Answer to Question #294757 in Statistics and Probability for Ravi

Question

Find number of students securing marks >104, between 155-165, <55 from the following class data.

Class(marks) frequency (students)

0-20 210

20-40 115

40-60 130

60-80 220

80-100 120

100-120 101

120-140 120

140-160 144

160-180 132

180-200 190

Answer 1

Marks frequency(f) cumulative frequency(cf)

0-20  210 210

20-40 115     325

40-60 130 455

60-80 220 675

80-100 120 795

100-120  101 896

120-140 120   1016

140-160 144 1160

160-180 132 1292

180-200   190 1482

We are required to find the number of students securing marks greater than 104. To do so, we need to determine the $i^{th}$ percentile such that $P_i=104$. First is to determine the value of $i$ which in turn will be used to find the position of the $104^{th}$ mark. The position of the $104^{th}$ mark is its cumulative frequency. We proceed as follows,

$P_i={l+({i\times n\over100}-cf)\times{c\over f}}$ where, $n=1482$

$l$ is the lower class boundary of the class with 104 marks

$cf$ is the cumulative frequency of the class preceding the class consisting of the mark, 104

$c$ is the width of the class with the 104^th mark

$f$ is the frequency of the class with the 104^th mark

Therefore,

$P_i=100+(14.82i-795)\times {20\over101}=104$

So,

$20.2=14.82i-795\implies 14.82i=815.2\implies i=55.0067476$

The mark, 104 is approximately the $55^{th}$ percentile. Its position is, ${i\times n\over 100}={55\times1482\over 100}=815.2\approx 816$

Now, the cumulative frequency of the $104^{th}$ mark is approximately 816.

The number of students securing marks greater than 104 is $1482-816=666$

Therefore, the number of students securing marks greater than 104 is 666.

$b)$

To find the number of students securing marks between 155 and 165, we find the cumulative frequencies for both scores and then determine the difference of their frequencies. That is, $cf_{165}-cf_{155}$.

The cumulative frequency of the $155^{th}$ mark.

We determine the value $i$ such that, $P_i=155$ where $P_i$ is the $i^{th}$ percentile given as,

$Pi=l+({i×n\over100}−cf)\times {c\over f}$ where, $n=1482$

$l$ is the lower class boundary of the class with the $155^{th}$ mark.

$cf$ is the cumulative frequency of the class preceding the class with the $155^{th}$ mark.

$c$ is the width of the class with the 155^th mark

$f$ is the frequency of the class with the $155^{th}$ mark

Now,

$P_i=140+(14.82i-1016)\times {20\over 144}=155$

So,

$108=14.82i-1016\implies i=75.8434548\approx76$

The score of 155 is the $76^{th}$ percentile. Its cumulative frequency is, ${76\times 1482\over 100}=1124$

Therefore, the number of students securing below 155 marks is 1124

 The cumulative frequency of the $165^{th}$ mark.

We determine the value $i$ such that, $P_i=165$ where $P_i$ is the $i^{th}$ percentile given as,

$Pi=l+({i×n\over100}−cf)\times {c\over f}$ where, $n=1482$

$l$ is the lower class boundary of the class with the $165^{th}$ mark.

$cf$ is the cumulative frequency of the class preceding the class with the $165^{th}$ mark.

$c$ is the width of the class with the $165^{th}$ mark

$f$ is the frequency of the class with the $165^{th}$ mark

Now,

$P_i=160+(14.82i-1160)\times {20\over 132}=165$

So,

$33=14.82i-1160\implies i=80.4993252\approx81$

The score of 165 is the $81^{st}$ percentile. Its cumulative frequency is, ${81\times 1482\over 100}=1193$

Therefore, the number of students securing below 165 marks is 1193

Therefore, $cf_{165}=1193$ and $cf_{155}=1124$ . The number of students securing marks between 155 and 165 is 1193-1124=69 students.

$c)$

The cumulative frequency of the $55^{th}$ mark.

We determine the value $i$ such that, $P_i=55$ where $P_i$ is the $i^{th}$ percentile given as,

$Pi=l+({i×n\over100}−cf)\times {c\over f}$ where, $n=1482$

$l$ is the lower class boundary of the class with the $55^{th}$ mark.

$cf$ is the cumulative frequency of the class preceding the class with the $55^{th}$ mark.

$c$ is the width of the class with the $55^{th}$ mark

$f$ is the frequency of the class with the $55^{th}$ mark

Now,

$P_i=40+(14.82i-325)\times {20\over 130}=55$

So,

$97.5=14.82i-325\implies i=28.5087719\approx29$

The score of 55 is the $29^{th}$ percentile. Its cumulative frequency is, ${29\times 1482\over 100}=422.5\approx 423$

Therefore, the number of students securing below 55 marks is 423.