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प्रश्न
The following is the distribution of 160 Workers according to the wages in a certain factory:
| Wages more than (in ₹) |
No. of workers |
| 8000 | 160 |
| 9000 | 155 |
| 10000 | 137 |
| 11000 | 91 |
| 12000 | 57 |
| 13000 | 23 |
| 14000 | 10 |
| 15000 | 1 |
| 16000 | 0 |
Determine the values of all quartiles and interpret the results.
उत्तर
The given table is a more than cumulative frequency. We transform the given table into less than cumulative frequency.
We construct the less than cumulative frequency table as given below:
| Wages (in ₹) |
No. of workers (f) | Less than Cumulative frequency (c.f.) |
| 8000 – 9000 | 160 – 155 = 5 | 5 |
| 9000 – 10000 | 155 – 137 = 18 | 23 |
| 10000 – 11000 | 137 – 91 = 46 | 69 ← Q1 |
| 11000 – 12000 | 91 – 57 = 34 | 103 ← Q2 |
| 12000 – 13000 | 57 – 23 = 34 | 137 ← Q3 |
| 13000 – 14000 | 23 – 10 = 13 | 150 |
| 14000 – 15000 | 10 – 1 = 9 | 159 |
| 15000 – 16000 | 1 – 0 = 1 | 160 |
| 16000 – 17000 | 0 | 160 |
| Total | 160 |
Here, N = 160
∴ Q1 class = class containing `("N"/4)^"th"` observation
∴ `"N"/4=160/4` = 40
Cumulative frequency which is just greater than (or equal to) 40 is 69.
∴ Q1 lies in the class 10000 – 11000
∴ L = 10000, f = 46, c.f. = 23, h = 1000
Q1 = `"L"+"h"/"f"("N"/4-"c.f.")`
= `10000 + (1000)/(46)(40 - 23)`
= `10000+1000/46(17)`
= `10000+17000/46`
= 10000 + 369.57
= 10369.57
Q2 class = class containing `((2"N")/4)^"th"` observation
∴ `(2"N")/4=(2xx160)/4` = 80
Cumulative frequency which is just greater than (or equal to) 80 is 103
∴ Q2 lies in the class 11000 – 12000
∴ L = 11000, f = 34, c.f. = 69, h = 1000
Q2 = `"L"+"h"/"f"("2N"/4 - "c.f.")`
= `11000 + (1000)/(34)(80 - 69)`
= `11000+1000/34(11)`
= `11000+11000/34`
= 11000 + 323.529
= 11323.529
Q3 class = class containing `(("3N")/4)^"th"` observation
`(3"N")/(4) = (3(160))/(4)` = 120
Cumulative frequency which is just greater than (or equal to) 120 is 137.
∴ Q3 lies in the class 12000 – 13000
∴ L = 12000, f = 34, c.f. = 103; h = 1000
Q3 = `"L"+"h"/"f"((3"N")/4 - "c.f.")`
= `12000 + (1000)/(34)(120 - 103)`
= `12000+1000/34(17)`
= `12000+1000/2`
= 12000 + 500
= 12500
∴ The quartiles are
Q1 = Rs.10369.57
Q2 = Rs. 11323.529
Q3 = Rs. 12500
Q1 < Q2 < Q3.
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संबंधित प्रश्न
Give the correct option:
Statements that do not apply to Quartiles.
- First, arrange the values in ascending or descending order.
- Observation can be divided into 4 parts.
- They are represented as Q1, Q2 and Q3.
- Q2 is also known as a median.
Give economic term:
Value that divides the whole set of observations into four equal parts.
State with reasons whether you agree or disagree with the following statement:
Median is also known as the second quartile.
Calculate Q3 for the following data.
| Sales (in lakhs ₹) | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| No. of firms | 20 | 30 | 70 | 48 | 32 | 50 |
Answer the following questions on the basis of the given data :
| Marks | 10 | 20 | 30 | 40 | 50 |
| No. of Students | 4 | 7 | 13 | 8 | 6 |
1) Write the formula of Q1 and Q3.
2) Find out the median of the above data?
3) Find out the cumulative frequency of the last value in the above data.
4) Find out the value of 'n' in the above data
Compute all the quartiles for the following series of observations:
16, 14.9, 11.5, 11.8, 11.1, 14.5, 14, 12, 10.9, 10.7, 10.6, 10.5, 13.5, 13, 12.6.
The heights (in cm) of 10 students are given below: 148, 171, 158, 151, 154, 159, 152, 163, 171, 145. Calculate Q1 and Q3 for above data.
For the following data of daily expenditure of families (in ₹), compute the expenditure below which 75% of families include their expenditure.
| Daily expenditure (in ₹) | 350 | 450 | 550 | 650 | 750 |
| No. of families | 16 | 19 | 24 | 28 | 13 |
Calculate all the quartiles for the following frequency distribution:
| No. of E-transactions per day | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| No. of days | 10 | 35 | 45 | 95 | 64 | 32 | 10 | 9 |
The following is the frequency distribution of heights of 200 male adults in a factory:
| Height (in cm.) | No. of male adults |
| 145 – 150 | 4 |
| 150 – 155 | 6 |
| 155 – 160 | 25 |
| 160 – 165 | 57 |
| 165 – 170 | 64 |
| 170 – 175 | 30 |
| 175 – 180 | 8 |
| 180 – 185 | 6 |
Find the central height.
Following is the grouped data for duration of fixed deposits of 100 senior citizens from a certain bank:
| Fixed deposit (in days) | 0 – 180 | 180 – 360 | 360 – 540 | 540 – 720 | 720 – 900 |
| No. of senior citizens | 15 | 20 | 25 | 30 | 10 |
Calculate the limits of fixed deposits of central 50% senior citizens.
| Weight (kg) | 40 – 45 | 45 – 50 | 50 – 55 | 55 – 60 | 60 – 65 | 65 –70 | 70 – 75 | 75 – 80 |
| No. of person | 4 | 15 | 20 | 30 | 20 | 10 | 8 | 4 |
For above data, find all quartiles and number of persons weighing between 57 kg and 72.
For the following data showing weights of 100 employees, find the maximum weight of the lightest 25% of employees.
| Weight (kg) | 45 – 50 | 50 – 55 | 55 – 60 | 60 – 65 | 65 – 70 | 70 – 75 | 75 – 80 |
| No. of employees | 6 | 8 | 15 | 26 | 20 | 14 | 11 |
