Answer
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Hint: The mean (or average) of observations, as we know is the sum of the values of all the observations divided by the total number of observations. If \[{x_1},{x_2},........,{x_n}\] are observation with respective frequencies \[{f_1},{f_2},........,{f_n}\], then their mean observation \[{x_1}\] occurs ${j_1}$ time, \[{x_2}\] occurs ${j_2}$ time and so on.
So, the mean $\overline x $ of the data is give by
\[\overline x = \dfrac{{{f_1}{x_1} + {f_2}{x_2} + ....... + {f_n}{x_n}}}{{{f_1} + {f_2} + ....... + {f_n}}}\]
Recall that we can write this in short form by using the Greek letter $\sum {\left( {CapitalSigma} \right)} $ which means summation.
\[\overline x = \sum\limits_{i = 1}^n {\dfrac{{{f_i}{x_i}}}{{\sum\limits_{i = 1}^n {{f_1}} }}} \] Varies from $1$ to $n$.
Complete step-by-step answer:
It is assumed that the frequency of each class – Internal is centered around its main point. So the midpoint (class marks) of each class can be chosen to represent the observation in the case.
Class mark = $\dfrac{{UpperClassLimit + LowerClassLimit}}{2}$
We have the sigma of frequencies is $80$and sigma is ${f_1}{x_1}$ is $8244$. So, the mean $\overline x $ of the given data is the give by
\[\overline x = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }} = \dfrac{{8244}}{{80}} = 103.05\]
The mean value is $103.05$
The choice of method to be used depends on the numerical value of ${x_i}$ and ${f_i}$.
If ${x_i}$ and ${f_i}$ are sufficiently small, then the direct method is an appropriate choice.
Note: If ${x_i}$ and ${f_i}$ are numerically large numbers, then we can go for the assumed mean method of step – deviation method. If the class size is unequal and ${x_i}$ are large. Numerically, we can still apply the step-deviation method for te mean by taking h to be a suitable divisor of all the
So, the mean $\overline x $ of the data is give by
\[\overline x = \dfrac{{{f_1}{x_1} + {f_2}{x_2} + ....... + {f_n}{x_n}}}{{{f_1} + {f_2} + ....... + {f_n}}}\]
Recall that we can write this in short form by using the Greek letter $\sum {\left( {CapitalSigma} \right)} $ which means summation.
\[\overline x = \sum\limits_{i = 1}^n {\dfrac{{{f_i}{x_i}}}{{\sum\limits_{i = 1}^n {{f_1}} }}} \] Varies from $1$ to $n$.
Complete step-by-step answer:
class Interval | Frequency(\[{f_i}\]) | Class mark(\[{x_1}\]) | \[{f_i}{x_i}\] |
$84 - 90$ | $8$ | $87$ | $696$ |
$90 - 96$ | $10$ | $93$ | $930$ |
$96 - 102$ | $16$ | $99$ | $1584$ |
$102 - 108$ | \[23\] | \[105\] | \[2415\] |
$108 - 114$ | \[12\] | \[111\] | \[1332\] |
$114 - 120$ | \[11\] | \[117\] | \[1287\] |
\[{\mathbf{80}}\] | \[\sum {{f_i}{x_i} = {\mathbf{8244}}} \] |
It is assumed that the frequency of each class – Internal is centered around its main point. So the midpoint (class marks) of each class can be chosen to represent the observation in the case.
Class mark = $\dfrac{{UpperClassLimit + LowerClassLimit}}{2}$
We have the sigma of frequencies is $80$and sigma is ${f_1}{x_1}$ is $8244$. So, the mean $\overline x $ of the given data is the give by
\[\overline x = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }} = \dfrac{{8244}}{{80}} = 103.05\]
The mean value is $103.05$
The choice of method to be used depends on the numerical value of ${x_i}$ and ${f_i}$.
If ${x_i}$ and ${f_i}$ are sufficiently small, then the direct method is an appropriate choice.
Note: If ${x_i}$ and ${f_i}$ are numerically large numbers, then we can go for the assumed mean method of step – deviation method. If the class size is unequal and ${x_i}$ are large. Numerically, we can still apply the step-deviation method for te mean by taking h to be a suitable divisor of all the
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