Mix Examples - Statistics Questions in English

(D) First,calculate the cumulative frequency $(cf)$ for each class:
$1$. Class $10-25$: $cf = 5$
$2$. Class $25-40$: $cf = 5 + 21 = 26$
$3$. Class $40-55$: $cf = 26 + 21 = 47$
$4$. Class $55-70$: $cf = 47 + 8 = 55$
$5$. Class $70-85$: $cf = 55 + 25 = 80$
$6$. Class $85-100$: $cf = 80 + 20 = 100$
The total frequency $n = 100$.
Calculate $\frac{n}{2} = \frac{100}{2} = 50$.
The median class is the class whose cumulative frequency is just greater than or equal to $\frac{n}{2} = 50$.
Looking at the cumulative frequencies: $5, 26, 47, 55, 80, 100$. The first cumulative frequency greater than $50$ is $55$,which corresponds to the class $55-70$.
Therefore,the median class is $55-70$.

(A) To find the median class,we first determine the total frequency $n = \sum f_i = 6 + 10 + 5 + 6 + 4 + 2 + 2 = 35$.
The median class is the class interval corresponding to the cumulative frequency just greater than $\frac{n}{2}$.
Here,$\frac{n}{2} = \frac{35}{2} = 17.5$.
Let's calculate the cumulative frequencies $(cf)$:
Class $10-20$: $cf = 6$
Class $20-30$: $cf = 6 + 10 = 16$
Class $30-40$: $cf = 16 + 5 = 21$
Since $21$ is the first cumulative frequency greater than $17.5$,the corresponding class interval $30-40$ is the median class.

(B) To find the median class,we first calculate the cumulative frequency $(cf)$ for each class:
$1$. For class $0-10$,$cf = 4$.
$2$. For class $10-20$,$cf = 4 + 5 = 9$.
$3$. For class $20-30$,$cf = 9 + 9 = 18$.
$4$. For class $30-40$,$cf = 18 + 10 = 28$.
$5$. For class $40-50$,$cf = 28 + 14 = 42$.
$6$. For class $50-60$,$cf = 42 + 8 = 50$.
The total number of observations is $n = 50$.
We need to find the class where the cumulative frequency is greater than or equal to $\frac{n}{2} = \frac{50}{2} = 25$.
Looking at the cumulative frequencies: $4, 9, 18, 28, 42, 50$.
The first cumulative frequency greater than $25$ is $28$,which corresponds to the class interval $30-40$.
Therefore,the median class is $30-40$.

(A) To construct a 'less than' type Ogive,we plot the cumulative frequencies against the upper class limits of the respective class intervals.
Therefore,the upper limits of the classes are represented on the $X$-axis and the corresponding cumulative frequencies are represented on the $Y$-axis.

(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 $\frac{18}{3} = 6$.

(C) The empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula:
$Z = 3M - 2\bar{x}$
Given values are $Z = 12$ and $M = 16$.
Substituting these values into the formula:
$12 = 3(16) - 2\bar{x}$
$12 = 48 - 2\bar{x}$
Rearranging the equation to solve for $\bar{x}$:
$2\bar{x} = 48 - 12$
$2\bar{x} = 36$
$\bar{x} = \frac{36}{2}$
$\bar{x} = 18$

(D) The empirical relationship between mean,median,and mode is given by the formula:
$Z = 3M - 2\bar{x}$
where $Z$ is the mode,$M$ is the median,and $\bar{x}$ is the mean.
Given: $Z = 25$ and $\bar{x} = 19$.
Substituting these values into the formula:
$25 = 3M - 2(19)$
$25 = 3M - 38$
$3M = 25 + 38$
$3M = 63$
$M = \frac{63}{3} = 21$
Therefore,the median is $21$.

(B) The empirical relationship between Mean,Median,and Mode is given by the formula:
$Z = 3M - 2\bar{x}$
Given that $M = 72$ and $\bar{x} = 70$,we substitute these values into the formula:
$Z = 3(72) - 2(70)$
$Z = 216 - 140$
$Z = 76$
Therefore,the value of $Z$ is $76$.

