Filter By

## All Questions

#### 50 students enter for a school javelin throw competition. The distance (in metres) thrown are recorded below : Distance (in m) 0-20 20-40 40-60 60-80 80-100 Number of students 6 11 17 12 4 (i) Construct a cumulative frequency table. (ii) Draw a cumulative frequency curve (less than type) and calculate the median distance thrown by using this curve. (iii) Calculate the median distance by using the formula for median.(iv) Are the median distance calculated in (ii) and (iii) same ?

(i) Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

 Distance fi CF 0-20 6 6 20-40 11 6+11 = 17 40-60 17 17+17 = 34 60-80 12 34 + 12 = 46 80-100 4 46 + 4 =50

Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables

 Distance Cumulative frequency (C.F) 0 less than 20 less than 40 less than 60 less than 80 less than 100 0 6 17 34 46 50

n = 50
$\frac{n}{2}= \frac{50}{2}= 25$

median = 49.41

Solution.  n = 50
$\frac{n}{2}= \frac{50}{2}= 25$

which lies in interval 40 – 60
l
= 40, h = 20, CF = 17 and f = 17
$median = l+\frac{\left ( \frac{n}{2}-cf \right )}{f}\times h$
=$40+\frac{\left ( 25-17 \right )}{17}\times 20$
=$40+\frac{8\times 20}{17}$

=40 + 9.41
= 49.41

(iv) Yes, the median distance calculated in (ii) and (iii) are same.

#### The following is the frequency distribution of duration for100 calls made on a mobile phone : Duration (in seconds) Number of calls 95-125 125-155 155-185 185-215 215-245 14 22 28 21 15 Calculate the average duration (in sec) of a call and also find the median from cumulative frequency curve.

Solution.

 Duration fi xi $u_{i}\frac{\left ( x_{i}-a \right )}{h}$ fiui 95-125 14 110 -2 -28 125-155 22 140 -1 -22 155=185 28 170 = a 0 0 185-215 21 200 1 21 215-245 21 230 2 30 $\sum f_{i}= 100$ $\sum f_{i}u_{i}= 1$

a = 170, h = 30
Average = $a+\frac{\sum f_{i}u_{i}}{\sum f_{i}}\times h= 170+\frac{1}{100}\times 30= 170+0\cdot 3= 170\cdot 3$

 less than type Duration Number of calls less than 95 less than 125 less than 155 less than 185 less than 215 less than 245 0 0+14 = 14 14+22 =36 36 + 28 = 64 64 + 21 = 85 85 + 15 =100

n = 100

$\frac{n}{2}= \frac{100}{2}= 50$
median is 170

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support

#### The annual rainfall record of a city for 66 days is given in the following table. Rainfall (in cm) 0-10 10-20 20-30 30-40 40-50 50-60 Number of days 22 10 8 15 5 6 Calculate the median rainfall using ogives (of more than type and of less than type)

Solution.

 (i) less than type (ii) more than type Rain fall No.of days Rain fall Number of days less than 0 0 more than or equal to 0 66 less than 10 0+22 = 22 more than or equal to 10 66-22 = 44 less than 20 22+10 = 32 more than or equal to 20 44-10 = 34 less than 30 32+8 = 40 more than or equal to 30 34-8 = 26 less than 40 40+15 = 55 more than or equal to 40 26-15 = 11 less than 50 55+5 =60 more than or equal to 50 11 - 5 =6 less than 60 60+6 =66 more than or equal to 60 6-6 =0

Now let us draw ogives of more than type and of less than type then find the median

$N= 66,\frac{N}{2}= \frac{66}{2}= 33$

Here median is 20

#### Size of agricultural holdings in a survey of 200 families is given in the following table: Size of agricultural holdings (in ha) Number of families 0-5 5-10 10-15 15-20 20-25 25-30 30-35 10 15 30 80 40 20 5 Compute median and mode size of the holdings

Solution.
$mode= \iota +\left [ \frac{f_{m}-f_{1}}{2f_{m}-f_{1}-f_{2}} \right ]\times h$

 Size of agricultural holdings fi cf 0-5 10 10 5-10 15 25 10-15 30 55 15-20 80 135 20-25 40 175 25-30 20 195 30-35 5 200

(i) Here n = 200

$\frac{n}{2}= \frac{200}{2}= 100$ which lies in interval  (15 – 20)

l = 15, h = 5, f = 80 and cf = 55

$median = l+\frac{\left ( \frac{n}{2}-cf \right )}{4}\times h= 15+\frac{\left ( 100-55 \right )\times 5}{80}$
=$15+\frac{45}{16}= 15+2\cdot 81= 17\cdot 81$
l = 15, fm = 80, f1 = 30, f2 = 40 and h = 5
$mode= \iota +\left [ \frac{f_{m}-f_{1}}{2f_{m}-f_{1}-f_{2}} \right ]\times h$

$= 15+\left [ \frac{80-30}{2\times 80-30-40} \right ]\times 5$
$= 15+\left [ \frac{50}{160-70} \right ]\times 5$
$= 15+\left [ \frac{50}{90} \right ]\times 5$

=$15+\frac{25}{9}$
=15 + 2.77 = 17.77

## Crack NEET with "AI Coach"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support

#### The distribution of heights (in cm) of 96 children is given below :                Height (in cm) Number of children 124-128 128-132 132-136 136-140 140-144 144-148 148-152 152-156 156-160 160-164 5 8 17 24 16 12 6 4 3 1 Draw a less than type cumulative frequency curve for this data and use it to compute median height of the children.

Solution.

