Sandeep Garg Class 11 Economics Solutions Chapter 5 – Measures of Central Tendency- Arithmetic Mean is illustrated by the professional economic educator from the contemporary edition of Sandeep Garg Economics Class 11 textbook solutions.
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Sandeep Garg Solutions Class 11 – Chapter 5
Question 1
From the following data find an average.
3 | 5 | 7 | 9 | 11 | 13 | 17 | 19 | 20 | 21 |
Answer:
Here, the average is calculated using a direct method.
\(\left ( \bar{X} \right ) =\, \frac{EX}{N}\)\(\bar{X}\, =\, \frac{3+5+7+9+11+13+17+19+20+21}{10}\)
\(\bar{X}\, =\, \frac{125}{10}\)
\(\bar{X}\, =\, 12.5\)
Question 2
Given below are the daily income of ten families. Evaluate the average daily income.
S. No. | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Daily Income | 100 | 120 | 80 | 85 | 95 | 130 | 200 | 250 | 225 | 275 |
Answer:
Serial No. | Daily Income (in Rs) (X) |
1 2 3 4 5 6 7 8 9 10 | 80 100 90 80 90 110 190 230 210 250 |
N =10 | ∑X= 1430 |
\(\bar{X}\, =\, \frac{80+100+90+80+90+110+190+230+210+255}{10}\)
\(\bar{X}\, =\, \frac{1430}{10}\)
\(\bar{X}\, =\, 143\)
Question 3
Find an average from the below series.
X | 2 | 4 | 5 | 8 | 9 |
f | 3 | 6 | 3 | 7 | 10 |
Answer:
X | f | fX |
2 4 5 8 9 | 3 6 3 7 10 | 6 24 15 56 90 |
N=Σf= 29 | ΣfX= 191 |
\(\bar{X}\, =\, \frac{191}{29}\)
\(\bar{X}\, =\, 6.59\)
Therefore, the average of the above series is 6.59
Question 4
Prepare arithmetic mean from the below frequency table.
Height (in cms.) | 55 | 58 | 60 | 62 | 64 | 65 |
Number of Flowers | 10 | 12 | 18 | 12 | 10 | 7 |
Answer:
Height (X) | Flowers (f) | fX |
55 58 60 62 64 65 | 10 12 18 12 10 7 | 550 696 1080 744 640 455 |
N=Σf= 69 | ΣfX=4165 |
Or, \(\bar{X}\, =\, \frac{3265}{69}\)
\(\bar{X}\, =\, 60.37\)
Question 5
From the following data arrange the mean marks acquired by the students using direct method.
Marks | 0−4 | 4−8 | 8−12 | 12−16 | 16−20 | 20−24 |
No. of Students | 7 | 9 | 16 | 8 | 6 | 4 |
Answer:
Class Interval (Marks) | Mid-Values (m) | Students (f) | fm |
0 − 4 4 − 8 8 − 12 12 − 16 16 − 20 20 − 24 | 2 6 10 14 18 22 | 7 9 16 8 6 4 | 14 54 160 112 108 88 |
Σf Σf=50 | Σfm Σfm=586 |
\(\bar{X}\, =\, \frac{536}{50}\)
\(\bar{X}\, =\, 10.72\)
Therefore, the mean marks acquired by the students are 10.72
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