Applied Maths Chapter 13 (Ex – 13.2) ML Aggarwal

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Applied Maths Chapter 13 Solutions

Descriptive Statistics

EXERCISE- 13.2

Q.1 Find the variance and the standard deviation for the following data:

3, 6, 11, 12, 18.

Ans. x̄ = Sum of observations/No. of observations

            = 50/5 = 10

Now,

xi – x̄	-7	-4	1	2	8
(xi – x̄)^2	49	16	1	4	64

Therefore, ∑|xi – x̄|^2 = 134

n = 5

σ^2 = ∑|xi – x̄|^2/n

= 134/5

= 26.8

So, σ = √26.8

= 5.15 (Approx)

Q.2 Find the mean, variance, and standard deviation for the following data:

1, 3, 7, 9, 10, 12.

Ans. Using, x̄ = Sum of observations/No. of observations

            = 42/6 = 7

Now,

(xi – x̄)	-6	-4	0	2	3	5
(xi – x̄)^2	36	16	0	4	9	25

Therefore, ∑|xi – x̄|^2 = 90

n = 6

σ^2 = ∑|xi – x̄|^2/n

= 90/6

= 15

So, σ = √15

= 3.9 (Approx)

Q.3 Find the mean, variance, and standard deviation of the following marks scored by 10 students:

45, 70, 62, 60, 50, 48, 67, 34, 65, 58.

Ans. n = 10

Using, x̄ = Sum of observations/No. of observations

            = 559/10 = 55.9

xi^2

2025

4900

3844

3600

2500

2304

4489

1156

4225

3364

Therefore, ∑xi^2 = 32407

σ^2 = ∑xi^2/n – (x̄)^2

= 32407/10 – 3124.81

= 3240.7 – 3124.81

= 115.89

σ = √155.89 = 10.78 (Approx)

Q.4 Find the mean and standard deviation for the following data:

xi	6	10	14	18	24	28	30
fi	2	4	7	12	8	4	3

Ans.

xi	6	10	14	18	24	28	30
fi	2	4	7	12	8	4	3
fixi	12	40	98	216	192	112	90
xi – x̄	-13	-9	-5	-1	5	9	11
(xi – x̄)^2	169	81	25	1	25	81	121
fi(xi – x̄)^2	338	324	175	12	200	324	363

∑fixi = 760, ∑fi = 40

x̄ = Sum of observations/No. of observations

            = 760/40 = 19

∑fi(xi – x̄)^2 = 1736

n = 40

σ^2 = ∑fi(xi – x̄)^2/n

= 1736/40

= 43.4

σ = √43.4

= 6.59 (Approx)

Q.5 Find the mean, variance, and standard deviation for the following data by using the shortcut method:

xi	60	61	62	63	64	65	66	67	68
fi	2	1	12	29	25	12	10	4	5

Ans. Assume Mean (a) = 64

xi	60	61	62	63	64	65	66	67	68
fi	2	1	12	29	25	12	10	4	5
(xi – a)	-4	-3	-2	-1	0	1	2	3	4
fi(xi – a)	-8	-3	-24	-29	0	12	20	12	20
(xi – x̄)^2	16	9	4	1	0	1	4	9	16
fi(xi – x̄)^2	32	9	48	29	0	12	40	36	80

fi(xi – a) = 0 , ∑fi = 100

x̄ = a + ∑fi(xi – a)/∑fi

= 64 + 0/100

= 64

∑fi(xi – x̄)^2 = 286 , ∑fi = 100

σ^2 = ∑fi(xi – x̄)^2/∑fi

= 286/100

= 2.86

σ = √2.86

= 1.69 (Approx)

Q.6 Find the mean, variance, and standard deviation for the following frequency distribution:

Classes	0-10	10-20	20-30	30-40	40-50
Frequency	5	8	15	16	6

Ans. Assumed Mean (a) = 25

Classes	0-10	10-20	20-30	30-40	40-50
Frequency	5	8	15	16	6
Class Mark (xi)	5	15	25	35	45
(xi – a)	-20	-10	0	10	20
fi(xi – a)	-100	-80	0	160	120
(xi – x̄)	-22	-12	-2	8	18
(xi – x̄)^2	484	144	4	64	324
fi(xi – x̄)^2	2420	1152	60	1024	1944

∑fi(xi – a) = 100, ∑fi = 50

Using, x̄ = a + ∑fi(xi – a)/∑fi

= 25 + 100/50

= 25 + 2 = 27

∑fi(xi – x̄)^2 = 6600 , ∑fi = 50

Using, σ^2 = ∑fi(xi – x̄)^2/∑fi

= 6600/50

= 132

σ = √132

= 11.49 (Approx)

Q.7 Calculate the mean, variance, and standard deviation for the following frequency distribution by using the step deviation method:

Classes	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Frequency	3	7	12	15	8	3	2

