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Applied Maths Chapter 13 Solutions
Descriptive Statistics
EXERCISE- 13.1
Q.1 Find a range of the following data:
4, 6, 7, 8, 3, 9, 15, 8, 25, 2
Ans. Ascending order = 2, 3, 4, 6, 7, 8, 8, 9, 15, 20
Range = Maximum range – Minimum range
= 25 – 2
= 23
Q.2 Calculate the quartile deviation of the following observations:
15, 20, 22, 28, 35, 27, 44, 48, 50, 55, and 60.
Ans. Ascending order = 15, 20, 22, 27, 28, 35, 44, 48, 50, 55, 60
No. of observation = 11 (odd)
Quartile 1 = [(n + 1)/4]Th
= [12/4]Th
= 3rd observation = 22
Quartile 3 = 3 [(n + 1)/4]Th
= 3 [12/4]Th
= (3 x 3)Th observation
= 9th observation = 50
Quartile deviation = (Q3 – Q1)/2
= (50 – 22)/2
= 28/2
= 14
Q.3 Find the mean deviation about the median of the following data:
3, 6, 11, 12, 18.
Ans. Ascending order = 3, 6, 11, 12, 18
n = 5
Median = [(n + 1)/2]Th observation
= 6/2 = 3rd observation = 11
|xi – Median| = 8, 5, 0, 1, 7
∑|xi – Median| = 8 + 5 + 0 + 1 + 7
= 21
Mean Deviation = 21/5
= 4.2
Q.4 Find the mean deviation about the median of the following data:
1, 3, 7, 9,10, 12.
Ans. x̄ = Sum of observations/No. of observations
= 42/6 = 7
|xi – x̄| = 6, 4, 0, 2, 3, 5
∑|xi – x̄| = 6 + 4 + 2 + 3 + 5
= 20
Mean Deviation = ∑|xi – x̄|/n
= 20/6
= 10/3
Q.5 Find the mean deviation about the median of the following data:
2, 7, 9, 11, 15, 16.
Ans. n = 6
Median = [(n/2)Th + (n/2 + 1)Th]/2
= (3th + 4th)/2
= (9 + 11)/2 = 10
|xi – M| = 8, 3, 1, 1, 5, 6
∑|xi – M| = 24
Mean Deviation = ∑|xi – M|/n
= 24/6
= 4
Q.6 Find the mean deviation about the mean, as well as the median for the following series:
(i) 6, 7, 10, 12, 13, 4, 8, 12
(ii) 20, 28, 40, 12, 30, 15, 50
Ans. (i) 6, 7, 10, 12, 13, 4, 8 , 12
Ascending order = 4, 6, 7, 8, 10, 12, 12, 13
x̄ = Sum of observations/No. of observations
= 72/8 = 9
|xi – x̄| = 5, 3, 2, 1, 1, 3, 3, 4
∑|xi – x̄| = 5 + 3 + 2 + 1 + 1 + 3 + 3 + 4 = 22
Mean Deviation (x̄) = ∑|xi – x̄|/n
= 22/8
= 2.75
Median = [(n/2)Th + (n/2 + 1)Th]/2
= (4th + 5th)/2
= (8 + 10)/2 = 9
|xi – M| = 5, 3, 2, 1, 1, 3, 3, 4
∑|xi – M| = 22
Mean Deviation = ∑|xi – M|/n
= 22/8
= 2.75
(ii) 20, 28, 40, 12, 30, 15, 50
Ascending order = 12, 15, 20, 28, 30, 40, 50
x̄ = Sum of observations/No. of observations
= 195/7 = 27.85
|xi – x̄| = 15.8, 12.85, 7.85, 0.15, 2.15, 12.15, 22.15
∑|xi – x̄| = 73.15
Mean Deviation (x̄) = ∑|xi – x̄|/n
= 73.15/7
= 10.45
n = 7 (odd)
Median = [(n + 1)/2]Th observation
= 8/2 = 4th observation = 28
|xi – M| = 16, 13, 8, 0, 2, 12, 22
∑|xi – M| = 73
Mean Deviation (M) = ∑|xi – M|/n
= 73/7
= 10.42
Q.7 Calculate the quartile deviation for the following frequency distribution.
| Class Interval | 33-34 | 35-39 | 40- 44 | 45-49 | 50- 54 | 55- 59 | 60- 64 | 65- 69 | 70- 74 | 75- 79 | 80- 84 |
| Frequency | 1 | 2 | 6 | 7 | 9 | 11 | 5 | 7 | 5 | 3 | 1 |
Ans.
