Calculate mean deviation from median for the following:
Answer
572.7k+ views
Hint:
Here, we have to use the concept of the median and mean deviation in this question. So firstly we have to convert the given cumulative frequency in the data into the normal frequency and then calculate the median of the data by using the formula of median. Then we have to calculate the mean deviation from the median by using the basic formula of the mean deviation.
Complete Complete Step by Step Solution:
Firstly we have to write the given data in the class interval and then we will convert the cumulative frequency given in the question into the normal frequency and also find out the middle point of the class intervals i.e. \[{{\text{x}}_{\text{i}}}\]. Therefore, we get
Now we can clearly see that \[\Sigma {{\text{f}}_{\text{i}}} = 100\]which means total number of students is equal to 100 i.e. \[{\text{N = 100}}\].
We can clearly see that from the above table that the median class is 40-50.
Now we have to calculate the median of the data by using the formula of the median.
Median\[{\text{ = L + }}\left( {\dfrac{{\dfrac{{\text{N}}}{2} - {\text{cf}}}}{{\text{f}}}} \right){\text{h}}\]where
L is the lower limit of the median class i.e. 40.
\[{\text{cf}}\]is the cumulative frequency i.e. 32.
f is the frequency of the median class i.e. 28.
h is the gap of class interval i.e. 10.
Now we have to put all the values in the formula of the median, we get
Median \[{\text{ = 40 + }}\left( {\dfrac{{\dfrac{{100}}{2} - {\text{32}}}}{{{\text{28}}}}} \right){\text{10 = 40 + }}\left( {\dfrac{{50 - {\text{32}}}}{{{\text{28}}}}} \right) = 40 + \left( {\dfrac{{18}}{{{\text{28}}}}} \right) = 40 + 6.43 = 46.43\]
Now, we have to simply calculate the mean deviation from median from the formula. Therefore,
Mean deviation from median \[ = \dfrac{{\Sigma {{\text{f}}_{\text{i}}}\left| {{{\text{x}}_{\text{i}}} - {\text{M}}} \right|}}{{\Sigma {{\text{f}}_{\text{i}}}}}\]where
M is the median of the data.
\[{{\text{x}}_{\text{i}}}\]is the median of the class interval.
\[{{\text{f}}_{\text{i}}}\]is the frequency of the class interval.
So, now we have to calculate the value of the\[\Sigma {{\text{f}}_{\text{i}}}\left| {{{\text{x}}_{\text{i}}} - {\text{M}}} \right|\]for all the classes. Then we get
From the data, we get\[\Sigma {{\text{f}}_{\text{i}}}\left| {{{\text{x}}_{\text{i}}} - {\text{M}}} \right| = 1428.6\]
So now we have to put the values in the formula of the mean deviation from median, we get
Mean deviation from median\[ = \dfrac{{1428.6}}{{100}} = 14.286\]
So, 14.286 is the mean deviation of the data from median.
Note:
1) Statistics is the science of collecting some data in the form of the number and studying it to forecast or predict its future possibility. Some definitions we should know.
2) Mean is equal to the ratio of sum of the total numbers and total count of the numbers. Mean is also known as the average of the numbers.
3) Mode is the most common or most repeating number.
4) Median is the middle value of the given list of numbers or it is the value which is separating the data into two halves i.e. upper half and lower half.
Here, we have to use the concept of the median and mean deviation in this question. So firstly we have to convert the given cumulative frequency in the data into the normal frequency and then calculate the median of the data by using the formula of median. Then we have to calculate the mean deviation from the median by using the basic formula of the mean deviation.
Complete Complete Step by Step Solution:
Firstly we have to write the given data in the class interval and then we will convert the cumulative frequency given in the question into the normal frequency and also find out the middle point of the class intervals i.e. \[{{\text{x}}_{\text{i}}}\]. Therefore, we get
Now we can clearly see that \[\Sigma {{\text{f}}_{\text{i}}} = 100\]which means total number of students is equal to 100 i.e. \[{\text{N = 100}}\].
We can clearly see that from the above table that the median class is 40-50.
Now we have to calculate the median of the data by using the formula of the median.
Median\[{\text{ = L + }}\left( {\dfrac{{\dfrac{{\text{N}}}{2} - {\text{cf}}}}{{\text{f}}}} \right){\text{h}}\]where
L is the lower limit of the median class i.e. 40.
\[{\text{cf}}\]is the cumulative frequency i.e. 32.
f is the frequency of the median class i.e. 28.
h is the gap of class interval i.e. 10.
Now we have to put all the values in the formula of the median, we get
Median \[{\text{ = 40 + }}\left( {\dfrac{{\dfrac{{100}}{2} - {\text{32}}}}{{{\text{28}}}}} \right){\text{10 = 40 + }}\left( {\dfrac{{50 - {\text{32}}}}{{{\text{28}}}}} \right) = 40 + \left( {\dfrac{{18}}{{{\text{28}}}}} \right) = 40 + 6.43 = 46.43\]
Now, we have to simply calculate the mean deviation from median from the formula. Therefore,
Mean deviation from median \[ = \dfrac{{\Sigma {{\text{f}}_{\text{i}}}\left| {{{\text{x}}_{\text{i}}} - {\text{M}}} \right|}}{{\Sigma {{\text{f}}_{\text{i}}}}}\]where
M is the median of the data.
\[{{\text{x}}_{\text{i}}}\]is the median of the class interval.
\[{{\text{f}}_{\text{i}}}\]is the frequency of the class interval.
So, now we have to calculate the value of the\[\Sigma {{\text{f}}_{\text{i}}}\left| {{{\text{x}}_{\text{i}}} - {\text{M}}} \right|\]for all the classes. Then we get
From the data, we get\[\Sigma {{\text{f}}_{\text{i}}}\left| {{{\text{x}}_{\text{i}}} - {\text{M}}} \right| = 1428.6\]
So now we have to put the values in the formula of the mean deviation from median, we get
Mean deviation from median\[ = \dfrac{{1428.6}}{{100}} = 14.286\]
So, 14.286 is the mean deviation of the data from median.
Note:
1) Statistics is the science of collecting some data in the form of the number and studying it to forecast or predict its future possibility. Some definitions we should know.
2) Mean is equal to the ratio of sum of the total numbers and total count of the numbers. Mean is also known as the average of the numbers.
3) Mode is the most common or most repeating number.
4) Median is the middle value of the given list of numbers or it is the value which is separating the data into two halves i.e. upper half and lower half.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

