Mix Examples - Statistics Questions in English

(A) In the median formula,$cf$ represents the cumulative frequency of the class preceding the median class. Here,$l$ is the lower limit of the median class,$n$ is the total number of observations,$f$ is the frequency of the median class,and $c$ is the class size (or class length).

(B) In the formula for the median,$M = l + \frac{(\frac{n}{2} - cf)}{f} \times h$,the term $f$ represents the frequency of the median class.

(C) In the formula for the median of grouped data,$M = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$,the term $h$ represents the class size or class length of the median class. Therefore,$h = \text{class length}$.

(D) The empirical relationship between the mean $(\bar{x})$,median $(M)$,and mode $(Z)$ for a moderately skewed distribution is given by the formula: $Z = 3M - 2\bar{x}$.
Rearranging this formula:
$Z - M = (3M - 2\bar{x}) - M$
$Z - M = 2M - 2\bar{x}$
$Z - M = 2(M - \bar{x})$
Comparing this with the given expression $Z - M = \ldots \ldots \ldots \quad (M - \bar{x})$,the missing value is $2$.

(A) 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 $Z = 16$ and $M = 18$.
Substituting these values into the formula:
$16 = 3(18) - 2\bar{x}$
$16 = 54 - 2\bar{x}$
$2\bar{x} = 54 - 16$
$2\bar{x} = 38$
$\bar{x} = 19$
Therefore,the value of the mean is $19$.

(B) The empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula: $Z = 3M - 2\bar{x}$.
Given: $Z = 95$ and $\bar{x} = 98$.
Substituting the values into the formula:
$95 = 3M - 2(98)$
$95 = 3M - 196$
$3M = 95 + 196$
$3M = 291$
$M = \frac{291}{3} = 97$.
Therefore,the value of $M$ is $97$.

(C) The empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula:
$Z = 3M - 2\bar{x}$
Given:
$M = 62.5$
$\bar{x} = 64$
Substituting these values into the formula:
$Z = 3(62.5) - 2(64)$
$Z = 187.5 - 128$
$Z = 59.5$
Therefore,the value of $Z$ is $59.5$.

(D) Given equations are:
$M + \bar{x} = 32$ --- $(1)$
$M - \bar{x} = 2$ --- $(2)$
Adding equation $(1)$ and $(2)$:
$(M + \bar{x}) + (M - \bar{x}) = 32 + 2$
$2M = 34$
$M = 17$
Substituting $M = 17$ in equation $(1)$:
$17 + \bar{x} = 32$
$\bar{x} = 32 - 17 = 15$
Using the empirical relationship between Mean,Median,and Mode:
$Z = 3M - 2\bar{x}$
Substituting the values of $M$ and $\bar{x}$:
$Z = 3(17) - 2(15)$
$Z = 51 - 30$
$Z = 21$
Therefore,the value of $Z$ is $21$.

(A) Given equations are:
$Z + M = 108$ --- $(1)$
$Z - M = -6$ --- $(2)$
Adding $(1)$ and $(2)$:
$(Z + M) + (Z - M) = 108 - 6$
$2Z = 102$
$Z = 51$
Substituting $Z = 51$ in $(1)$:
$51 + M = 108$
$M = 108 - 51 = 57$
Using the empirical relationship between Mean,Median,and Mode:
$Mode = 3 \times Median - 2 \times Mean$
$Z = 3M - 2\bar{x}$
$51 = 3(57) - 2\bar{x}$
$51 = 171 - 2\bar{x}$
$2\bar{x} = 171 - 51$
$2\bar{x} = 120$
$\bar{x} = 60$

(B) Given equations are:
$(1)$ $\bar{x} + Z = 37.5$
$(2)$ $\bar{x} - Z = 1.5$
Adding $(1)$ and $(2)$: $2\bar{x} = 39 \implies \bar{x} = 19.5$
Subtracting $(2)$ from $(1)$: $2Z = 36 \implies Z = 18$
Using the empirical relationship between Mean,Median,and Mode: $Z = 3M - 2\bar{x}$
$18 = 3M - 2(19.5)$
$18 = 3M - 39$
$3M = 18 + 39 = 57$
$M = 57 / 3 = 19$
Therefore,the median $M$ is $19$.

(C) The empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula: $Z = 3M - 2\bar{x}$.
Given,$Z - M = 2$,which implies $M = Z - 2$.
Also,$\bar{x} = 33.5$.
Substituting these values into the formula:
$Z = 3(Z - 2) - 2(33.5)$
$Z = 3Z - 6 - 67$
$Z = 3Z - 73$
$73 = 3Z - Z$
$73 = 2Z$
$Z = 36.5$.

