 |
| 25 | 200 | 0 | 238 | -228 | 4006 |
x̄ = Sum/No.
= 200/25 = 8
Using, μ = Σfi(xi – x̄)/Σfi
μ1 = 0
μ2 = 238/25 = 9.52
μ3 = – 228/25 = – 9.12
μ4 = 4006/25 = 160.24
Q.3 The mean, mode and standard deviation of a frequency distribution are 27.5, 30 and 7.2 respectively. Calculate Karl Pearson’s coefficient of skewness of the distribution.
Ans. S(kp) = (Mean – Mode)/Standard Deviation
= (27.5 – 30)/7.2
= -2.5/7.2
= -0.347
Since, 0.5>S(kp)>-0.5,
Distance is fairly symmetric.
Q.4 The mean, mode and standard deviation of a frequency distribution are 49, 45 and 23 respectively. Calculate Karl Pearson’s coefficient of skewness of the distribution.
Ans.
S(kp) = 3(Mean – Median)/Standard Deviation
= 3(49 – 45)/23
= (3 x 4)/23
= 12/23
= 0.52
Since, S(kp) > 0.5,
Distance is carrying moderate skewness.
Q.5 The mean, mode and Karl Pearson’s coefficient of skewness of a frequency distribution are 32, 35 and -0.56 respectively. Calculate the standard deviation of the distribution.
Ans. S(kp) = (Mean – Mode)/Standard Deviation
-0.56 = (32 – 35)/SD
SD = -3/ -0.56
= 5.36
Q.6 The mean, standard deviation and Karl Pearson’s coefficient of a frequency distribution are 15, 2.5 and -0.6 respectively. Find the medium of the frequency distribution.
Ans. S(kp) = 3(Mean – Median)/Standard Deviation
-0.6 = 3(15 – Me)/2.5
-0.6 x 2.5 = 3(15 – Me)
-1.5 =3(15 – Me)
-1.5/3 = 15 – Me
-0.5 = 15 – Me
Me = 15 + 0.5
= 15.5
Q.7 The marks obtained by 60 students in a class test are given as under:
| Marks | 0 | 1 | 2 | 3 | 4 | 5 |
| No. of students | 2 | 4 | 16 | 25 | 8 | 5 |
Find the Karl Pearson’s coefficient of skewness of the given data:
Ans. Mode = 3
| Marks (xi) | 0 | 1 | 2 | 3 | 4 | 5 |
| No. of students (fi) | 2 | 4 | 16 | 25 | 8 | 5 |
| xi fi | 0 | 4 | 32 | 75 | 32 | 25 |
| xi^2 | 0 | 1 | 4 | 9 | 16 | 25 |
| fi xi^2 | 0 | 4 | 64 | 225 | 128 | 125 |
Σfi xi = 168 , Σfi = 60 , Σfi xi^2 = 546
x̄ = 168/60 = 2.8
SD = 1/n √nΣfi xi^2 – (Σfi xi)^2
= 1/60 √60 x 546 – (168)^2
= 1/60 √32760 – 28224
= 1/60 √4536
= 1/60 x 67.35
= 1.12
S(kp) = (Mean – Mode)/Standard Deviation
= (2.8 – 3)/1.12
= -0.2/1.12
= -0.178
Therefore, fairly symmetric.
