NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | Statistics for Economics IMPORTANT QUESTIONS | CHAPTER –5 Measures of Central Tendency |EDUGROWN|
NCERT Most important question:
Q1.Define median.
Answer: Median is a value located centrally of a series in such a way that half of the value of the series is above it and the other half is below.
Q2.What is the mode?
Answer: The mode is a value that frequently occurs in the series. Which means the modal value has the highest frequency in the series.
Q3.Define the partition value.
Answer: The value that divides the series into more than two parts is known as a partition value.
Q4.Explain quartile.
Answer: The end value of the statistical series when divided into four parts is known as quartile.
Q5.What is positional average?
Answer: Positional average are those averages whose value is worked out on the basis of their position in the statistical series.
Q6.Define the central tendency.
Answer: All the methods of statistical analysis by which the average of the statistical series are analysed is known as a central tendency.
Q7.What are the purpose of average is the statistical method?
Answer: The purpose of the average is the statistical method are
- Brief description
- Comparison
- Formulation of policies
- Statistical analysis
- One value of all
Q8.What are the different kinds of statistical average?
Answer: The different kinds of statistical average are.
- Mathematical average
- Positional average
Q9.What are the two methods that can calculate the simple arithmetic mean in case of individual series?
Answer: The two methods that can calculate the simple arithmetic mean in the case of individual series are.
- Direct method
- Short-cut method
Q10.What are the methods calculating simple arithmetic mean?
Answer: The methods of calculating simple arithmetic mean are.
- Individual series
- Discrete series
- Frequency distribution
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Q11. If the arithmetic mean of the data given below is 28, find (a) the missing frequency, and (b) the median of the series:
| Profit per retail shop (in Rs) | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Number of retail shops | 12 | 18 | 27 | – | 17 | 6 |
Answer
(a) Let the missing frequency be x
Arithmetic mean = 28 (given)
| Profit per retail shop (in Rs) Class Interval | No. of retail shops Frequency (f) | Mid Value (m) | fm |
| 0-10 | 12 | 5 | 60 |
| 10-20 | 18 | 15 | 270 |
| 20-30 | 27 | 25 | 675 |
| 30-40 | x | 35 | 35x |
| 40-50 | 17 | 45 | 765 |
| 50-60 | 6 | 55 | 330 |
| Σf = 80 + x | Σfx = 2100 + 35x |
Mean = Σfx/Σf
⇒ 28 = 2100 + 35x/80 + x⇒ 2240 + 28x = 2100 + 35
⇒ 2240 – 2100 = 35x – 25x⇒ 140 = 7x⇒ x = 140/7 = 20Missing frequency = 20
(b)
| Class Interval | Frequency (f) | Cumulative frequency (CF) |
| 0-10 | 12 | 12 |
| 10-20 | 18 | 30 |
| 20-30 | 27 | 57 |
| 30-40 | x | 77 |
| 40-50 | 17 | 94 |
| 50-60 | 6 | 100 |
| Total | Σf = 100 |
Median = Size of (N/2)th item
= 100/2 = 50th item
It lies in class 20-30.
Q12. The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.
| 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 |
Answer
| Workers | Daily Income (in Rs)X |
| A | 120 |
| B | 150 |
| C | 180 |
| D | 200 |
| E | 250 |
| F | 300 |
| G | 220 |
| H | 350 |
| I | 370 |
| J | 260 |
| Total | ΣX = 2400 |
N = 10
Arithmetic Mean = ΣX/N
= 2400/10
= 240
Arithmetic Mean = 240
Q13. Following information pertains to the daily income of 150 families. Calculate the arithmetic mean.
