Question

The following frequency distribution table shows the heights of 36 students of a group.
Height (in cm)
Tally Marks
141-145
146-150
151-155
156-160
All All I
161-165
Frequency (Number of Students)
6
3
14
11
2
Total
36
If two new students of heights 150.5 cm and 155.5 cm joined the group, construct the
new frequency distribution table for the group.​

Answer 1

Answer:

Minimum height =131cm

Maximum height =156cm

Range =156−131=25

Number of classes =4

Class range =  

4

25

​  

=6.25 approx 6

class 1=131+6=137

class 2=137+6=143

class 3=143+6=149

class 4=149+6155

class 5=155+6=161

Class interval                     Frequency

131−137                              9

137−143                              3

143−149                              2

149−155                              4

155−161                              2

Step-by-step explanation:

Answer 2

Correct Question

The following frequency distribution table shows the heights of 36 students in a group.

$\begin{tabular}{|c|c|c|} \underline{Height (in cm)} &\underline{Tally Marks} &\underline{Frequency (Number of Students)}\\141-145& IIIII {} I& 6 \\146-150& III &3\\151-155& IIIII {} IIIII {} IIII & 14\\156-160& IIIII {} IIIII {} I &11 \\ 161-165 &II& 2\\\overline{\hspace{1.4 cm}Total} & \overline{\hspace{2 cm}36 \hspace{2cm}} & \overline{\hspace{2 cm}36 \hspace{2cm}} \end{tabular}$

If two new students of heights 150.5 cm and 155.5 cm joined the group, construct the new frequency distribution table for the group.​

Answer

This is the new frequency distribution table

$\begin{tabular}{|c|c|C|} \underline{Height (in cm)} & \underline{Tally Marks} & \underline{Frequency (Number of Students)}\\140.5-145.5&IIIII {} I & 6\\145.5-150.5& III& 3\\150.5-155.5 & IIIII {} IIIII {} IIIII &15\\155.5-160.5& IIIII {} IIIII {} II &12 \\ 160.5-165.5 & II & 2\\\overline{\hspace{1.4 cm}Total} & \overline{\hspace{2 cm}38 \hspace{2cm}} & \overline{\hspace{2 cm}38 \hspace{2cm}}\end{tabular}$

Given

Frequency Distribution table

The heights of the 2 new students are 150.5 cm and 155.5 cm

To Find

The new frequency distribution table

Solution

Here we have,

$\begin{tabular}{|c|c|c|} \underline{Height (in cm)} &\underline{Tally Marks} &\underline{Frequency (Number of Students)}\\141-145& IIIII {} I& 6 \\146-150& III &3\\151-155& IIIII {} IIIII {} IIII & 14\\156-160& IIIII {} IIIII {} I &11 \\ 161-165 &II& 2\\\overline{\hspace{1.4 cm}Total} & \overline{\hspace{2 cm}36 \hspace{2cm}} & \overline{\hspace{2 cm}36 \hspace{2cm}} \end{tabular}$

The heights of the 2 new students are 150.5 cm and 155.5 cm

The reason we cannot include them in the present table is because of the discontinuous classes.

Here if we convert the classes to continuous we can easily fit the two students.

To convert them to continuous classes we need to change the limits by the formula

new lower limit

= (current lower limit - the upper limit of the previous class)/2

new upper limit

= (the lower limit of the next class - current upper limit)/2

Hence we get,

$\begin{tabular}{|c|c|} \underline{Height (in cm)} & \underline{Frequency (Number of Students)}\\140.5-145.5 & 6\\145.5-150.5& 3\\150.5-155.5 & 14\\155.5-160.5 &11 \\ 160.5-165.5 & 2\\\overline{\hspace{1.4 cm}Total} & \overline{\hspace{2 cm}36 \hspace{2cm}}\end{tabular}$

Now we can see that the second and the third class both have 150.5 cm as their upper and lower limits respectively.

Since in a continuous frequency table, the upper limit is not counted when grouping the frequency,

the student with 150.5 cm height will belong to the third class.

Similarly, the student with a height of 155.5 cm will belong to the 4th class.

Hence we will increase the frequency count of these classes to get

$\begin{tabular}{|c|c|C|} \underline{Height (in cm)} & \underline{Tally Marks} & \underline{Frequency (Number of Students)}\\140.5-145.5&IIIII {} I & 6\\145.5-150.5& III& 3\\150.5-155.5 & IIIII {} IIIII {} IIIII &15\\155.5-160.5& IIIII {} IIIII {} II &12 \\ 160.5-165.5 & II & 2\\\overline{\hspace{1.4 cm}Total} & \overline{\hspace{2 cm}38 \hspace{2cm}} & \overline{\hspace{2 cm}38 \hspace{2cm}}\end{tabular}$

Hence this is the new frequency distribution table.

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