Q.8 Calculate the Karl Pearson’s coefficient of skewness for the following frequency distribution:
| xi | 14 | 15 | 16 | 17 | 18 | 19 | 20 |
| fi | 14 | 26 | 18 | 8 | 2 | 1 | 1 |
Ans. Mode = 15
| xi | 14 | 15 | 16 | 17 | 18 | 19 | 20 |
| fi | 14 | 26 | 18 | 8 | 2 | 1 | 1 |
| fi xi | 196 | 390 | 288 | 136 | 36 | 19 | 20 |
| xi^2 | 196 | 225 | 256 | 289 | 324 | 361 | 400 |
| fi xi^2 | 2744 | 5850 | 4608 | 2312 | 648 | 361 | 400 |
Σfi xi = 1085 , Σfi = 70 , Σfi xi^2 = 16923
x̄ = 1085/70 = 15.5
SD = 1/n √nΣfi xi^2 – (Σfi xi)^2
= 1/70 √70 x 16923 – (1085)^2
= 1/70 √1184610 – 1177225
= 1/70 √7385
= 1/70 x 85.93
= 1.22
S(kp) = (Mean – Mode)/Standard Deviation
= (15.5 – 15)/1.22
= 0.5/1.22
= 0.409
Therefore, distribution is fairly symmetric.
Q.9 The following table shows the ages of the patients admitted in a hospital during a year:
| Age(in years) | 5-15 | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 |
| No. of patients | 6 | 11 | 21 | 23 | 14 | 5 |
Find the Karl Pearson’s coefficient of skewness:
(i) using mode (ii) using median
Ans.
| Age (in years) | 5-15 | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 |
| No. of patients (fi) | 6 | 11 | 21 | 23 | 14 | 5 |
| xi | 10 | 20 | 30 | 40 | 50 | 60 |
| xi fi | 60 | 220 | 630 | 920 | 700 | 300 |
| xi^2 | 100 | 400 | 900 | 1600 | 2500 | 3600 |
| fi xi^2 | 600 | 4400 | 18900 | 36800 | 35000 | 18000 |
Σfi xi = 2830 , Σfi = 80 , Σfi xi^2 = 103700
x̄ = 2830/80 = 35.375
Sd = 1/n √nΣfi xi^2 – (Σfi xi)^2
= 1/80 √80 x 103700 – (2830)^2
= 1/80 √8296000 – 8008900
= 1/80 √287100
= 1/80 x 1042.64
= 13.03
(i) Mode:
Mode = L + [(f1 – f0)/(2f1 – f0 – f2) x h]
= 35 + [(23 – 21)/(46 – 21 – 14) x 10]
= 35 + 20/11
= 35 + 1.81 = 36.81
S(kp) = (Mean – Mode)/Standard Deviation
= (35.37 – 36.81)/13.03
= -1.44/13.03
= -0.111
Therefore, fairly symmetric.
(ii) Median
| CF | 6 | 7 | 38 | 61 | 75 | 80 |
n = 80, n/2 = 40
Median class = 35 – 45
L = 35, h = 10, f = 23, CF = 38
Median = L + [(n/2 – CF)/f x h]
= 35 + [(40 – 38)/23 x 10]
= 35 – 20/23
= 35.9
S(kp) = 3(Mean – Median)/Standard Deviation
= 3(35.37 – 35.89)/13.03
= 3 x -0.52/13.03
= -1.56/13.03
= -0.114
Therefore, fairly symmetric.
Q.10 The following table shows the height of plants (in cm) in a nursery:
| Height (in cm) | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 |
| No. of plants | 4 | 3 | 8 | 12 | 6 | 2 |
Find the Karl Pearson’s coefficient of skewness.
Ans.
| Height (in cm) | 30-40 | 40-50 | 5-60 | 60-70 | 70-80 | 80-90 |
| No. of plants | 4 | 3 | 8 | 12 | 6 | 2 |
| xi | 35 | 45 | 55 | 65 | 75 | 85 |
| xi^2 | 1225 | 2025 | 3025 | 4225 | 5625 | 7225 |
| fi xi^2 | 4900 | 6075 | 24400 | 50700 | 33750 | 14450 |
Σfi xi = 2115 , Σfi = 35 , Σfi xi^2 = 134275
x̄ = 2115/35 = 60.42
Sd = 1/n √nΣfi xi^2 – (Σfi xi)^2
= 1/35 √35 x 134275 – (2115)^2
= 1/35 √4699625 – 4473225
= 1/35 √226400
= 1/35 x 475.81
= 13.59
(i) Mode:
Mode = L + [(f1 – f0)/(2f1 – f0 – f2) x h]
= 60 + [(12 – 8)/(24 – 8 – 6) x 10]
= 60 + 4/10 x 10
= 64
S(kp) = (Mean – Mode)/Standard Deviation
= (60.42 – 64)/13.59
= -3.58/13.59
= -0.266
Therefore, fairly symmetric.
