Advertisements
Advertisements
Question
From the following data, draw the two types of cumulative frequency curves and determine the median:
Marks | Frequency |
140 – 144 | 3 |
144 – 148 | 9 |
148 – 152 | 24 |
152 – 156 | 31 |
156 – 160 | 42 |
160 – 164 | 64 |
164 – 168 | 75 |
168 – 172 | 82 |
172 – 176 | 86 |
176 – 180 | 34 |
Solution
(i) Less than series:
Marks | Number of students |
Less than 144 | 3 |
Less than 148 | 12 |
Less than 152 | 36 |
Less than 156 | 67 |
Less than 160 | 109 |
Less than 164 | 173 |
Less than 168 | 248 |
Less than 172 | 230 |
Less than 176 | 416 |
Less than 180 | 450 |
We plot the points A(144, 3), B(148, 12), C(152, 36), D(156, 67), E(160, 109), F(164, 173), G(168, 248) and H(172, 330), I(176, 416) and J(180, 450).
Join AB, BC, CD, DE, EF, FG, GH, HI, IJ and JA with a free hand to get the curve representing the ‘less than type’ series.
(ii) More than series:
Marks | Number of students |
More than 140 | 450 |
More than 144 | 447 |
More than 148 | 438 |
More than 152 | 414 |
More than 156 | 383 |
More than 160 | 341 |
More than 164 | 277 |
More than 168 | 202 |
More than 172 | 120 |
More than 176 | 34 |
Now, on the same graph paper, we plot the points A1(140, 450), B1(144, 447), C1(148, 438), D1(152, 414), E1(156, 383), F1(160, 277), H1(168, 202), I1(172, 120) and J1(176, 34).
Join A1B1, B1C1, C1D1, D1E1, E1F1, F1G1, G1H1, H1I1 and I1J1 with a free hand to get the ‘more than type’ series.
The two curves intersect at point L. Draw LM ⊥ OX cutting the x-axis at M. Clearly, M = 166cm
Hence, median = 166cm
APPEARS IN
RELATED QUESTIONS
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.
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 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 median of the distribution given below is 14.4 . Find the values of x and y , if the total frequency is 20.
Class interval : | 0-6 | 6-12 | 12-18 | 18-24 | 24-30 |
Frequency : | 4 | x | 5 | y | 1 |
For a frequency distribution, mean, median and mode are connected by the relation
If the median of the following frequency distribution is 32.5. Find the values of f1 and f2.
The marks obtained by 100 students of a class in an examination are given below.
Mark | No. of Student |
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 'a less than' type cumulative frequency curves (ogive). Hence find the median.
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:
Look at the following table below.
Class interval | Classmark |
0 - 5 | A |
5 - 10 | B |
10 - 15 | 12.5 |
15 - 20 | 17.5 |
The value of A and B respectively are?
Form the frequency distribution table from the following data:
Marks (out of 90) | Number of candidates |
More than or equal to 80 | 4 |
More than or equal to 70 | 6 |
More than or equal to 60 | 11 |
More than or equal to 50 | 17 |
More than or equal to 40 | 23 |
More than or equal to 30 | 27 |
More than or equal to 20 | 30 |
More than or equal to 10 | 32 |
More than or equal to 0 | 34 |