Use graph paper for this question. A survey regarding height (in cm ) of 60 boys belonging to Class 10 of a school was conducted. The following data was recorded: Height in cm 135 - 140 140 - 145 145 - 150 150 - 155 155 - 160 160 - 165 165 - 170 No. of boys 4 8 20 14 7 6 1 Taking 2 cm = height of 10 cm along one axis and 2 cm = 10 boys along the other axis draw an ogive of the above distribution. Use the graph to estimate the following: The median

The cumulative frequency distribution for the given data is as follows: Height (in cm ) No. of boys Cumulative frequency 135 - 140 4 4 140 - 145 8 12 145 - 150 20 32 150 - 155 14 46 155 - 160 7 53 160 - 165 6 59 165 - 170 1 60 To draw the Ogive: Plot the points ( 140 , 4 ) , ( 145 , 12 ) , ( 150 , 32 ) , ( 155 , 46 ) , ( 160 , 53 ) , ( 165 , 59 ) , ( 170 , 60 ) and ( 135 , 0 ) after drawing a kink on the x - axis near to the origin, representing that it starts from 135 instead of 0 . The Ogive for the given data is as follows: Here, n (total number of boys) = 60 , which is even. To find the Median: ( i ) Let A be the point on y - axis representing frequency = n 2 = 30 ( i i ) Through A , we draw a horizontal line to meet the cumulative frequency curve at P . ( i i i ) Through P , draw a vertical line to meet the x - axis at Q . The abscissa of the point Q represents 149 . 5 . ∴ The required median height = 149 . 5 cm .

Use graph paper for this question. A survey regarding height (in cm ) of 60 boys belonging to Class 10 of a school was conducted. The following data was recorded: Height in cm 135 - 140 140 - 145 145 - 150 150 - 155 155 - 160 160 - 165 165 - 170 No. of boys 4 8 20 14 7 6 1 Taking 2 cm = height of 10 cm along one axis and 2 cm = 10 boys along the other axis draw an ogive of the above distribution. Use the graph to estimate the following: If above 158 cm is considered as the tall boys of the class. Find the number of boys in the class who are tall.

The cumulative frequency distribution for the given data is as follows: Height (in cm ) No. of boys Cumulative frequency 135 - 140 4 4 140 - 145 8 12 145 - 150 20 32 150 - 155 14 46 155 - 160 7 53 160 - 165 6 59 165 - 170 1 60 To draw the Ogive: Plot the points ( 140 , 4 ) , ( 145 , 12 ) , ( 150 , 32 ) , ( 155 , 46 ) , ( 160 , 53 ) , ( 165 , 59 ) , ( 170 , 60 ) and ( 135 , 0 ) after drawing a kink on the x - axis near to the origin, representing that it starts from 135 instead of 0 . The Ogive for the given data is as follows: To find the number of tall boys in the class mark a point D be the point on x - axis representing 158 Through D , we draw a vertical line to meet the cumulative frequency curve at E . Through E , draw a horizontal line to meet the y - axis at F . The ordinate of the point F represents 51 . Number of tall boys with height more than 158 cm = Total number of boys - Number of boys with boys less than or equal to 158 cm . ∴ The number of tall boys with height more than 158 cm = 60 - 51 = 9 .

Use graph paper for this question. The marks obtained by 120 students in an English test are given below: Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 80 - 90 90 - 100 Number of students 5 9 16 22 26 18 11 6 4 3 Draw the ogive and hence, estimate the upper quartile marks.

If we arrange the data into cumulative frequency table, we get Marks Number of students ( f ) Cumulative Frequency Points 0 - 10 5 5 10 , 5 10 - 20 9 5 + 9 = 14 20 , 14 20 - 30 16 14 + 16 = 30 30 , 30 30 - 40 22 30 + 22 = 52 40 , 52 40 - 50 26 52 + 26 = 78 50 , 78 50 - 60 18 78 + 18 = 96 60 , 96 60 - 70 11 96 + 11 = 107 70 , 107 70 - 80 6 107 + 6 = 113 80 , 113 80 - 90 4 113 + 4 = 117 90 , 117 90 - 100 3 117 + 3 = 120 100 , 120 n = 120 Upper quartile is 3 4 n th term So, here upper quartile = 3 × 120 4 = 90 th term. Now, take marks on the x - axis and lower class limit on the y - axis and plot the graph. We get, So, from the graph we can see that the upper quartile mark is 72 .

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter -1 - Statistics>Exercise>Q 39

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