Table of Contents
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
Stay tuned to CoolGyan’S for more CBSE Class 11 Statistics important questions, question papers, sample papers, syllabus and Commerce notifications.
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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