Advertisements
Advertisements
Question
The median of the following data is 525. Find the values of x and y, if the total frequency is 100.
Class interval | Frequency |
0 – 100 | 2 |
100 – 200 | 5 |
200 – 300 | x |
300 – 400 | 12 |
400 – 500 | 17 |
500 – 600 | 20 |
600 – 700 | y |
700 – 800 | 9 |
800 – 900 | 7 |
900 – 1000 | 4 |
Solution
Class interval | Frequency | Cumulative Frequency |
0 – 100 | 2 | 2 |
100 –200 | 5 | 7 |
200 – 300 | x | 7 + x |
300 –400 | 12 | 19 + x |
400 – 500 | 17 | 36 + x |
500 – 600 | 20 | 56 + x |
600 – 700 | y | 56 + x + y |
700 – 800 | 9 | 65 + x + y |
800 – 900 | 7 | 72 + x + y |
900 – 1000 | 4 | 76 + x + y |
Median = 525, so Median Class = 500 – 600
76 + x + y = 100
⇒ x + y = 24 ......(i)
Median = `l + (n/2 - cf)/f xx h`
Since, l = 500, h = 100, f = 20, cf = 36 + x and n = 100
Therefore, putting the value in the Median formula, we get;
525 = `500 + (50 - (-36 + x))/20 xx 100`
So x = 9
y = 24 – x ......[From equation (i)]
y = 24 – 9 = 15
Therefore, the value of x = 9 and y = 15.
APPEARS IN
RELATED QUESTIONS
For a certain frequency distribution, the value of Mean is 101 and Median is 100. Find the value of Mode.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
Number of letters | Number of surnames |
1 - 4 | 6 |
4 − 7 | 30 |
7 - 10 | 40 |
10 - 13 | 6 |
13 - 16 | 4 |
16 − 19 | 4 |
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
The following table gives the frequency distribution of married women by age at marriage:
Age (in years) | Frequency |
15-19 | 53 |
20-24 | 140 |
25-29 | 98 |
30-34 | 32 |
35-39 | 12 |
40-44 | 9 |
45-49 | 5 |
50-54 | 3 |
55-59 | 3 |
60 and above | 2 |
Calculate the median and interpret the results.
If the median of the following data is 32.5, find the missing frequencies.
Class interval: | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 | 60 - 70 | Total |
Frequency: | f1 | 5 | 9 | 12 | f2 | 3 | 2 | 40 |
Given below is the number of units of electricity consumed in a week in a certain locality:
Class | 65 – 85 | 85 – 105 | 105 – 125 | 125 – 145 | 145 – 165 | 165 – 185 | 185 – 200 |
Frequency | 4 | 5 | 13 | 20 | 14 | 7 | 4 |
Calculate the median.
Find the correct answer from the alternatives given.
Distance Covered per litre (km) | 12 - 14 | 14 - 16 | 16 - 18 | 18 - 20 |
No. of cars | 11 | 12 | 20 | 7 |
The median of the distances covered per litre shown in the above data is in the group . . . . . .
Which measure of central tendency can be determine graphically?
The median of the following frequency distribution is 25. Find the value of x.
Class: | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
Frequency: | 6 | 9 | 10 | 8 | x |
Consider the following frequency distribution:
Class | 0 – 6 | 6 – 12 | 12 – 18 | 18 – 24 | 24 – 30 |
Frequency | 12 | 10 | 15 | 8 | 11 |
The median class is:
A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household:
Family size | 1 – 3 | 3 – 5 | 5 – 7 | 7 – 9 | 9 – 11 |
Numbers of Families | 7 | 8 | 2 | 2 | 1 |
Find the median of this data.