| Class Interval | 29.5-34.5 | 34.5-39.5 | 39.5-44.5 | 44.5-49.5 | 49.5-54.5 | 54.5-59.5 | 59.5-64.5 | 64.5-69.5 | 69.5-74.5 | 74.5-79.5 | 79.5-84.8 |
| Frequency | 1 | 2 | 6 | 7 | 9 | 11 | 5 | 7 | 5 | 3 | 1 |
| Class Frequency | 1 | 3 | 9 | 16 | 25 | 36 | 44 | 51 | 56 | 59 | 60 |
For Quartile (Q1) = n/4 = 60/4 = 15
Q1 = L + [(n/4 – CF)/f x h]
= 44.5 [(15 – 9)/7 x 5]
= 44.5 + 4.28
= 48.78
For Quartile (Q3) = 3n/4 = 3 x 60/4 = 45
Q1 = L + [(3n/4 – CF)/f x h]
= 64.5 [(45 – 44)/7 x 5]
= 64.5 + 0.71
= 65.21
Quartile Deviation = (Q3 -Q1)/2
= (65.21 – 48.78)/2
= 16.43/2
= 8.215
Q.8 Find the mean deviation about the mean for the following data:
| Size (xi) | 1 | 3 | 5 | 7 | 9 | 11 | 13 | 15 |
| Frequency (fi) | 3 | 3 | 4 | 14 | 7 | 4 | 3 | 4 |
Ans.
| Size (xi) | 1 | 3 | 5 | 7 | 9 | 11 | 13 | 15 |
| Frequency (fi) | 3 | 3 | 4 | 14 | 7 | 4 | 3 | 4 |
| fi xi | 3 | 9 | 20 | 98 | 63 | 44 | 39 | 60 |
| |xi – x̄| | 7 | 5 | 3 | 0 | 1 | 3 | 5 | 7 |
| fi |xi – x̄| | 21 | 15 | 12 | 0 | 7 | 12 | 15 | 28 |
∑fi xi = 336 , ∑fi = 42
x̄ = ∑fi xi/∑fi
x̄ = 336/42
= 8
∑fi |xi – x̄| = 124 , ∑fi = 42
Mean Deviation = ∑fi |xi – x̄|/ ∑fi
= 124/42
= 2.95 (Approx)
Q.9 Find the mean deviation about the median for the following data:
| xi | 15 | 21 | 27 | 30 | 35 |
| fi | 3 | 5 | 6 | 7 | 8 |
Ans.
| xi | 15 | 21 | 27 | 30 | 35 |
| fi | 3 | 5 | 6 | 7 | 8 |
| CF | 3 | 8 | 14 | 21 | 29 |
| |xi – Median| | 15 | 9 | 3 | 0 | 5 |
| fi |xi – Median| | 45 | 45 | 18 | 0 | 40 |
n = 29 (odd)
Median = [(n + 1)/2]Th
= [30/2]Th
= 15Th observation = 30
Therefore, median = 30
∑fi |xi – Median| = 148, ∑fi = 29
Mean Deviation = ∑fi |xi – Median|/∑fi
= 148/29
= 5.1 (Approx)
Q.10 Find the mean deviation about the mean for the following data:
| Marks obtained | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Numbers of students | 2 | 3 | 8 | 14 | 8 | 3 | 2 |
Ans. Let, assumed mean (a) = 45
| Marks obtained | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Numbers of students | 2 | 3 | 8 | 14 | 8 | 3 | 2 |
| Class Mark (xi) | 15 | 25 | 35 | 45 | 55 | 65 | 75 |
| di = xi – a | -30 | -20 | -10 | 0 | 10 | 20 | 30 |
| fi di | -60 | -60 | -80 | 0 | 80 | 60 | 60 |
| |xi – x̄| | 30 | 20 | 10 | 0 | 10 | 20 | 30 |
| fi |xi – x̄| | 60 | 60 | 80 | 0 | 80 | 60 | 60 |
∑fi di = 0 , ∑fi = 40
Therefore, x̄ = a + (∑fi di/∑fi)
= 45 + (0/41)
= 45
Mean Deviation = ∑fi |xi – x̄|/∑fi
= 400/40
= 10
Q.11 Find the mean deviation about the median for the following data:
| Class Interval | 0-6 | 6-12 | 12-18 | 18-24 | 24-30 |
| Frequency | 4 | 5 | 3 | 6 | 2 |
Ans.
| Class Interval | 0-6 | 6-12 | 12-18 | 18-24 | 24-30 |
| Frequency | 4 | 5 | 3 | 6 | 2 |
| Class Frequency | 4 | 9 | 12 | 18 | 20 |
| Class Mark (xi) | 3 | 9 | 15 | 21 | 27 |
| |xi – Median| | 11 | 5 | 1 | 7 | 13 |
| fi |xi – Median| | 44 | 25 | 3 | 42 | 26 |
n = 20 , n/2 = 20/2 = 10
Median class = 12 – 18
Lower limit (L) 140/= 12
Class Frequency = 9
Frequency = 3
Height = 6
Median = L + [(n/2 – CF)/F x h]
= 12 + (10 – 9)/3 x 6
= 12 + 2 = 14
∑fi |xi – Median| = 140 , ∑fi = 20
Mean Deviation = ∑fi |xi – Median|/ ∑fi
= 140/20
= 7
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.