(A) We know the empirical relationship between mean,median,and mode is given by: $Z = 3M - 2\bar{x}$.
We can rewrite this equation as:
$Z - M = 3M - 2\bar{x} - M$
$Z - M = 2M - 2\bar{x}$
$Z - M = 2(M - \bar{x})$
Given that $Z - M = 4$,we substitute this value into the equation:
$4 = 2(M - \bar{x})$
Dividing both sides by $2$:
$M - \bar{x} = \frac{4}{2}$
$M - \bar{x} = 2$

(C) The empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula:
$Z = 3M - 2\bar{x}$
Given values are $Z = 36.8$ and $M = 33.6$.
Substituting these values into the formula:
$36.8 = 3(33.6) - 2\bar{x}$
Calculate $3 \times 33.6$:
$36.8 = 100.8 - 2\bar{x}$
Rearrange the equation to solve for $\bar{x}$:
$2\bar{x} = 100.8 - 36.8$
$2\bar{x} = 64$
Divide by $2$:
$\bar{x} = \frac{64}{2} = 32$
Thus,the value of $\bar{x}$ is $32$.

(A) Given equations are:
$Z+M=88$ --- (Equation $1$)
$Z-M=2$ --- (Equation $2$)
To find the value of $M$,subtract Equation $2$ from Equation $1$:
$(Z+M) - (Z-M) = 88 - 2$
$Z + M - Z + M = 86$
$2M = 86$
$M = \frac{86}{2}$
$M = 43$

(C) We know the empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by: $Z = 3M - 2\bar{x}$.
Given equations:
$1) Z + M = 34 \implies Z = 34 - M$
$2) M + \bar{x} = 40 \implies \bar{x} = 40 - M$
Substitute these into the empirical formula:
$34 - M = 3M - 2(40 - M)$
$34 - M = 3M - 80 + 2M$
$34 - M = 5M - 80$
$34 + 80 = 5M + M$
$114 = 6M$
$M = \frac{114}{6} = 19$
Therefore,the value of $M$ is $19$.

(A) The median of a frequency distribution can be determined graphically by drawing both the 'less than' type and 'more than' type Ogives.
The point of intersection of these two Ogives provides the median value.
The $x$-coordinate of this intersection point represents the median of the data,while the $y$-coordinate represents the cumulative frequency corresponding to the median.
Given that the intersection point is $(20, 25)$,the $x$-coordinate is $20$.
Therefore,the median of the data is $20$.

(B) The empirical relationship between Mean $( \bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula: $Z = 3M - 2\bar{x}$.
Given that $\bar{x} = Z - 3$,we can express the mode as $Z = \bar{x} + 3$.
Substituting these values into the empirical formula:
$\bar{x} + 3 = 3(22) - 2\bar{x}$
$\bar{x} + 3 = 66 - 2\bar{x}$
Adding $2\bar{x}$ to both sides:
$3\bar{x} + 3 = 66$
Subtracting $3$ from both sides:
$3\bar{x} = 63$
Dividing by $3$:
$\bar{x} = 21$.
Therefore,the mean is $21$.

(A) The empirical relationship between Mean $( \bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula: $Z = 3M - 2\bar{x}$.
Given that $M - \bar{x} = 2$,we can express the mean as $\bar{x} = M - 2$.
Substituting this into the empirical formula:
$Z = 3M - 2(M - 2)$
$20.5 = 3M - 2M + 4$
$20.5 = M + 4$
$M = 20.5 - 4$
$M = 16.5$
Therefore,the median $M$ is $16.5$.