Ans. Assumed Mean (a) = 65

Classes	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Frequency	3	7	12	15	8	3	2
Class Mark	35	45	55	65	75	85	95
di = xi – a	-30	-20	-10	0	10	20	30
ui = di/h (h=10)	-3	-2	-1	0	1	2	3
fi ui	-9	-14	-12	0	8	6	6
ti = di/c (c=10)	-3	-2	-1	0	1	2	3
fi.ti	-9	-14	-12	0	8	6	6
ti^2	9	4	1	0	1	4	9
fi.ti^2	27	28	12	0	8	12	18

∑fi ui = -15 , h = 10, ∑fi = 50

Using, x̄ = a + [∑fi.ui/∑fi x h]

= 65 + [-15/50 x 10]

= 65 – 3

= 62

∑fi = 50, ∑fi.ti = -15, ∑fi.ti^2 = 105

Using, σ^2 = c^2 [(∑fi.ti^2/∑fi) – (∑fi.ti/∑fi)^2]

= 100 [(105/50) – (-15/50)^2]

= 100 [(105 x 50 – 225)/50^2]

= 100 [ (5250 – 225)/2500]

= 5025/25 = 201

σ = √201

= 14.2 (Approx)

Q.8 Find the mean, variance, and standard deviation for the following data by using step deviation method :

Classes	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Frequency	3	4	7	7	15	9	6	6	3

Ans. Assumed Mean (a) = 55, c =10

Classes	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Frequency	3	4	7	7	15	9	6	6	3
Class Mark(xi)	15	25	35	45	55	65	75	85	95
ti = (xi-a)/c	-4	-3	-2	-1	0	1	2	3	4
ti^2	16	9	4	1	0	1	4	9	16
fi.ti	-12	-12	-14	-7	0	9	12	18	12
fi.ti^2	48	36	28	7	0	9	24	54	48

∑fi = 60, ∑fi.ti = 6, ∑fi.ti^2 = 254

Using, x̄ = a + [∑fi.ti/∑fi x c]

= 55 + [ 6/60 x 10]

= 56

Using, σ^2 = c^2 [(∑fi.ti^2/∑fi) – (∑fi.ti/∑fi)^2]

= 10^2 [(254/60) – (61/60)^2]

= 100 [(254/6) – (1/100)]

= 100 [(2540 – 6)/600]

= 2534/6 = 422.33

σ = √422.33

= 20.55 (Approx)

Q.9 The heights, to the nearest cm, of 30 men are 159, 170, 174, 173, 175, 160, 161, 164, 163, 165, 164, 171, 162, 170, 177, 185, 181, 180, 175, 165, 186, 174, 168, 168, 176, 176, 165, 175, 167 and 180. Using class intervals 155-160, and 160-165, … draw up a grouped frequency distribution and use this to estimate the arithmetic mean and the standard deviation.

Ans. Assumed Mean (a) = 172.5, c = 5

Height	155-160	160-165	165-170	170-175	175-180	180-185	185-190
Frequency	1	6	6	6	6	3	2
Class Mark(xi)	157.5	162.5	167.5	172.5	177.5	182.5	187.5
ti = (xi-a)/c	-3	-2	-1	0	1	2	3
ti^2	9	4	1	0	1	4	9
fi.ti	-3	-12	-6	0	6	6	6
fi.ti^2	9	24	6	0	6	12	18

∑fi = 30, ∑fi.ti = -3, ∑fi.ti^2 = 75

Using, x̄ = a + [∑fi.ti/∑fi x c]

= 172.5 + [-3/30 x 5]

= 172.5 – 0.5 = 172

Using, σ^2 = c^2 [(∑fi.ti^2/∑fi) – (∑fi.ti/∑fi)^2]

= 5^2 [(75/30) – (-3/30)^2]

= 25 [(25/10) – (1/100)]

= 25 [(250 – 1)/100]

= 25 x 249/100 = 62.25

σ = √62.25

= 7.9 (Approx)

Q.10 The mean of 5 observations is 6 and the standard deviation is 2. If the three observations are 5,7 and 9, find the other two observations.

Ans. Let, x & y be the two unknown values of the data set.

x̄ (5 observations) = 6

Sum of observation/5 = 6

5 + 7 + 9 + x + y = 6 x 5

21 + x + y = 30

x + y = 30 – 21

x + y = 9

Therefore, y = 9 – x…….(i)

σ = 2

σ^2 = 4

∑xi^2/n – (x̄)^2 = 4

(5^2 + 7^2 + 9^2 + x^2 + y^2)/5 – 6^2 = 4

(25 + 49 + 81 + x^2 + (9-x)^2/5 = 4 + 6^2

(155 + x^2 + 81 + x^2 – 18x)/5 = 40

2x^2 – 18x + 81 + 155 = 40 x 5

2x^2 – 18x + 236 -200 = 0

2x^2 – 18x + 36 = 0

x^2 – 9x +18 = 0

x^2 – 6x- 3x + 18 = 0

x(x – 6) – 3(x – 6) = 0

(x – 6) (x – 3) = 0

Therefore,

x = 6 or 3

Hence, y = 3 or 6 [From i]

Q.11 The mean and the variance of 5 observations are 4.4 and 8.24 respectively. If the three observations are 1, 2, and 6, find the other two observations.

