Question-1
Solution:
(a)
Profit (in Rs) |
No of shops Frequency (f) |
Mid value x |
fx |
0-10 10-20 20-30 30-40 40-50 50-60 |
12 18 27 P 17 6 |
5 15 25 35 45 55 |
60 270 675 35x 765 330 |
80+p |
2100+35p |
Given that Arithmetic mean of the data is 28.
Therefore,
â‡’ 2100+35p = 2240+28p
(i.e) 7p = 140
â‡’ p = 20
Thus the missing frequency is 20.
(b)
Profit (in Rs) |
No of shops f |
Cumulative frequency |
0-10 10-20 20-30 30-40 40-50 50-60 |
12 18 27 20 17 6 |
12 30 57 77 94 100 |
Median =
= 20 + = = 27.407
Therefore Median = 27.41
Question-2
Workers | A | B | C | D | E | F | G | H | I | J |
Daily Income (in Rs) | 120 | 150 | 180 | 200 | 250 | 300 | 220 | 350 | 370 | 260 |
Total Daily income for ten workers = = 2400
Therefore Arithmetic mean = = 240
Question-3
Following information pertains to the daily income of 150 families. Calculate the arithmetic mean.
Solution:
Income |
No of families f |
Mid class x |
fx |
75 - 85 85 - 95 95-105 105-115 115-125 125-135 135-145 145-155 |
10 25 20 25 10 20 15 25 |
80 90 100 110 120 130 140 150 |
800 2250 2000 2750 1200 2600 2100 3750 |
150 |
17450 |
Arithmetic Mean = = 116.33
Question-4
4. The size of land holdings of 380 families in a village is given below. Find the median size of land holdings.
Solution:
Size of land (in acres) |
No of families f |
Mid value x |
c.f |
0 â€“ 100 100-200 200-300 300-400 400-500 |
40 89 148 64 39 |
50 150 250 350 450 |
40 129 277 341 380 |
380 |
Median =
= 200 + = 200 + 41.217 = 241.22
Thus, the median size of land holdings = 241.22 acres.
Question-5
5. The following series relates to the daily income of workers employed in a firm. Compute
(a) Highest income of lowest 50% workers
(b) Minimum income earned by the top 25% workers and
(c) Maximum income earned by lowest 25% workers.
Daily Income (in Rs) 10â€“14 15â€“19 20â€“24 25â€“29 30â€“34 35â€“39
Number of workers 5 10 15 20 10 5
(Hint: compute median, lower quartile and upper quartile.)
[Ans. (a) Rs 25.11 (b) Rs 19.92 (c) Rs 29.19]
Solution:
Daily income (in Rs) |
No of workers f |
Mid- value x |
fx |
Cumulative Frequency (c.f) |
9.5-14.5 14.5-19.5 19.5-24.5 24.5-29.5 29.5-34.5 34.5-39.5 |
5 10 15 f_{1} 20 f_{2} 10 f_{3} 5 |
12 17 22 27 32 37 |
60 170 330 540 320 185 |
5 15 30 50 60 65 |
65 |
1605 |
= 24.5 + = = 25.12
Lower Quartile Q_{1} =
=
= 19.5 + = 19.5 + 0.42 = 19.92
Upper Quartile Q_{3} =
=
= 29.5 + = 29.5 -
= = 28.87
Thus,
Highest income of lowest 50% workers = 25.12
Minimum income earned by the top 25% workers = 19.92
Maximum income earned by lowest 25% workers = 28.87
Question-6
Production yield 50â€“53 53â€“56 56â€“59 59â€“62 62â€“65 65â€“68 68â€“71 71â€“74 74â€“77
Number of farms 3 8 14 30 36 28 16 10 5
Solution:
Production yield (1 kg/hectare) |
No of farms f |
Mid class x |
fx |
Cumulative frequency c.f |
50-53 53-56 56-59 59-62 62-65 65-68 68-71 71-74 74-77 |
3 8 14 30 f_{0} 36 f_{1} 28 f_{2} 16 10 5 |
51.5 54.5 57.5 60.5 63.5 66.5 69.5 72.5 75.5 |
154.5 436 805 1815 2286 1862 1112 725 377.5 |
3 11 25 55 91 119 135 145 150 |
150 |
9573 |
Mean = = 63.82
Median =
= 62 + = = 63.667
Mode =
=
= 62 + = = 63.286
Thus, Mean = 63.82 kg/hectare,
Median = 63.667 kg/hectare and Mode = 63.29 kg/hectare.