| Income (in Rs) | Number of families |
| More than 75 | 150 |
| More than 85 | 140 |
| More than 95 | 115 |
| More than 105 | 95 |
| More than 115 | 70 |
| More than 125 | 60 |
| More than 135 | 40 |
| More than 145 | 25 |
Answer
| Income | No. of families Frequency (f) | Mid Class |
urn:uuid:1076b3c7-1f66-96c0-b5d2-96c01f661076(x)fx75-85108080085-952590225095-105201002000105-115251102750115-125101201200125-135201302600 135-145 151402100 145-155 251503750
150
17450 Arithmetic Mean = Σfx/Σf
= 17450/150
= 116.33
Q14. The size of land holdings of 380 families in a village is given below. Find the median size of land holdings.
| Size of Land Holdings (in acres) | Less than 100 | 100-200 | 200-300 | 300-400 | 400 and above |
| Number of families | 40 | 89 | 148 | 64 | 39 |
Answer
| Size of Land Holdings Class Interval | Number of families(f) | Cumulative frequency (CF) |
| 0-100 | 40 | 40 |
| 100-200 | 89 | 129 |
| 200-300 | 148 | 277 |
| 300-400 | 64 | 341 |
| 400-500 | 39 | 380 |
| Total | Σf = 380 |
Σf = N = 380
Median = Size of (N/2)th item
= 380/2 = 190th item
It lies in class 200-300.
Median size of land holdings = 241.22 acres
Q15. 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) | 10-14 | 15-19 | 20-24 | 25-29 | 30-34 | 35-39 |
| Number of workers | 5 | 10 | 15 | 20 | 10 | 5 |
(Hint: Compute median, lower quartile and upper quartile.)
Answer
| Daily Income (in Rs) Class Interval | No of workers(f) | Cumulative frequency (CF) |
| 9.5-14.5 | 5 | 5 |
| 14.5-19.5 | 10 | 15 |
| 19.5-24.5 | 15 | 30 |
| 24.5-29.5 | 20 | 50 |
| 29.5-34.5 | 10 | 60 |
| 34.5-39.5 | 5 | 65 |
| Total | Σf = 65 |
(a) Σf = N = 65
Median = Size of (N/2)th item
= 65/2 = 32.5th item
It lies in class 24.5-29.5.
Highest income of lowest 50% workers = Rs 25.12
(b) First, we need to find Q1
Class interval of Q1 = (N/4)th items
= (65/4)th items = 16.25th item
It lies in class 19.5-24.5.
Minimum income earned by the top 25% workers = Rs 19.92
(c) First, we need to find Q3
Class interval of Q3 = 3 (N/4)th items
= 3 (65/4)th items = 3 × 16.25th item
= 48.75th item
It lies in class 24.5-29.5.
Maximum income earned by lowest 25% workers = Rs 29.19
Q16. The following table gives production yield in kg. per hectare of wheat of 150 farms in a village. Calculate the mean, median and mode values.
| Production yield (kg. per hectare) | 50-53 | 53-56 | 56-59 | 59-62 | 62-65 | 65-68 | 68-71 | 71-74 | 74-77 |
| Number of workers | 3 | 8 | 14 | 30 | 36 | 28 | 16 | 10 | 5 |
Answer
| Production Yield (kg. per hectare) | No. of farms Frequency (f) | Mid Class (x) | fx | Cumulative frequency (CF) |
| 50-53 | 3 | 51.5 | 154.5 | 3 |
| 53-56 | 8 | 54.5 | 436 | 11 |
| 56-59 | 14 | 57.5 | 805 | 25 |
| 59-62 | 30 | 60.5 | 1815 | 55 |
| 62-65 | 36 | 63.5 | 2286 | 91 |
| 65-68 | 28 | 66.5 | 1862 | 119 |
| 68-71 | 16 | 69.5 | 1112 | 135 |
| 71-74 | 10 | 72.5 | 725 | 145 |
| 74-77 | 5 | 75.5 | 377.5 | 150 |
| Σf = 150 | Σfx = 9573 |
Mean = Σfx/Σf = 9573/150 = 63.82 hectare
Modal Class = 62-65
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