(D) 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 $M = 15.4$ and $\bar{x} = 14.5$.
Substituting these values into the formula:
$Z = 3(15.4) - 2(14.5)$
$Z = 46.2 - 29.0$
$Z = 17.2$.
Therefore, the value of $Z$ is $17.2$.

(A) Given equations are:
$(1)$ $M + \bar{x} = 165$
$(2)$ $M - \bar{x} = 1$
Adding equations $(1)$ and $(2)$:
$2M = 166 \implies M = 83$
Subtracting equation $(2)$ from $(1)$:
$2\bar{x} = 164 \implies \bar{x} = 82$
Using the empirical relationship between Mean,Median,and Mode:
$Z = 3M - 2\bar{x}$
$Z = 3(83) - 2(82)$
$Z = 249 - 164$
$Z = 85$

(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 $Z - M = 12$,we can write $Z = M + 12$.
Substituting this into the empirical formula:
$M + 12 = 3M - 2\bar{x}$
$12 = 3M - M - 2\bar{x}$
$12 = 2M - 2\bar{x}$
$12 = 2(M - \bar{x})$
Dividing both sides by $2$:
$M - \bar{x} = 6$.

(C) The modal class is the class interval with the highest frequency.
From the given table,the frequencies are:
$0-10: 7$
$10-20: 16$
$20-30: 25$
$30-40: 32$
$40-50: 20$
The maximum frequency is $32$,which corresponds to the class interval $30-40$.
Therefore,the modal class is $30-40$.

(D) To find the median class,we first calculate the cumulative frequency $(cf)$ for the given data:
$1$. Class $0-10$: Frequency $12$,$cf = 12$
$2$. Class $10-20$: Frequency $18$,$cf = 12 + 18 = 30$
$3$. Class $20-30$: Frequency $20$,$cf = 30 + 20 = 50$
$4$. Class $30-40$: Frequency $17$,$cf = 50 + 17 = 67$
$5$. Class $40-50$: Frequency $13$,$cf = 67 + 13 = 80$
The total number of observations $N = 80$.
We find $N/2 = 80/2 = 40$.
The median class is the class interval whose cumulative frequency is greater than or equal to $N/2 = 40$.
Looking at the cumulative frequencies $(12, 30, 50, 67, 80)$,the first value greater than $40$ is $50$.
Therefore,the corresponding class interval is $20-30$.

(B) The mean (average) of a set of $n$ observations $x_1, x_2, ..., x_n$ is defined as the sum of all observations divided by the total number of observations.
Mathematically,this is represented as $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$,where $\sum x_i$ is the sum of all values and $n$ is the total count of observations.

(A) The mean $(\bar{x})$ of grouped data using the direct method is calculated by the formula: $\bar{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}$,where $f_{i}$ is the frequency of the $i^{th}$ class interval and $x_{i}$ is the class mark (midpoint) of the $i^{th}$ class interval. The sum $\sum f_{i}$ is often denoted by $N$ or $n$,representing the total number of observations.

(B) In the given formula for the mean of grouped data,$\bar{x} = \frac{\Sigma f_{i} x_{i}}{n}$:
$\bar{x}$ represents the mean.
$f_{i}$ represents the frequency of the $i^{th}$ class.
$x_{i}$ represents the class mark (midvalue) of the $i^{th}$ class.
$n$ represents the total frequency,which is equal to $\Sigma f_{i}$.
Therefore,$f_{i}$ represents the frequency of the $i^{th}$ class.

(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 $= 30 - 20 = 10$.

(D) In an exclusive class interval (or continuous series),the lower limit is included in the class,but the upper limit is excluded.
For the class interval $30-45$:
- The lower limit is $30$,which is included in this class.
- The upper limit is $45$,which is excluded from this class and is instead included in the next class interval (e.g.,$45-60$).
Therefore,the value $45$ is not included in the class $30-45$.

(C) cumulative frequency curve is a graphical representation of the cumulative frequency distribution.
It is plotted by taking the upper class limits on the $x$-axis and the corresponding cumulative frequencies on the $y$-axis.
This curve is commonly referred to as an $ogive$.

(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 mode of a grouped data is calculated using the formula:
$Z = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Where:
$l$ = lower limit of the modal class
$h$ = size of the class interval
$f_1$ = frequency of the modal class
$f_0$ = frequency of the class preceding the modal class
$f_2$ = frequency of the class succeeding the modal class.

(B) In the formula for the mode of grouped data,$Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$,the variables are defined as follows:
$l$: lower limit of the modal class.
$f_{1}$: frequency of the modal class.
$f_{0}$: frequency of the class preceding the modal class.
$f_{2}$: frequency of the class succeeding the modal class.
$c$: size of the class interval.
Therefore,$f_{1}$ represents the frequency of the modal class.