Ans. Let, x & y be the two unknown observations of the data set.

Using, x̄ = Sum of observations/No. of observations

4.4 = (1 + 2 + 6 + x + y)/5

4.4 x 5 = 9 + x + y

22 – 9 = x + y

13 = x + y

13 – x = y……..(i)

σ^2 = 8.24

∑xi^2/n – (x̄)^2 = 4

(1^2 + 2^2 + 6^2 + x^2 + y^2)/5 – (4.4)^2 = 8.24

(1 + 4 + 36 + x^2 + (13-x)^2)/5 – 19.36 = 8.24

(41 + x^2 + 169 + x^2 – 26x)/5 = 8.24 + 19.36

2x^2 – 26x + 210 = 27.6 x 5

2x^2 – 26x + 210 = 138

2x^2 – 26x + 210 – 138 = 0

2x^2 – 26x + 72 = 0

x^2 – 13x + 36 = 0

x^2 – 9x – 4x 7 + 36 = 0

x(x – 9) – 4(x – 9) = 0

(x – 9 ) (x – 4) = 0

Therefore,

x = 9 , 4

Hence. y = 4 or 9 [From i]

Q.12 The variance of 20 observations is 5. If each observation is multiplied by 2, find the new variance of the resulting observations.

Ans. Given, n = 20, σ^2 = 5

Multiplying Factor = 2

Let, the original data set be,

x1, x2, x3,……….xn

For this data set,

x̄ = ∑xi/n

x̄ = (x1 + x2 + x3….+ xn)/n

nx̄ = x1 + x2 + x3….+ xn

σ^2 = 5

∑(xi)^2/n – (x̄)^2 = 5….(i)

Now, new data set

2×1 + 2×2 + 2×3….+ 2xn

x̄ (N) = (2×1 + 2×2 + 2×3….+ 2xn)/n

x̄ (N) = 2(x1 + x2 + x3….+ xn)/n

x̄ (N) = (2 x n x x̄)/n

x̄ (N) = 2x̄

New Variance (σ^2)N :

(σ^2)N = ∑x Ni/n – (x̄ N)^2

= [(2×1)^2 + (2×2)^2 + (2×3)^2….+ (2xn)^2]/n -(2x̄)^2

= [4×1^2 + 4×2^2 + 4×3^2….+ 4xn^2]/n – 4(x̄)^2

= 4[x1^2 + x2^2 + x3^2….+ xn^2]/n – 4(x̄)^2

= 4 ∑xi^2/n – 4(x̄)^2

= 4 [∑xi^2/n – (x̄)^2]

= 4 x 5……….(i)

= 20

Q.13 While calculating the mean and variance of 10 readings, a student wrongly uses reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and variance.

Ans. n = 10, incorrect x̄ = 45

x̄ = Sum of observations/No. of observations

Using, incorrect values given in the question,

we get,

45 = Sum/10

Sum = 450 (incorrect)

Now, correct sum = 450 – 52 + 25

= 423

Correct x̄ = Correct Sum/10

= 423/10 = 42.3

Now, finding variance

n = 10, Incorrect variance = 16, Incorrect x̄ = 45, Correct x̄ = 42.3

We know that,

σ^2 = ∑xi^2/n – (x̄)^2

Using, incorrect values given in the question, we get

16 = ∑xi^2/10 – (45)^2

16 + 2025 = ∑xi^2/10

20410 = ∑xi^2 (incorrect)

Correct ∑xi = 20410 – (52)^2 + (25)^2

= 20410 – 2704 + 625

= 18331

Correct x̄ = 42.3

Correct σ^2 = ∑xi^2/n – (x̄)^2

= 18331/10 – (42.3)^2

= 1833.1 – 1789.29

= 43.81

Q.14 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 correct standard deviation in each of the following cases:

(a) If the wrong observation is omitted.

(b) If it is replaced by 12.

Ans. n = 20, incorrect x̄ = 10

x̄ = Sum of observations/No. of observations

Using, incorrect values given in the question,

we get,

10 = Sum/20

Sum = 200 (incorrect)

(a) Now, Correct sum = 200 – 8 = 192

New no. of observations = 20 – 1 = 19

New x̄ = 192/19 = 10.11

(b) Now, Correct sum = 200 – 8 + 12 = 204

No. of observations = 20

Correct x̄ = 204/20

= 10.2

FAQ’s related to 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 Applied Maths Chapter 13

In 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, Applied Maths Chapter 13 promises an enriching learning experience that will set you on the path to success. Applied Maths Chapter 13, we delve deep into advanced mathematical concepts that are crucial for understanding.