Q.11 For a certain frequency distribution Q1 = 4, Q2 = 15 and Q3 = 19. Find the Bowley’s coefficient of skewness.
Ans. Skewness by Bowley’s coefficient = (Q3 +Q1 -2Q2)/(Q3 – Q1)
= [19 + 4 – (2 X 15)]/[19 – 4]
= (23 – 30)/15
= -7/15
= -0.467
Q.12 The lower and upper quartile for a certain frequency distribution are 7 and 15 respectively. If the Bowley’s coefficient of skewness is 0.25, find the median of the distribution.
Ans. Q1 = 7, Q2 = ?, Q3 = 15
Skewness by Bowley’s coefficient = (Q3 +Q1 -2Q2)/(Q3 – Q1)
0.25 = [15 + 7 – 2Q2]/[15 – 7]
0.25 = (22 – 2Q2)/8
0.25 x 8 = 22 – 2Q2
2 = 22 – 2Q2
2Q2 = 22 – 2
Q2 = 20/2
Q2 = 10
Q.13 In a frequency distribution, the coefficient of skewness based on quartiles is 0.25. If the sum of upper quartile and lower quartile is 80 and the median is 35, find the values of lower and upper quartiles.
Ans. Skewness by Bowley’s coefficient = 0.25
Q3 + Q1 = 80, Q2 = 35
Let, upper and lower quartile be x & y respectively,
Skewness by Bowley’s coefficient = (Q3 +Q1 -2Q2)/(Q3 – Q1)
0.25 = [80 – 2(35)]/[x – y]
x – y = (80 – 70)/0.25
= 10/0.25
= 40
Here,
x + y = 80
x – y = 40
2x = 120
Therefore,
x = 60 [Q3]
y = 120 [Q1]
Q.14 Calculate the Bowley’s coefficient of skewness for the following data:
| xi | 15 | 18 | 20 | 22 | 25 | 27 | 30 |
| fi | 4 | 6 | 8 | 9 | 7 | 8 | 6 |
Ans.
| xi | 15 | 18 | 20 | 22 | 25 | 27 | 30 |
| fi | 4 | 6 | 8 | 9 | 7 | 8 | 6 |
| CF | 4 | 10 | 18 | 27 | 34 | 42 | 48 |
For Q1: [(n/4) + (n/4 + 1)th/2]
= [12th + 13th]/2
= (20 + 20)/2
= 20
For Q2: [(n/2) + (n/2 + 1)Th/2]
Therefore, Q2 = 22
For Q3: [(3n/4) + (3n/4 + 1)Th/2]
= [36th + 37th]/2
Therefore, Q3 = 27
Skewness by Bowley’s coefficient = (Q3 +Q1 -2Q2)/(Q3 – Q1)
= (27 + 20 – 2 x 22)/(27 – 20)
= (47 -44)/7
= 3/7
= 0.428
Q.15 The following shows the monthly income (in Rs.) of workers in a factory.
| Income (in Rs.) | 7000-8000 | 8000-9000 | 9000-10000 | 10000-11000 | 11000-12000 |
| No. of workers | 4 | 8 | 9 | 6 | 3 |
Calculate the Bowley’s coefficient of skewness.
