Advertisements
Advertisements
प्रश्न
Calculate the missing frequency form the following distribution, it being given that the median of the distribution is 24
Age (in years) | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
Number of persons |
5 | 25 | ? | 18 | 7 |
उत्तर
Let the missing frequency be x.
To find the median let us put data in the table given below:
Age (in years) | Number of persons (f) | Cumulative frequency (cf) |
0-10 | 5 | 5 |
10-20 | 25 | 30 |
20-30 | x | 30+x |
30-40 | 18 | 48+x |
40-50 | 7 | 55+x |
The given median is 24,
∴ the median class is 20-30.
∴ / = 20, ℎ = 10, 𝑁 = 55 + 𝑥, 𝑓 = 𝑥 𝑎𝑛𝑑 𝑐𝑓 = 30
𝑀𝑒𝑑𝑖𝑎𝑛 = 𝑙 +`((N/2 -cf)/f) xx h`
⇒ 24= 20 + `(((55+x)/2-30)/x) xx 10`
⇒`24-20 = ((55+x-60)/(2x))xx 10`
`⇒ 4= ((x-5)/(2x)) xx 10`
⟹ 8𝑥 = 10𝑥 − 50
⟹ 2𝑥 = 50
⟹ 𝑥 = 25
Thus, the missing frequency is 25.
APPEARS IN
संबंधित प्रश्न
During the medical check-up of 35 students of a class, their weights were recorded as follows:
Weight (in kg | Number of students |
Less than 38 | 0 |
Less than 40 | 3 |
Less than 42 | 5 |
Less than 44 | 9 |
Less than 46 | 14 |
Less than 48 | 28 |
Less than 50 | 32 |
Less than 52 | 35 |
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph verify the result by using the formula.
The following table gives production yield per hectare of wheat of 100 farms of a village.
Production yield (in kg/ha) | 50 − 55 | 55 − 60 | 60 − 65 | 65 − 70 | 70 − 75 | 75 − 80 |
Number of farms | 2 | 8 | 12 | 24 | 38 | 16 |
Change the distribution to a more than type distribution and draw ogive.
The marks obtained by 100 students of a class in an examination are given below:
Marks | Number of students |
0 – 5 | 2 |
5 – 10 | 5 |
10 – 15 | 6 |
15 – 20 | 8 |
20 – 25 | 10 |
25 – 30 | 25 |
30 – 35 | 20 |
35 – 40 | 18 |
40 – 45 | 4 |
45 – 50 | 2 |
Draw cumulative frequency curves by using (i) ‘less than’ series and (ii) ‘more than’ series.Hence, find the median.
What is the lower limit of the modal class of the following frequency distribution?
Age (in years) | 0 - 10 | 10- 20 | 20 -30 | 30 – 40 | 40 –50 | 50 – 60 |
Number of patients | 16 | 13 | 6 | 11 | 27 | 18 |
What is the cumulative frequency of the modal class of the following distribution?
Class | 3 – 6 | 6 – 9 | 9 – 12 | 12 – 15 | 15 – 18 | 18 – 21 | 21 – 24 |
Frequency |
7 | 13 | 10 | 23 | 54 | 21 | 16 |
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 | 6 | 42 | 55 | 70 | 53 | 20 |
Form a ‘less than type’ cumulative frequency distribution.
The following table gives the life-time (in days) of 100 electric bulbs of a certain brand.
Life-tine (in days) | Less than 50 |
Less than 100 |
Less than 150 |
Less than 200 |
Less than 250 |
Less than 300 |
Number of Bulbs | 7 | 21 | 52 | 9 | 91 | 100 |
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 following is the cumulative frequency distribution ( of less than type ) of 1000 persons each of age 20 years and above . Determine the mean age .
Age below (in years): | 30 | 40 | 50 | 60 | 70 | 80 |
Number of persons : | 100 | 220 | 350 | 750 | 950 | 1000 |
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 |