In an interview with maximum marks 250 , the marks of top 200 students were distributed as below Marks 100 - 114 115 - 129 130 - 144 145 - 159 160 - 174 175 - 189 190 - 204 205 - 219 220 - 235 f 4 5 5 18 18 27 44 40 39 Calculate 12 t h percentile to the nearest integer.

The distribution is not continuous, to make it continuous, we use adjustment factor = upper class - preceeding lower class 2 = 115 - 114 2 = 0 . 5 The cumulative frequency distribution for given data is as under Marks 99 . 5 - 114 . 5 114 . 5 - 129 . 5 129 . 5 - 144 . 5 144 . 5 - 159 . 5 159 . 5 - 174 . 5 174 . 5 - 189 . 5 189 . 5 - 204 . 5 204 . 5 - 219 . 5 219 . 5 - 235 . 5 f 4 5 5 18 18 27 44 40 39 cumulative frequency 4 9 14 32 50 77 121 161 200 Here, n = Σ f i = 200 For P 12 = 12 n 100 t h value i.e., 12 × 200 100 t h value = 24 t h value. which lies in the class 144 . 5 - 159 . 5 ∴ P 12 = l + 12 n 100 - C f × i = 144 . 5 + 24 - 14 18 × 15 = 152 . 83 Thus, P 12 = 153 (nearest to integer)

In an interview with maximum marks 250 , the marks of top 200 students were distributed as below Marks 100 - 114 115 - 129 130 - 144 145 - 159 160 - 174 175 - 189 190 - 204 205 - 219 220 - 235 f 4 5 5 18 18 27 44 40 39 Calculate Median to the nearest integer

The distribution is not continuous, to make it continuous, we use adjustment factor = 115 - 114 2 = 0 . 5 The cumulative frequency distribution for given data is as under Marks 99 . 5 - 114 . 5 114 . 5 - 129 . 5 129 . 5 - 144 . 5 144 . 5 - 159 . 5 159 . 5 - 174 . 5 174 . 5 - 189 . 5 189 . 5 - 204 . 5 204 . 5 - 219 . 5 219 . 5 - 235 . 5 f 4 5 5 18 18 27 44 40 39 cumulative frequency 4 9 14 32 50 77 121 161 200 For Median = n 2 t h value i.e., = 200 2 t h value = 100 t h value. which lies in the class 189 . 5 - 204 . 5 l = 189 . 5 ; c = 77 ; f = 44 ; i = 15 ∴ M d = l + n 2 - C f × i = 189 . 5 + 100 - 77 44 × 15 = 189 . 5 + 23 × 15 44 = 197 . 34 Thus, M d = 197 (nearest to integer)

In an interview with maximum marks 250 , the marks of top 200 students were distributed as below Marks 100 - 114 115 - 129 130 - 144 145 - 159 160 - 174 175 - 189 190 - 204 205 - 219 220 - 235 f 4 5 5 18 18 27 44 40 39 Calculate Upper Quartile to the nearest integer

The distribution is not continuous, to make it continuous, we use adjustment factor = 115 - 114 2 = 0 . 5 The cumulative frequency distribution for given data is as under Marks 99 . 5 - 114 . 5 114 . 5 - 129 . 5 129 . 5 - 144 . 5 144 . 5 - 159 . 5 159 . 5 - 174 . 5 174 . 5 - 189 . 5 189 . 5 - 204 . 5 204 . 5 - 219 . 5 219 . 5 - 235 . 5 f 4 5 5 18 18 27 44 40 39 cumulative frequency 4 9 14 32 50 77 121 161 200 Upper quartile Q 3 = 3 n 4 t h value = 3 × 200 4 th value = 150 th value which lies in class 204 . 5 - 219 . 5 . l = 204 . 5 ; C = 121 ; f = 40 ; i = 15 ∴ Q 3 = 1 + 3 n 4 - C f × i = 204 . 5 + 150 - 121 40 × 15 = 204 . 5 + 10 . 875 = 215 . 375 = 215 (nearest integer)

In an interview with maximum marks 250 , the marks of top 200 students were distributed as below Marks 100 - 114 115 - 129 130 - 144 145 - 159 160 - 174 175 - 189 190 - 204 205 - 219 220 - 235 f 4 5 5 18 18 27 44 40 39 What was the percentile rank of the student who scored 200 marks?

