Advertisements
Advertisements
Question
If the median of the following frequency distribution is 32.5. Find the values of f1 and f2.
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | Total |
| Frequency | f1 | 5 | 9 | 12 | f2 | 3 | 2 | 40 |
Solution
Given: Median = 32.5
We prepare the cumulative frequency table, as given below.
| Class interval: | Frequency: (fi) | Cumulative frequency (c.f.) |
| 0-10 | f1 | f1 |
| 10-20 | 5 | 5 + f1 |
| 20-30 | 9 | 14 + f1 |
| 30-40 | 12 | 26 +f1 |
| 40-50 | f2 | 26 +f1 + f2 |
| 50-60 | 3 | 29 + f1 + f2 |
| 60-70 | 2 | 31 + f1 + f2 |
| N = 40 = 31 + f1 +f2 |
Now, we have
N = 40
31 + f1 + f2 = 40
f2 = 9 - f1 ........(1)
Also, `("N")/(2) = 20`
Since median = 32.5 so the median class is 30 - 40.
Here, l = 30, f = 12 , F = 14 + f1 and h = 10
We know that
Median = `"l" + {{("N")/(2) -"F")/("f")} xx "h"`
`32.5 = 30 + {(20-(14+"f"_1))/(12)} xx 10`
`2.5 = ((6 - "f"_1) xx 10)/(12)`
2.5 x 12 = 60 - 10`"f"_1`
`"f"_1 = (30)/(10)`
= 3
Putting the value of `f_1` in (1), we get
`"f"_2` = 9 - 3
= 6
Hence, the missing frequencies are 3 and 6.
APPEARS IN
RELATED QUESTIONS
The following distribution gives the daily income of 50 workers of a factory.
| Daily income (in Rs | 100 − 120 | 120 − 140 | 140 − 160 | 160 − 180 | 180 − 200 |
| Number of workers | 12 | 14 | 8 | 6 | 10 |
Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
The monthly profits (in Rs.) of 100 shops are distributed as follows:
| Profits per shop: | 0 - 50 | 50 - 100 | 100 - 150 | 150 - 200 | 200 - 250 | 250 - 300 |
| No. of shops: | 12 | 18 | 27 | 20 | 17 | 6 |
Draw the frequency polygon for it.
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.
The monthly pocket money of 50 students of a class are given in the following distribution
| Monthly pocket money (in Rs) | 0 - 50 | 50 – 100 | 100 – 150 | 150 -200 | 200 – 250 | 250 - 300 |
| Number of Students | 2 | 7 | 8 | 30 | 12 | 1 |
Find the modal class and give class mark of the modal class.
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.
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 |
Consider the following frequency distributions
| Class | 65 - 85 | 85 - 105 | 105 - 125 | 125 - 145 | 145 - 165 | 165 - 185 | 185-205 |
| Frequency | 4 | 5 | 13 | 20 | 14 | 7 | 4 |
The difference of the upper limit of the median class and the lower limit of the modal class is?
Consider the following distribution:
| Marks obtained | Number of students |
| More than or equal to 0 | 63 |
| More than or equal to 10 | 58 |
| More than or equal to 20 | 55 |
| More than or equal to 30 | 51 |
| More than or equal to 40 | 48 |
| More than or equal to 50 | 42 |
The frequency of the class 30 – 40 is:
Find the unknown entries a, b, c, d, e, f in the following distribution of heights of students in a class:
| Height (in cm) |
Frequency | Cumulative frequency |
| 150 – 155 | 12 | a |
| 155 – 160 | b | 25 |
| 160 – 165 | 10 | c |
| 165 – 170 | d | 43 |
| 170 – 175 | e | 48 |
| 175 – 180 | 2 | f |
| Total | 50 |
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: Less than type cumulative frequency distribution.