 Height Number of children less than 124 less than 128 less than  132 less than 136 less than  140 less than  144 less than  148 less than  152 less than  156 less than  160 less than  164 0 5 13 30 54 70 82 88 92 95 96

$\frac{n}{2}= \frac{96}{2}= 48$
Hence the median is = 139

#### The median of the following data is 50. Find the values of p and q, if the sum of all the frequencies is 90. Marks Frequency 20-30 30-40 40-50 50-60 60-70 70-80 80-90 p 15 25 20 q 8 10

Solution.

 marks Frequency Cummulative frequency 20-30 1 p 30-40 15 15+p 40-50 25 40+p = cf 50-60 20=f 60+p 60-70 q 68+p+q 70-80 8 68+p+q 80-90 10 78+p+q

n = 90, $\frac{n}{2}= 45$
l = 50, f = 20, cf = 40 + p, h = 10
median = $l+\frac{\left ( \frac{n}{2}-cf \right )}{f}\times h$

$50= 50+\frac{\left ( 45-40-p \right )}{20}\times 10$

$0= \frac{5-p}{2}$
5 – p = 0
p = 5
78 + 5 + q = 90
q = 90 – 83
q = 7

## Crack JEE Main with "AI Coach"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support

#### The mean of the following frequency distribution is 50, but the frequencies f1 and f2 in classes 20-40 and 60-80, respectively are not known. Find these frequencies, if the sum of all the frequencies is 120. Class 0-20 20-40 40-60 60-80 80-100 Frequency 17 fi 32 f2 19

Solution.

 Class (fi) xi $\mu _{c}= \frac{\left ( x_{i}-a \right )}{h}$ fi$\mu _{i}$ 0-20 17 10 -2 -34 20-40 f1 30 -1 -f1 40-60 32 50=a 0 0 60-80 f2 70 1 f2 80-100 19 90 2 38 $\sum f_{i}= 68+f_{i}+f_{2}$

Sum of all frequencies = 120
$\Rightarrow$68 + f1 + f2 = 120
f1 + f2 = 52                  …(1)
a = 50, h = 20
mean = $a+\frac{\sum f_{i}{\mu _{i}}}{\sum f_{i}}\times h$
50= 50 +  $\frac{\left ( 4+f_{2}-f_{1} \right )\times 20}{20}$
0= (4 + f2 – f1)
–f2 + f1 = 4                  …(2)
add (1) and (2) we get
2f1 = 56  $\Rightarrow f_{1}= 28$
Put f1 = 28 in equation (1)
f2 = 52 – 28  $\Rightarrow f_{2}= 24$

#### The table below shows the salaries of 280 persons. Salary(in thousand (Rs)) Number of persons 5-10 10-15 15-20 20-25 25-30 30-35 35-40 40-45 45-50 49 133 63 15 6 7 4 2 1 Calculate the median and mode of the data.

Solution.

 Salary fi cf 5-10 49 49 10-15 133 49+133=182 15-20 63 182+63=245 20-25 15 245+15 = 260 25-30 6 260+6 = 266 30-35 7 266+7 = 273 35-40 4 273+4 = 277 40-45 2 277+2 = 279 45-50 1 279+1 = 280

$n= 280,\frac{n}{2}= \frac{280}{2}= 140$
f1 = 49, fm= 133, f2= 63, cf = 49, f = 133
l = 10, h = 5
median = $\iota +\frac{\left ( \frac{n}{2}-cf \right )}{f}\times h$
=$10+\frac{\left ( 140-49 \right )}{133}\times 5$
=$10+\frac{91\times 5}{133}$
=$10+\frac{455}{133}= 10+3\cdot 421$
$= 13\cdot 421$

In thousands = 13.421 × 1000 = 13421 Rs.
Mode = $\iota +\left [ \frac{f_{m}-f_{i}}{2f_{m}-f_{i}-f_{2}} \right ]\times h$

=$10+\left [ \frac{133-49}{2\times 133-49-63} \right ]\times 5$

=$10+\frac{84\times 5}{266-112}=10+\frac{84\times 5}{154}$
=10 + 2.727
=12.727
In thousands = 12.727 × 1000 = 12727 Rs.

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support

#### The weights of tea in 70 packets are shown in the following table :  Weight (in gram) Number of packets 200-201 201-202 202-203 203-204 204-205 205-206 13 27 18 10 1 1  Draw the less than type and more than type ogives for the data and use them to find the median weight.

Solution.

 Less than type More than type Weight Number of packets Number of packets Number of students Less than 200 0 More than or equal to 200 70 Less than 201 13 More than or equal to 201 70-13 = 57 Less than 202 40 More than or equal to 202 57-27 =30 Less than 203 58 More than or equal to 203 30-18 =12 Less than 204 68 More than or equal to 204 12-10 = 2 Less than 205 69 More than or equal to 205 2-1 = 1 Less than 206 70 More than or equal to 206 1-1 = 0

$Here\, \, n = 70,\frac{n}{2}= \frac{70}{2}= 35$
Hence median = 201.8

#### The weights of tea in 70 packets are shown in the following table :   Weight (in gram) Number of packets 200-201 201-202 202-203 203-204 204-205 205-206 13 27 18 10 1 1 Draw the less than type ogive for this data and use it to find the median weight.

Solution.

 Weight cf Less than 201 13 Less than 202 27+13=40 Less than 203 40+18=58 Less than 204 58+10=68 Less than 205 68+1 =69 Less than 206 69+1 = 70

$n = 70,\frac{n}{2}= \frac{70}{2}= 35$
Hence the median is 201.8