Mean of Grouped Data Questions in English

(C) The formula $\bar{x} = \frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}$ is used to calculate the arithmetic mean of grouped data.
In this formula,$f_{i}$ represents the frequency of the $i$-th class interval,and $x_{i}$ represents the class mark (mid-value) of the $i$-th class interval.
The symbol $\Sigma$ denotes the summation.
Therefore,$\Sigma f_{i}$ represents the sum of the frequencies of all the classes,which is often denoted by $N$.

(A) The given formula $\bar{x} = A + \frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}$ is the Assumed Mean Method for calculating the mean of grouped data.
In this formula,$A$ represents the assumed mean.
$x_{i}$ represents the class mark (midpoint) of the $i^{th}$ class interval.
$d_{i}$ is the deviation of the class mark from the assumed mean,which is defined as $d_{i} = x_{i} - A$.
Therefore,the correct option is $A$.

(C) The given formula $\bar{x} = A + \frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}} \times c$ represents the Step-Deviation Method for calculating the mean of grouped data.
In this formula,$A$ is the assumed mean,$c$ is the class size (width),and $x_{i}$ is the class mark.
The variable $u_{i}$ is defined as the deviation of the class mark from the assumed mean,divided by the class size.
Therefore,$u_{i} = \frac{x_{i} - A}{c}$.

(C) In the formula for the arithmetic mean of grouped data,$\bar{x} = \frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}$:
$\bar{x}$ represents the mean of the data.
$f_{i}$ represents the frequency of the $i$th class.
$x_{i}$ represents the class mark or the mid-value of the $i$th class interval,calculated as $\frac{\text{Upper limit} + \text{Lower limit}}{2}$.
Therefore,$x_{i}$ represents the mid-value of the $i$th class.

(D) The midpoint (or class mark) of a class interval is calculated using the formula: $\text{Midpoint} = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.
For the class interval $15-30$, the lower limit is $15$ and the upper limit is $30$.
$\text{Midpoint} = \frac{15 + 30}{2} = \frac{45}{2} = 22.5$.

(B) In an excluding type of class interval,the class length is calculated by subtracting the lower limit from the upper limit.
For the class $20-30$,the lower limit is $20$ and the upper limit is $30$.
Therefore,the class length $= 30 - 20 = 10$.

(B) The formula for the mean $\bar{x}$ of a frequency distribution is given by:
$\bar{x} = \frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}$
Given $\Sigma f_{i} x_{i} = 1790$ and $\Sigma f_{i} = 50$.
Substituting the values into the formula:
$\bar{x} = \frac{1790}{50} = 35.8$
Therefore,the mean is $35.8$.

(C) The formula for the mean using the step-deviation method is given by:
$\bar{x} = A + \left( \frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}} \right) \times c$
Given values:
$A = 62.5$,$\Sigma f_{i} u_{i} = -13$,$\Sigma f_{i} = 100$,$c = 15$
Substituting these values into the formula:
$\bar{x} = 62.5 + \left( \frac{-13}{100} \right) \times 15$
$\bar{x} = 62.5 + (-0.13) \times 15$
$\bar{x} = 62.5 - 1.95$
$\bar{x} = 60.55$

(D) The total frequency is given as $\Sigma f_{i} = 48$.
It is also given that $\Sigma f_{i} = 43 + f$.
Equating the two expressions for the total frequency:
$43 + f = 48$
Subtracting $43$ from both sides:
$f = 48 - 43$
$f = 5$
Therefore,the missing frequency $f$ is $5$.

(A) The formula for the mean using the assumed mean method is $\bar{x} = A + \frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}$.
Given values are $A = 62.5$,$\Sigma f_{i} d_{i} = -50$,and $\Sigma f_{i} = 200$.
Substituting these values into the formula:
$\bar{x} = 62.5 + \frac{-50}{200}$
$\bar{x} = 62.5 - 0.25$
$\bar{x} = 62.25$.

