मराठी

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 - Mathematics

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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:

  1. the median
  2. the number of student who obtained more than 75% in test.
  3. the number of students who did not pass in the test if the pass percentage was 40.
  4. the lower quartile.
आलेख

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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पाठ 24: Measure of Central Tendency(Mean, Median, Quartiles and Mode) - Exercise 24 (E) [पृष्ठ ३७५]

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सेलिना Mathematics [English] Class 10 ICSE
पाठ 24 Measure of Central Tendency(Mean, Median, Quartiles and Mode)
Exercise 24 (E) | Q 7 | पृष्ठ ३७५

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:

  1. The median height.
  2. The interquartile range.
  3. 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:

  1. Median 
  2. Lower quartile 
  3. 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?


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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

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Find the mean of: 5, 2.4, 6.2, 8.9, 4.1 and 3.4


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