Question-7
Solution:
Arithmetic mean, Assumed mean and Weighted mean.
Arithmetic mean:
Arithmetic mean is the most commonly used measure of central tendency. It is defined as the sum of the values of all observations divided by the number of observations and is usually denoted by x . In general, if there are N observations as X1, X2, X3, ..., XN, then the Arithmetic Mean is given by
=
Where, = sum of all observations and N = total number of observations.
Assumed mean:
As in case of individual series the calculations can be simplified by using assumed mean method, as described earlier, with a simple modification. Since frequency (f) of each item is given here, we multiply each deviation (d) by the frequency to get fd. Then we get S fd. The next step is to get the total of all frequencies i.e. S f. Then find out S fd/S f. Finally the arithmetic mean is calculated by
using assumed mean method.
Weighted mean:
Sometimes it is important to assign weights to various items according to their importance, when you calculate the arithmetic mean. For example, there are two commodities, mangoes and potatoes. You are interested in finding the average price of mangoes (p1) and potatoes (p2).
The arithmetic mean will be. However, you might want to give more importance to the rise in price of potatoes (p2). To do this, you may use as â€˜weightsâ€™ the quantity of mangoes (q1) and the quantity of potatoes (q2). Now the arithmetic mean weighted by the quantities would be.
Question-8
Solution:
Suppose we express, Arithmetic Mean = M_{e} Median = M_{i} Mode = M_{o} so that e, i and o are the suffixes. The relative magnitude of the three are M_{e}>M_{i}>M_{o} or M_{e}<M_{i}<M_{o} (suffixes occurring in alphabetical order). The median is always between the arithmetic mean and the mode.
Question-9
Solution:
Measures of central tendency or averages are used to summarise the data with a single value. Which can represent the entire data. It specifies a single most representative value to describe the data set. Arithmetic mean is the most commonly used average. It is simple to calculate and is based on all the observations. But it is unduly affected by the presence of extreme items. Median is a better summary for such data. Mode is generally used to describe the qualitative data. Median and mode can be easily computed.
Question-10
Solution:
The first Quartile (denoted by Q1) or lower quartile has 25% of the items of the distribution below it and 75% of the items are greater than it.
The second Quartile (denoted by Q2) or median has 50% of items below it and 50% of the observations above it.
The third Quartile (denoted by Q3) or upper Quartile has 75% of the items of the distribution below it and 25% of the items above it.
Thus, Q1 and Q3 denote the two limits within which central 50% of the data lies.
Question-11
11. Which average would be suitable in the following cases?
(i) Average size of readymade garments.
(ii) Average intelligence of students in a class.
(iii) Average production in a factory per shift.
(iv) Average wages in an industrial concern.
(v) When the sum of absolute deviations from average is least.
(vi) When quantities of the variable are in ratios.
(vii) In case of open-ended frequency distribution.
Solution:
(i) Direct method.
(ii) Assumed mean method.
(iii) Step deviation method.
(iv) Step deviation method.
(v) Direct method.
(vi) Assumed mean method.
(vii) Step deviation method.
Question-12
Solution:
Arithmetic mean.
Question-13
Solution:
Hundred equal parts.
Question-14
Solution:
Mode.
Question-15
Solution:
2median = mode â€“ 3mean.
Question-16
Solution:
Median.