(B) The formula for the mean using the step-deviation method is $\bar{x} = A + \left( \frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}} \right) \times c$.
Given values are $A = 200$,$\Sigma f_{i} = 45$,$\Sigma f_{i} u_{i} = -216$,and $c = 10$.
Substituting these values into the formula:
$\bar{x} = 200 + \left( \frac{-216}{45} \right) \times 10$
$\bar{x} = 200 + \left( \frac{-2160}{45} \right)$
$\bar{x} = 200 - 48$
$\bar{x} = 152$.

(C) The formula for the mean using the step-deviation method is:
$\bar{x} = A + \left( \frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}} \right) \times c$
Given values are $A = 49.5$,$\Sigma f_{i} = 40$,$\Sigma f_{i} u_{i} = -5$,and $c = 20$.
Substituting these values into the formula:
$\bar{x} = 49.5 + \left( \frac{-5}{40} \right) \times 20$
$\bar{x} = 49.5 + (-0.125) \times 20$
$\bar{x} = 49.5 - 2.5$
$\bar{x} = 47$

(A) The formula for the mean using the step-deviation method is given by:
$\bar{x} = A + \left( \frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}} \right) \times c$
Substituting the given values:
$A = 325, c = 50, \Sigma f_{i} u_{i} = 28, \Sigma f_{i} = 200$
$\bar{x} = 325 + \left( \frac{28}{200} \right) \times 50$
$\bar{x} = 325 + \left( \frac{28}{4} \right)$
$\bar{x} = 325 + 7$
$\bar{x} = 332$

(D) The formula for the mean using the assumed mean method is given by: $\bar{x} = A + \frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}$.
Given values are $A = 25$,$\Sigma f_{i} d_{i} = 120$,and $\Sigma f_{i} = 140$.
Substituting these values into the formula:
$\bar{x} = 25 + \frac{120}{140}$
$\bar{x} = 25 + \frac{12}{14} = 25 + \frac{6}{7}$
$\bar{x} = 25 + 0.85714...$
$\bar{x} \approx 25.857$.

(B) The formula for the mean using the step-deviation method is $\bar{x} = A + \left( \frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}} \right) \times c$.
Given values are $A = 450$,$c = 100$,$\Sigma f_{i} u_{i} = -20$,and $\Sigma f_{i} = 20$.
Substituting these values into the formula:
$\bar{x} = 450 + \left( \frac{-20}{20} \right) \times 100$
$\bar{x} = 450 + (-1) \times 100$
$\bar{x} = 450 - 100$
$\bar{x} = 350$.

(B) The sum of the initial $10$ observations is given by $\Sigma x_{i} = n \times \bar{x} = 10 \times 15.7 = 157$.
When a new observation $19$ is added,the new sum of observations becomes $\Sigma x_{i}' = 157 + 19 = 176$.
The total number of observations now becomes $n' = 10 + 1 = 11$.
The new mean is calculated as $\bar{x}' = \frac{\Sigma x_{i}'}{n'} = \frac{176}{11} = 16$.

(A) The sum of $5$ observations is given by $\Sigma x_{i} = n \times \bar{x} = 5 \times 16 = 80$.
When the observation $5$ is replaced by $-5$,the corrected sum of observations is $\Sigma x_{i}' = 80 - 5 + (-5) = 80 - 10 = 70$.
The new mean is calculated as $\text{New Mean} = \frac{\text{Corrected } \Sigma x_{i}}{n} = \frac{70}{5} = 14$.

(D) Let the observations be $x_1, x_2, ..., x_{15}$.
The mean is given by $\frac{\sum x_i}{15} = 16$,which implies $\sum x_i = 16 \times 15 = 240$.
When $2$ is added to each observation,the new sum becomes $\sum (x_i + 2) = \sum x_i + (15 \times 2) = 240 + 30 = 270$.
The new mean after adding $2$ is $\frac{270}{15} = 18$.
When each result is divided by $3$,the new mean becomes $\frac{18}{3} = 6$.

(B) The initial sum of $15$ observations is given by $\Sigma x_{i} = n \times \bar{x} = 15 \times 25 = 375$.
The corrected sum is calculated by adding the correct value and subtracting the incorrect value: $\text{Corrected } \Sigma x_{i} = 375 + 50 - 20 = 405$.
The correct mean is the corrected sum divided by the total number of observations: $\text{Correct mean} = \frac{405}{15} = 27$.