Ans.
| Income (in Rs.) | 7000-8000 | 8000-9000 | 9000-10000 | 10000-11000 | 11000-12000 |
| No. of workers | 4 | 8 | 9 | 6 | 3 |
| CF | 4 | 12 | 21 | 27 | 30 |
For Q1: n/4 = 30/4 = 7.5
Q1 class = 8000-9000
l = 8000, CF = 4, f = 8, h = 1000
Using, Q1 = l + [(n/4 – CF)/f x h]
= 8000 + [(7.5 – 4)/8 x 1000]
= 8000 + [3500/8]
= 8000 + 437.5
= 8437.5
For Q2: n/2 = 15
Q2 class = 9000-10000
l = 9000, CF = 12, f = 9, h = 1000
Using, Q2 = l + [(n/2 – CF)/f x h]
= 9000 + [(15 – 12)/9 x 1000]
= 9000 + [1000/3]
= 9000 + 333.3
= 9333.3
For Q3: 3n/4 = 3x 30/4 = 22.5
Q3 class = 10000-11000
l = 10000, CF = 21, f = 6, h = 1000
Using, Q3 = l + [(n/4 – CF)/f x h]
= 10000 + [(22.5 – 21)/6 x 1000]
= 10000 + [1500/6]
= 10000 + 250
= 10250
Skewness by Bowley’s coefficient = (Q3 +Q1 -2Q2)/(Q3 – Q1)
= (10250 + 8437.5 – 2 x 9333.3)/(10250 – 8437.5)
= (18687.5 – 18666.6)/1812.5
= 20.9/1812.5
= 0.011
Q.16 The first three central moments of a distribution are 0, 3 and 0.8 respectively. Find the moment coefficients of skewness β1 and γ1.
Ans. μ1 = 0, μ2 = 3, μ3 = 0.8
β1 = (μ3)^2/(μ2)^3
= (0.8)^2/3^3
= 0.64/27
= 0.0237
γ1 = ±√β1
= √0.0237
= 0.154
Q.17 In a distribution second and third central moments are 10 and -9 respectively. In another distribution second second and third central moments are 18 and -12 respectively. Which distribution is more skewed to the left?
Ans. In distribution 1 = μ2 = 10, μ3 = -9
β1 = (μ3)^2/(μ2)^3
= (-9)^2/10^3
= 81/1000
= 0.081
In distribution 2 = μ2 = 18, μ3 = -12
β1 = (μ3)^2/(μ2)^3
= (-12)^2/18^3
= 144/5832
= 0.0247
Now,
0.081 > 0.0247
So, distribution 1 is more skewed towards the left.
Q.18 In a frequency distribution, variance is 5, and moment coefficient of skewness γ1 = -0.5. Find the third central moment of the distribution.
Ans. σ^2 = 5, μ2 = 5
γ1 = – 0.5
γ1 = √ β1
-0.5 = √μ3^2/μ2^3
-0.5 = μ3/√μ2^3
-0.5 = μ3/√5^3
-0.5 = μ2/√125
-0.5 x √125 = μ3
-0.5 x 11.18 = μ3
-5.59 = μ3
Q.19 Compute the moment coefficient of skewness β1 for the following distribution:
| Marks obtained | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| Frequency | 6 | 12 | 22 | 24 | 16 | 12 | 8 |
Ans.
| Marks obtained | Frequency(fi) | xi | fi xi | (xi – x̄)^2 | (xi – x̄)^3 | fi(xi – x̄)^2 | fi(xi – x̄)^3 |
| 0-10 | 6 | 5 | 30 | 900 | -27000 | 5400 | -16200 |
| 10-20 | 12 | 15 | 180 | 400 | -8000 | 4800 | -96000 |
| 20-30 | 22 | 25 | 550 | 100 | -1000 | 2200 | -22000 |
| 30-40 | 24 | 35 | 840 | 0 | 0 | 0 | 0 |
| 40-50 | 16 | 45 | 720 | 100 | 1000 | 1600 | 16000 |
| 50-60 | 12 | 55 | 660 | 400 | 8000 | 4800 | 96000 |
| 60-70 | 8 | 65 | 520 | 900 | 27000 | 7200 | 21600 |
| 100 | 3500 | 26000 | 48000 |
x̄ = 3500/100 = 35
μ2 = Σfi(xi – x̄)^2/Σfi
= 26000/100 = 260
μ3 = Σfi(xi – x̄)^3/Σfi
= 48000/100 = 480
Now, β1 = (μ3)^2/(μ2)^3
= (480 x 480)/(260 x 260 x 260)
= 0.013
Q.20 The first four central moments of a frequency distribution are 0, 6, 19 and 42 respectively. Examine the skewness and kurtosis of the distribution.