Here the distribution is not continuous, to make it continuous, we use adjustment factor = 115 - 114 2 = 0 . 5 The cumulative frequency distribution for given data is as under Marks 99 . 5 - 114 . 5 114 . 5 - 129 . 5 129 . 5 - 144 . 5 144 . 5 - 159 . 5 159 . 5 - 174 . 5 174 . 5 - 189 . 5 189 . 5 - 204 . 5 204 . 5 - 219 . 5 219 . 5 - 235 . 5 f 4 5 5 18 18 27 44 40 39 cumulative frequency 4 9 14 32 50 77 121 161 200 Percentile rank of any particular item x is the percentage of observations equal to or less than x . Here, P x = 200 We know that, percentile rank P x = l + x × n 100 - C f × i ⇒ 200 = l + x × n 100 - C f × i Since 200 lies in the class 189 . 5 - 204 . 5 . ∴ 200 = 189 . 5 + x × 200 100 - 77 44 × 15 ⇒ 10 . 5 × 44 = ( 2 x - 77 ) 15 ⇒ 462 15 = 2 x - 77 ⇒ 30 . 8 + 77 = 2 x ⇒ x = 107 . 8 2 = 53 . 9 Hence, the required percentile rank of the student who scored 200 marks be 53 . 9 percentile.

Draw cumulative frequency curve for the following data and estimate the median Saving in ₹ (less than) 10 20 30 40 50 60 70 80 Cumulative frequency 15 35 64 84 96 120 192 256

Given that, The given less than type cumulative freq. distribution is as under Saving in ₹ (less than) 10 20 30 40 50 60 70 80 Cumulative frequency 15 35 64 84 96 120 192 256 Taking 1 cm along x - -axis = ₹ 10 and 1 cm along y -axis = 20 Plot the points ( 0 , 0 ) , ( 10 , 15 ) , ( 20 , 35 ) , ( 30 , 64 ) , ( 40 , 84 ) , ( 50 , 96 ) , ( 60 , 120 ) , ( 70 , 192 ) and ( 80 , 256 ) and join all these points by free hand curve. To get the required less than ogive. Median: M d is n 2 t h value i.e., 256 2 t h th value i.e. 128 t h value. Let P be the point on y -axis represents 128 . From P , draw a horizontal line meets the ogive at Q . From Q , draw Q R ⊥ on x -axis meeting x -axis at R . Then abscissa of point R gives the required median. ∴ M d = 61

Draw cumulative frequency curve for the following data and estimate the semi-interquartile range Saving in ₹ (less than) 10 20 30 40 50 60 70 80 Cumulative frequency 15 35 64 84 96 120 192 256

Given that, The given less than type cumulative freq. distribution is as under Saving in ₹ (less than) 10 20 30 40 50 60 70 80 Cumulative frequency 15 35 64 84 96 120 192 256 Taking 1 cm along x - -axis = ₹ 10 and 1 cm along y -axis = 20 Plot the points ( 0 , 0 ) , ( 10 , 15 ) , ( 20 , 35 ) , ( 30 , 64 ) , ( 40 , 84 ) , ( 50 , 96 ) , ( 60 , 120 ) , ( 70 , 192 ) and ( 80 , 256 ) and join all these points by free hand curve. To get the required less than ogive. To find Q 1 : Q 1 is n 4 t h item i.e., 256 4 t h item i.e., 64 t h value. Let A be the point on y -axis representing 64 , through A draw a horizontal line meeting ogive at B , from B draw B C perpendicular to x -axis meets x -axis at C . Then abscissa of point C gives the required value of Q 1 . ∴ Q 1 = 30 To find Q 3 : Q 3 3 n 4 t h item i.e. 3 × 256 4 t h item i.e. 192 t h item. Let L be the point on y -axis representing 192 , through L , draw a horizontal line meeting ogive at M , draw ⊥ M N on x -axis. Then abscissa of point N gives the required value of Q 3 . ∴ Q 3 = 70 Thus, semi interquartile range = Q 3 - Q 1 2 = 70 - 30 2 = 20 .

MEDIUM

11th ICSE

IMPORTANT

Earn 100

The marks obtained by $70$ students in an examination are given below. Calculate $Q_{1}$ , $D_{3}$ and $P_{95}$ for the distribution.