(A) The formula for the mean $\bar{x}$ is given by $\bar{x} = \frac{\sum x_i}{n}$.
The first ten natural numbers are $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$.
The sum of these numbers is $\sum x_i = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55$.
The number of observations is $n = 10$.
Therefore,the mean $\bar{x} = \frac{55}{10} = 5.5$.

(D) The mean of a set of observations is defined as $\bar{x} = \frac{\Sigma x_{i}}{n}$,where $n$ is the number of observations.
Therefore,$\Sigma x_{i} = n\bar{x}$.
Now,consider the expression $\Sigma(x_{i} - \bar{x})$.
This can be expanded as $\Sigma x_{i} - \Sigma \bar{x}$.
Since $\bar{x}$ is a constant,$\Sigma \bar{x} = n\bar{x}$.
Substituting the values,we get $\Sigma x_{i} - n\bar{x} = n\bar{x} - n\bar{x} = 0$.
Thus,the sum of deviations of observations from their mean is always $0$.

(B) The formula for the mean using the assumed mean method is given by: $\bar{x} = A + \frac{\Sigma f_{i} d_{i}}{n}$.
Given values are $n = 100$,$A = 15$,and $\bar{x} = 15$.
Substituting these values into the formula:
$15 = 15 + \frac{\Sigma f_{i} d_{i}}{100}$.
Subtracting $15$ from both sides:
$15 - 15 = \frac{\Sigma f_{i} d_{i}}{100}$.
$0 = \frac{\Sigma f_{i} d_{i}}{100}$.
Multiplying both sides by $100$,we get:
$\Sigma f_{i} d_{i} = 0$.

(A) The formula for the mean using the step-deviation method is given by: $\bar{x} = A + \left( \frac{\Sigma f_{i} u_{i}}{n} \right) \times c$
Given values are: $\bar{x} = 20$,$\Sigma f_{i} u_{i} = -50$,$n = 100$,and $c = 10$.
Substituting these values into the formula:
$20 = A + \left( \frac{-50}{100} \right) \times 10$
Simplify the expression:
$20 = A + (-0.5) \times 10$
$20 = A - 5$
Solving for $A$:
$A = 20 + 5$
$A = 25$
Thus,the assumed mean $A$ is $25$.

(A) The formula for the mean using the step-deviation method is given by: $\bar{x} = A + \left( \frac{\Sigma f_{i} u_{i}}{n} \right) \times c$
Given values are: $\bar{x} = 54.3$,$\Sigma f_{i} u_{i} = 2$,$n = 25$,and $c = 10$.
Substituting these values into the formula:
$54.3 = A + \left( \frac{2}{25} \right) \times 10$
Simplify the expression:
$54.3 = A + \left( \frac{20}{25} \right)$
$54.3 = A + 0.8$
Solving for $A$:
$A = 54.3 - 0.8$
$A = 53.5$

(B) The formula for the mean using the assumed mean method is given by $\bar{x} = A + \frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}$,where $\Sigma f_{i} = n$.
Given that $\bar{x} = 20$,$A = 20$,and $n = 100$.
Substituting these values into the formula: $20 = 20 + \frac{\Sigma f_{i} d_{i}}{100}$.
Subtracting $20$ from both sides,we get $0 = \frac{\Sigma f_{i} d_{i}}{100}$.
Therefore,$\Sigma f_{i} d_{i} = 0 \times 100 = 0$.

(B) The formula for the mean $(\bar{x})$ of a frequency distribution is given by $\bar{x} = \frac{\Sigma f_{i} x_{i}}{n}$.
Given that $\Sigma f_{i} x_{i} = 245$ and $n = 100$.
Substituting these values into the formula:
$\bar{x} = \frac{245}{100} = 2.45$.

(A) The formula for the mean of a frequency distribution is given by $\bar{x} = \frac{\Sigma f_{i} x_{i}}{N}$,where $N$ is the total frequency.
Given that $\Sigma f_{i} x_{i} = 120$ and $N = 25$.
Substituting these values into the formula:
$\bar{x} = \frac{120}{25} = 4.8$.
Therefore,the mean is $4.8$.