(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 $6$:
$96 = 80 + x$
$x = 96 - 80$
$x = 16$

(C) Given the relation $3 \bar{x} = 2 M = 60.$
From this,we can find the values of mean $(\bar{x})$ and median $(M)$:
$\bar{x} = \frac{60}{3} = 20$
$M = \frac{60}{2} = 30$
The empirical relationship between mean,median,and mode $(Z)$ is given by the formula:
$Z = 3 M - 2 \bar{x}$
Substituting the values of $M$ and $\bar{x}$ into the formula:
$Z = 3(30) - 2(20)$
$Z = 90 - 40$
$Z = 50$

(B) The empirical relationship between mean,median,and mode is given by the formula: $Mode = 3 \times \text{Median} - 2 \times \text{Mean}$.
Given:
Mean $(\bar{x})$ = $72.5$
Median $(M)$ = $73.9$
Substituting these values into the formula:
$Mode = 3(73.9) - 2(72.5)$
$Mode = 221.7 - 145.0$
$Mode = 76.7$
Therefore,the mode of the data is $76.7$.

(D) The empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula: $Z = 3M - 2\bar{x}$.
To find the value of the missing term in $Z - M = \ldots \times (M - \bar{x})$,we rearrange the formula:
$Z = 3M - 2\bar{x}$
Subtract $M$ from both sides:
$Z - M = 3M - 2\bar{x} - M$
$Z - M = 2M - 2\bar{x}$
Factor out $2$:
$Z - M = 2(M - \bar{x})$
Comparing this with the given expression,the missing value is $2$.

(C) We know the empirical relationship between Mean,Median,and Mode is given by: $Z = 3M - 2\bar{x}$.
Subtract $M$ from both sides of the equation:
$Z - M = 3M - 2\bar{x} - M$
$Z - M = 2M - 2\bar{x}$
Factor out $2$ on the right side:
$Z - M = 2(M - \bar{x})$
Divide both sides by $(M - \bar{x})$:
$\frac{Z - M}{M - \bar{x}} = 2$.

(B) We know the empirical relationship between mean,median,and mode is given by: $Z = 3M - 2\bar{x}$.
To find the value in the blank,we rearrange the equation:
$Z - \bar{x} = 3M - 2\bar{x} - \bar{x}$
$Z - \bar{x} = 3M - 3\bar{x}$
$Z - \bar{x} = 3(M - \bar{x})$
Comparing this with the given expression,the missing value is $3$.

(B) The cumulative frequency distribution is a statistical tool used to determine the median of a dataset.
To calculate the median,we first find the cumulative frequencies of the given data.
Then,we identify the median class by finding the value $N/2$,where $N$ is the total frequency.
The median is then calculated using the formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$,where $cf$ is the cumulative frequency of the class preceding the median class.
Therefore,the correct option is $B$.

(C) measure of central tendency is a summary statistic that represents the center point or typical value of a dataset. The three most common measures of central tendency are the Mean, Median, and Mode.
Range is a measure of dispersion (or variability), not central tendency. It is calculated as the difference between the maximum and minimum values in a dataset $(Range = \text{Maximum Value} - \text{Minimum Value})$.
Therefore, Range is not a measure of central tendency.

(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 $\frac{(45 \times 6 - 30)}{5}$.

(D) First,calculate the total frequency $\Sigma f_{i} = 10 + 12 + 13 + 16 + 9 = 60$.
Thus,option $D$ is false because $\Sigma f_{i} = 60$,not $50$.
Now,let's verify other options:
$1$. Cumulative frequencies: $10, 22, 35, 51, 60$.
$2$. Median class: $N/2 = 60/2 = 30$. The cumulative frequency just greater than $30$ is $35$,which corresponds to the class $20-30$. So,option $A$ is true.
$3$. Modal class: The highest frequency is $16$,which corresponds to the class $30-40$. So,option $B$ is true.
$4$. Cumulative frequency of class $20-30$ is $10 + 12 + 13 = 35$. So,option $C$ is true.

(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 $\frac{(30 \times 16) + (10 \times 12)}{30 + 10}$.

(B) The empirical relationship between the three measures of central tendency (Mean,Median,and Mode) is given by the formula:
$Mode = 3 \times Median - 2 \times Mean$
In symbolic notation,this is represented as:
$Z = 3M - 2\bar{x}$
Where:
$Z$ = Mode
$M$ = Median
$\bar{x}$ = Mean
Therefore,the correct option is $B$.