Marks more than $80$ $70$ $60$ $50$ $40$ $30$ $20$

No. od students $0$ $7$ $18$ $40$ $40$ $63$ $70$

Important Questions on Statistics

EASY

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 2>Chapter 1 - Statistics>CHAPTER TEST>Q 16

In an interview with maximum marks $= 21^{t h}$ , the marks of top $200$ students were distributed as below

Marks	$100 - 114$	$115 - 129$	$130 - 144$	$145 - 159$	$160 - 174$	$175 - 189$	$190 - 204$	$205 - 219$	$220 - 235$
$f$	$4$	$5$	$5$	$18$	$18$	$27$	$P_{95} : P_{95} = {(\frac{95 n}{100})}^{th}$	$40$	$39$

Calculate $12 t h$ percentile to the nearest integer.

EASY

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 2>Chapter 1 - Statistics>CHAPTER TEST>Q 16

In an interview with maximum marks $250$ , the marks of top $200$ students were distributed as below

Marks	$100 - 114$	$115 - 129$	$130 - 144$	$145 - 159$	$160 - 174$	$175 - 189$	$190 - 204$	$205 - 219$	$220 - 235$
$f$	$4$	$5$	$5$	$18$	$18$	$27$	$44$	$40$	$39$

Calculate Median to the nearest integer

MEDIUM

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 2>Chapter 1 - Statistics>CHAPTER TEST>Q 16

In an interview with maximum marks $250$ , the marks of top $200$ students were distributed as below

Marks	$100 - 114$	$115 - 129$	$130 - 144$	$145 - 159$	$160 - 174$	$175 - 189$	$190 - 204$	$205 - 219$	$220 - 235$
$f$	$4$	$5$	$5$	$18$	$18$	$27$	$44$	$40$	$39$

Calculate Upper Quartile to the nearest integer

EASY

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 2>Chapter 1 - Statistics>CHAPTER TEST>Q 16

In an interview with maximum marks $250$ , the marks of top $200$ students were distributed as below

Marks	$100 - 114$	$115 - 129$	$130 - 144$	$145 - 159$	$160 - 174$	$175 - 189$	$190 - 204$	$205 - 219$	$220 - 235$
$f$	$4$	$5$	$5$	$18$	$18$	$27$	$44$	$40$	$39$

What was the percentile rank of the student who scored $200$ marks?

EASY

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 2>Chapter 1 - Statistics>CHAPTER TEST>Q 17

Draw an ogive for the following frequency distribution and from it estimate the median marks and the number of students who obtained more than $75 %$ marks.

Marks	$0 - 9$	$10 - 19$	$20 - 29$	$30 - 39$	$40 - 49$	$50 - 59$	$60 - 69$	$70 - 79$	$80 - 89$	$90 - 99$
No. of students	$5$	$9$	$16$	$22$	$26$	$18$	$11$	$6$	$4$	$3$

EASY

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 2>Chapter 1 - Statistics>CHAPTER TEST>Q 18

Draw cumulative frequency curve for the following data and estimate the median

Saving in $₹$ (less than)	$10$	$20$	$30$	$40$	$50$	$60$	$70$	$80$
Cumulative frequency	$15$	$35$	$64$	$84$	$96$	$120$	$192$	$256$

MEDIUM

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 2>Chapter 1 - Statistics>CHAPTER TEST>Q 18

Draw cumulative frequency curve for the following data and estimate the semi-interquartile range

Saving in $₹$ (less than)	$10$	$20$	$30$	$40$	$50$	$60$	$70$	$80$
Cumulative frequency	$15$	$35$	$64$	$84$	$96$	$120$	$192$	$256$

Marks more than	$80$	$70$	$60$	$50$	$40$	$30$	$20$
No. od students	$0$	$7$	$18$	$40$	$40$	$63$	$70$

The marks obtained by 70 students in an examination are given below. Calculate Q1, D3 and P95 for the distribution. Marks more than 80 70 60 50 40 30 20 No. od students 0 7 18 40 40 63 70

Important Questions on Statistics

Learn and Prepare for any exam you want !

The marks obtained by $70$ students in an examination are given below. Calculate $Q_{1}$ , $D_{3}$ and $P_{95}$ for the distribution.

Marks more than $80$ $70$ $60$ $50$ $40$ $30$ $20$

No. od students $0$ $7$ $18$ $40$ $40$ $63$ $70$