(D) The formula for the mean using the assumed mean method is given by $\bar{x} = A + \frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}$.
Given that $n = \Sigma f_{i} = 200$,$\Sigma f_{i} d_{i} = 0$,and $A = 25$.
Substituting these values into the formula:
$\bar{x} = 25 + \frac{0}{200}$
$\bar{x} = 25 + 0$
$\bar{x} = 25$.
Therefore,the mean is $25$.

(D) The sum of deviations of observations from their mean is always zero.
Mathematically,$\sum_{i=1}^{n} (x_i - \bar{x}) = \sum_{i=1}^{n} x_i - \sum_{i=1}^{n} \bar{x}$.
Since $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$,it follows that $\sum_{i=1}^{n} x_i = n \bar{x}$.
Substituting this,we get $\sum_{i=1}^{n} (x_i - \bar{x}) = n \bar{x} - n \bar{x} = 0$.
For $n = 10$,$\sum_{i=1}^{10} (x_i - \bar{x}) = 0$.

(C) The given formula $\bar{x} = A + \frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}} \times c$ is the step-deviation method used to calculate the mean of grouped data.
In this formula:
- $\bar{x}$ is the mean.
- $A$ is the assumed mean.
- $f_{i}$ is the frequency of the $i^{th}$ class.
- $u_{i} = \frac{x_{i} - A}{c}$,where $x_{i}$ is the class mark.
- $c$ represents the class length (or class size),which is the difference between the upper limit and the lower limit of a class interval.

(D) The formula for the mean of $n$ observations is $\bar{x} = \frac{\sum x_i}{n}$.
Given observations are $12, 13, x, 17, 18, 20$ and the number of observations $n = 6$.
The mean is given as $16$.
Substituting the values into the formula:
$16 = \frac{12 + 13 + x + 17 + 18 + 20}{6}$
$16 = \frac{80 + x}{6}$
Multiplying both sides by $6$:
$96 = 80 + x$
$x = 96 - 80$
$x = 16$

(A) The formula for the mean using the assumed mean method is given by: $\bar{x} = A + \frac{\Sigma f_{i} d_{i}}{n}$.
Given values are $\bar{x} = 19.7$,$A = 20$,and $n = 50$.
Substituting these values into the formula:
$19.7 = 20 + \frac{\Sigma f_{i} d_{i}}{50}$.
Subtracting $20$ from both sides:
$19.7 - 20 = \frac{\Sigma f_{i} d_{i}}{50}$.
$-0.3 = \frac{\Sigma f_{i} d_{i}}{50}$.
Multiplying both sides by $50$:
$\Sigma f_{i} d_{i} = -0.3 \times 50$.
$\Sigma f_{i} d_{i} = -15$.

(C) The average score of $15$ girls is $10$. Therefore,the sum of the scores of the girls = $15 \times 10 = 150$.
The average score of $20$ boys is $10$. Therefore,the sum of the scores of the boys = $20 \times 10 = 200$.
The total sum of the scores of all $35$ students = $150 + 200 = 350$.
Since we only have the average values,we cannot determine individual scores,such as the highest or lowest scores.
Thus,only the sum of the scores of the whole class can be calculated.

(C) The initial average score of $6$ tests is $45$.
Therefore,the total sum of the scores for $6$ tests is $45 \times 6 = 270$.
After dropping the lowest score of $30$,the new sum of the scores becomes $270 - 30 = 240$.
The number of remaining tests is $6 - 1 = 5$.
The new average score is calculated by dividing the new sum by the number of remaining tests: $\frac{240}{5} = 48$.
Representing this in the given format,the expression is $\frac{(45 \times 6 - 30)}{5}$.

(B) The average score of a group is calculated by dividing the total sum of scores by the total number of students.
Total score of $30$ boys = $30 \times 16 = 480$.
Total score of $10$ girls = $10 \times 12 = 120$.
Total score of the whole class = $480 + 120 = 600$.
Total number of students = $30 + 10 = 40$.
Average score of the whole class = $\frac{\text{Total score}}{\text{Total students}} = \frac{480 + 120}{40} = \frac{(30 \times 16) + (10 \times 12)}{30 + 10}$.
Thus,the correct expression is $\frac{(30 \times 16) + (10 \times 12)}{30 + 10}$.