Ans. μ1 = 0, μ2 = 6, μ3 = 19, μ4 = 42
For skewness β1 = (μ3)^2/(μ2)^3
= (19 x 19)/6^3
= 361/216
= 1.67
β1 > 0 Therefore, Distribution is positively skewed.
For Kurtosis β2 = μ4/(μ2)^2
= 42/36
= 1.167
β2 < 0 Therefore, distribution is platykurtic.
Q.21 For a mesokurtic distribution, the standard deviation is 0.4. calculate the value of fourth central moment.
Ans. Given σ = 0.4
σ^2 = 0.16(μ2)
Since, Distance is mesokurtic,
Therefore, β2 = 3
Using, β2 = μ4/(μ2)^2
3 = μ4/(0.16)^2
3 x 0.0256 = μ4
0.0768 = μ4
Q.22 For a leptokurtic distribution β2 = 4.3 and μ = 10. Calculate the variance of distribution.
Ans. β2 = 4.3, μ4 = 10, σ^2 = ? (μ2)
β2 = μ4/(μ2)^2
4.3 = 10/(μ2)^2
(μ2)^2 = 10/4.3
(μ2)^2 = 100/43
μ2 = √100/43
μ2 = 10/6.55
μ2 = 1.52
Q.23 For a central distribution, mean is 7, standard deviation is 3, β1 is 1.5 and β2 is 3.4. Find the first four central moments.
Ans. x̄ = 7, SD = 3 (σ), β1 = 1.5, β2 = 3.4
μ1 = 0
μ2 = (σ)^2 = 3^2 = 9
Using β1 = (μ3)^2/(μ2)^3
1.5 = (μ3)^2/9^3 [ From above]
1.5 x 729 = (μ3)^2
μ3 = √1093.5
= 33.07
Using β2 = μ4/(μ2)^2
3.4 = μ4/9^2
3.4 x 81 = μ4
275.4 = μ4
Q.24 Calculate the first four central moments for the following distribution. Also calculate β1 and β2. Comment upon the nature of skewness and kurtosis.
Ans.
| xi | fi | fi xi | (xi – x)^2 | (xi – x̄)^3 | (xi – x̄)^2 | fi(xi – x̄)^2 | fi(xi – x̄)^3 | fi(xi – x̄)^3 |
| 4 | 2 | 80 | 36 | -216 | 1296 | 72 | -432 | 2592 |
| 6 | 5 | 30 | 16 | -64 | 256 | 80 | -320 | 1280 |
| 10 | 8 | 80 | 0 | 0 | 0 | 0 | 0 | 0 |
| 12 | 7 | 84 | 4 | 8 | 16 | 28 | 56 | 112 |
| 16 | 3 | 48 | 36 | 216 | 1296 | 108 | 648 | 3888 |
| 25 | 250 | 288 | -48 | 7872 |
μ1 = 0, x̄ = 250/25 = 10
μ2 = 288/25 = 11.52
μ3 = -48/25 = -1.92
μ4 = 7872/25 = 314.88
β1 = (μ3)^2/(μ2)^3
= (-1.91)^2/(11.52)^3
= 0.0024
γ1 = -√β1
=-√0.0024 < 0 [Negatively skewed]
β2 = μ4/(μ2)^2
= 314.88/(11.52)^
= 2.37 < 3 [It is platykurtic]
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.
Excellent article. I am going through a few of these issues as well.. https://menbehealth.wordpress.com/