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प्रश्न
The marks obtained by 120 students in a mathematics test is given below:
Marks | No. of students |
0 – 10 | 5 |
10 – 20 | 9 |
20 – 30 | 16 |
30 – 40 | 22 |
40 – 50 | 26 |
50 – 60 | 18 |
60 – 70 | 11 |
70 – 80 | 6 |
80 – 90 | 4 |
90 – 100 | 3 |
Draw an ogive for the given distributions on a graph sheet. Use a suitable scale for your ogive. Use your ogive to estimate:
- the median
- the number of student who obtained more than 75% in test.
- the number of students who did not pass in the test if the pass percentage was 40.
- the lower quartile.
उत्तर
Marks | No. of students | c.f. |
0 – 10 | 5 | 5 |
10 – 20 | 9 | 14 |
20 – 30 | 16 | 30 |
30 – 40 | 22 | 52 |
40 – 50 | 26 | 78 |
50 – 60 | 18 | 96 |
60 – 70 | 11 | 107 |
70 – 80 | 6 | 113 |
80 – 90 | 4 | 117 |
90 – 100 | 3 | 120 |
First of all, we plot the points (0, 0), (10, 5), (20, 14), (30, 30), (40, 52), (50, 78), (60, 96), (70, 107), (80, 113), (90, 117), (100, 120) on graph paper and join them by free hand curve to give the required ogive.
Median = `120/2` = 60.5th term.
i. Through 60.5th marks, draw a line segment parallel to x-axis which meets the curve at A.
From A, draw a line segment perpendicular to, x-axis meeting at B.
∴ B is the median = 43 ...(approx.)
ii. On x-axis, a point P representing 75, drawn vertical line meeting ogive at Q. From Q draw a ⊥ on y-axis meeting y-axis at R, the ordinate of y be 110.
No. of students who obtained upto 75% marks in the test = 110
∴ No. of students who obtained more than 75% = 120 – 110 = 10
iii. No. of students who obtained less than 40% marks in the test = 52 ...(∴ in the graph x = 40, y = 52)
iv. The lower quartile = (Q1)
= `120 xx 1/4`
= 30th term
= 30
From a point B (30) on y-axis, draw a parallel to x-axis meeting the curve at Q and from Q. Draw a line parallel to x-axis meeting it at 30.
∴ Lower Quartile = 30 = Q1.
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संबंधित प्रश्न
Calculate the mean of the distribution given below using the shortcut method.
Marks | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 |
No. of students | 2 | 6 | 10 | 12 | 9 | 7 | 4 |
The following distribution represents the height of 160 students of a school.
Height (in cm) | No. of Students |
140 – 145 | 12 |
145 – 150 | 20 |
150 – 155 | 30 |
155 – 160 | 38 |
160 – 165 | 24 |
165 – 170 | 16 |
170 – 175 | 12 |
175 – 180 | 8 |
Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine:
- The median height.
- The interquartile range.
- The number of students whose height is above 172 cm.
Find the mean of the natural numbers from 3 to 12.
From the following cumulative frequency table, draw ogive and then use it to find:
- Median
- Lower quartile
- Upper quartile
Marks (less than) | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100 |
Cumulative frequency | 5 | 24 | 37 | 40 | 42 | 48 | 70 | 77 | 79 | 80 |
Attempt this question on graph paper.
Age (yrs ) | 5-15 | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 | 65-75 |
No.of casualties | 6 | 10 | 15 | 13 | 24 | 8 | 7 |
(1) Construct the 'less than' Cumulative frequency curve for the above data. using 2 cm =10 years on one axis and 2 cm =10 casualties on the other.
(2)From your graph determine :
(a)the median
(b)the lower quartile
What is the median of 7, 10, 7, 5, 9, 10?
Find the mean of first 10 prime numbers.
Find the mean of the following frequency distribution :
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
Frequency | 4 | 4 | 7 | 10 | 12 | 8 | 5 |
The mean of a certain number of observations is 32. Find the resulting mean, if the observation is, divided by 0.5
Find the mean of: 5, 2.4, 6.2, 8.9, 4.1 and 3.4