Class 11 Applied Maths Chapter 13 (Ex – 13.5)

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Class 11 Applied Maths Chapter 13

Class 11 Applied Maths Chapter 13 Solutions

Descriptive Statistics

EXERCISE- 13.5

Q.1 Find Covariance(X, Y) between x and y when Σxi = 50, Σyi = 30, Σxiyi = -115 and n = 1.

Ans. Using Covariance(X, Y) = 1/N [ΣXY – 1/N ΣX.ΣY]

Covariance (X, Y) = 1/10 [-115 – 1/10 x 50 x 30]

Covariance(X, Y) = 1/10[-115 – 150]

Covariance(X, Y) = 1/10[-265]

Covariance (X, Y) = – 26.5

Q.2 find Σxi = 100, Σyi = 140, Σ(xi – 10) (yi – 15) = 60 and n = 10, find covariance between x and y.

Ans. Σxi = 100

n = 10

Therefore, x̄ = 10

Taking, Σyi = 150

n = 10

ȳ = 15

Using, covariance (x, y) = Σ(xi – x̄)(yi – ȳ)]/N

Here, x̄ = 10 & ȳ = 15,

Covariance (X, Y) = Σ(xi – 10)(yi -15)]/10

Covariance (X, Y) = 60/10

= 6

Q.3 Find the covariance of the data given below:

(1,5), (2,7), (3,9), (4,11), (5,10), (6,9), (7,8), (8,7), (9,6), (10,5)

Ans. ΣXY = 5 + 14 + 27 + 44 + 50 + 54 + 56 + 56 + 54 + 50 = 410

 ΣX = 1 + 2 + 3 …. + 10 = [n x (n+1)]/2

= (10 x 11)/2

= 55

ΣY = 77

Using Covariance(X, Y) = 1/N [ΣXY – 1/N ΣX.ΣY]

Covariance (X, Y) = 1/10 [410 – 1/10 x 55 x 77]

Covariance(X, Y) = 1/10 [410 – 847/2]

Covariance (X, Y) = 1/10 [410 – 423.5]

Covariance(X, Y) = 1/10 [-13.5]

Covariance (X, Y) = -1.35

Q.4 Calculate the covariance of the following bivariate data:

X456789101112131415
Y787266605448423630241812

Ans. A = 10, B = 45

X456789101112131415
Y787266605448423630241812
U = X – A-6-5-4-3-2-1012345
V = Y – B3327211593-3-9-15-21-27-33
UV-198-135-84-45-18-30-9-30-63-108-165

n = 12 , ΣU = -6, ΣV = 0, ΣUV = -858

Using Covariance(X, Y) = 1/N [ΣUV – 1/N ΣU.ΣV]

Covariance (X, Y) = 1/12 [-858 – 1/12 x -6 x 0]

Covariance(X, Y) = -858/12

Covariance (X, Y) = -71.5

Q.5 Calculate the covariance of the observations (3,5), (6,7), (9,9), (12,11), (15,13), (18,15), (21,17), (24,19) using assumed means A = 13 and B = 12.

Ans.

U = X – A-10-7-4-125811
V = Y – B-7-5-3-11357
UV70351212154077

ΣU = 4, ΣV = 0, ΣUV = 252

Using Covariance(X, Y) = 1/N [ΣUV – 1/N ΣU.ΣV]

Covariance (X, Y) = 1/8 [252 – 1/8 x 4 x 0]

Covariance(X, Y) = 252/8

Covariance (X, Y) = 31.5

Q.6 Calculate the covariance of the following data:

X12345678
Y1694114916

Ans.

X12345678
Y1694114916
XY161812462872144

ΣX = 40, ΣY = 60, ΣXY = 300

Using Covariance(X, Y) = 1/N [ΣXY – 1/N ΣX.ΣY]

Covariance (X, Y) = 1/8 [300 – 1/8 x 40 x 60]

Covariance(X, Y) = 1/8 x 0

Covariance (X, Y) = 0

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.


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2 responses to “Class 11 Applied Maths Chapter 13 (Ex – 13.5)”

  1. REPUBLICAN Avatar
    REPUBLICAN

    THANKS FOR DOING THIS NOBLE WORK

    1. My pleasure….!!

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