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प्रश्न
Find the mode of the following frequency distribution.
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| Frequency | 8 | 10 | 10 | 16 | 12 | 6 | 7 |
उत्तर
| Class | Frequency |
| 0-10 | 8 |
| 10-20 | 10 |
| 20-30 | 10 →f0 |
| 30-40 | 16 →f1 |
| 40-50 | 12 →f2 |
| 50-60 | 6 |
| 60-70 | 7 |
ere, 30-40 is the modal class and I = 30, h = 10
∴ Mode = I + `((f_0 - f_1)/(2f_1 - f_0 - f_2)) xx "h"`
= 30 + `((16 - 10)/(2xx16-10-12)) xx 10`
= 30 + `16/10xx10 = 30 + 6 = 36`
संबंधित प्रश्न
The heights of 50 girls of Class X of a school are recorded as follows:
| Height (in cm) | 135 - 140 | 140 – 145 | 145 – 150 | 150 – 155 | 155 – 160 | 160 – 165 |
| No of Students | 5 | 8 | 9 | 12 | 14 | 2 |
Draw a ‘more than type’ ogive for the above data.
From the following frequency, prepare the ‘more than’ ogive.
| Score | Number of candidates |
| 400 – 450 | 20 |
| 450 – 500 | 35 |
| 500 – 550 | 40 |
| 550 – 600 | 32 |
| 600 – 650 | 24 |
| 650 – 700 | 27 |
| 700 – 750 | 18 |
| 750 – 800 | 34 |
| Total | 230 |
Also, find the median.
The following table, construct the frequency distribution of the percentage of marks obtained by 2300 students in a competitive examination.
| Marks obtained (in percent) | 11 – 20 | 21 – 30 | 31 – 40 | 41 – 50 | 51 – 60 | 61 – 70 | 71 – 80 |
| Number of Students | 141 | 221 | 439 | 529 | 495 | 322 | 153 |
(a) Convert the given frequency distribution into the continuous form.
(b) Find the median class and write its class mark.
(c) Find the modal class and write its cumulative frequency.
The median of a given frequency distribution is found graphically with the help of
If \[u_i = \frac{x_i - 25}{10}, \Sigma f_i u_i = 20, \Sigma f_i = 100, \text { then }\]`overlineX`
The arithmetic mean of the following frequency distribution is 53. Find the value of k.
| Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
| Frequency | 12 | 15 | 32 | k | 13 |
Calculate the mean of the following frequency distribution :
| Class: | 10-30 | 30-50 | 50-70 | 70-90 | 90-110 | 110-130 |
| Frequency: | 5 | 8 | 12 | 20 | 3 | 2 |
For the following distribution:
| C.I. | 0 - 5 | 6 - 11 | 12 - 17 | 18 - 23 | 24 - 29 |
| f | 13 | 10 | 15 | 8 | 11 |
the upper limit of the median class is?
If the sum of all the frequencies is 24, then the value of z is:
| Variable (x) | 1 | 2 | 3 | 4 | 5 |
| Frequency | z | 5 | 6 | 1 | 2 |
The following are the ages of 300 patients getting medical treatment in a hospital on a particular day:
| Age (in years) | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 |
| Number of patients | 60 | 42 | 55 | 70 | 53 | 20 |
Form: More than type cumulative frequency distribution.