(A) The mode is defined as the value that appears most frequently in a data set.
First,let us list the frequency of each observation:
$2$ appears $1$ time.
$3$ appears $4$ times.
$4$ appears $2$ times.
$5$ appears $2$ times.
$6$ appears $1$ time.
Since the observation $3$ has the highest frequency of $4$,the mode of the given data is $3$.

(B) In the formula for calculating the mode of grouped data,the modal class is the class with the highest frequency. This maximum frequency is denoted by $f_1$. The frequency of the class preceding the modal class is denoted by $f_0$,and the frequency of the class succeeding the modal class is denoted by $f_2$.

(C) In the formula for calculating the mode of a grouped frequency distribution,$l$ represents the lower limit of the modal class.
Given the modal class is $70-85$.
The lower limit of this class is $70$.
Therefore,$l = 70$.

(B) The modal class is defined as the class interval with the highest frequency in a given frequency distribution.
By observing the provided table,we compare the frequencies:
$9, 8, 12, 7, 15, 1$.
The maximum frequency is $15$.
The class interval corresponding to the frequency $15$ is $20-24$.
Therefore,the modal class is $20-24$.

(C) In the formula for calculating the mode of grouped data,$f_{1}$ represents the frequency of the modal class.
The modal class is the class interval with the highest frequency.
Looking at the given frequency distribution:
- The frequencies are $12, 18, 27, 20, 17, 6$.
- The maximum frequency is $27$,which corresponds to the class interval $200-300$.
Therefore,the frequency of the modal class is $f_{1} = 27$.

(A) In the given frequency distribution,the maximum frequency is $8$,which corresponds to the class interval $5-7$.
Therefore,the modal class is $5-7$.
In the formula for mode,$f_{1}$ is the frequency of the modal class,$f_{0}$ is the frequency of the class preceding the modal class,and $f_{2}$ is the frequency of the class succeeding the modal class.
Here,the modal class is $5-7$,so its frequency $f_{1} = 8$.
The class preceding the modal class is $3-5$,and its frequency is $f_{0} = 3$.
Thus,$f_{0} = 3$.

(A) The modal class is the class interval with the highest frequency.
Given that the modal class is $20-24$.
In the formula for calculating the mode,$l$ represents the lower limit of the modal class.
Therefore,$l = 20$.

(D) In the formula for the mode of a grouped frequency distribution,$f_{1}$ represents the frequency of the modal class,$f_{0}$ represents the frequency of the class preceding the modal class,and $f_{2}$ represents the frequency of the class succeeding the modal class.
Given that the modal class is $70-85$ with frequency $25$,we have $f_{1} = 25$.
The frequency of the class preceding the modal class is $8$,so $f_{0} = 8$.
The frequency of the class succeeding the modal class is $20$,so $f_{2} = 20$.
Therefore,the values of $f_{0}, f_{1}$ and $f_{2}$ are $8, 25$ and $20$ respectively.

(B) The median of grouped data is calculated using the formula:
$M = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
Where:
$l$ = lower limit of the median class
$n$ = total number of observations
$cf$ = cumulative frequency of the class preceding the median class
$f$ = frequency of the median class
$h$ = class size (width of the class interval).

(C) The cumulative frequency of a class is the sum of the frequencies of all classes up to that class.
Cumulative frequency of the third class = (Frequency of first class) + (Frequency of second class) + (Frequency of third class)
Cumulative frequency of the third class = $8 + 15 + 18 = 41$.

(D) The cumulative frequency of a class is the sum of the frequencies of all classes up to that class.
Let $cf_n$ be the cumulative frequency of the $n^{th}$ class and $f_n$ be the frequency of the $n^{th}$ class.
We know that $cf_n = cf_{n-1} + f_n$.
Therefore,$cf_{n-1} = cf_n - f_n$.
Given that the cumulative frequency of the fourth class $(cf_4)$ is $25$ and the frequency of the fourth class $(f_4)$ is $10$,we have:
$cf_3 = cf_4 - f_4$
$cf_3 = 25 - 10 = 15$.
Thus,the cumulative frequency of the third class is $15$.

(A) First,calculate the cumulative frequency $(cf)$ for each class:
- $20-25$: $cf = 2$
- $25-30$: $cf = 2 + 5 = 7$
- $30-35$: $cf = 7 + 8 = 15$
- $35-40$: $cf = 15 + 10 = 25$
- $40-45$: $cf = 25 + 7 = 32$
- $45-50$: $cf = 32 + 10 = 42$
- $50-55$: $cf = 42 + 3 = 45$
The total frequency $n = 45$.
We need to find $\frac{n}{2} = \frac{45}{2} = 22.5$.
The median class is the class whose cumulative frequency is just greater than $22.5$.
Looking at the cumulative frequencies,$25$ is the first value greater than $22.5$,which corresponds to the class interval $35-40$.
Therefore,the median class is $35-40$.