Mix Examples - Statistics and Probability Questions in English

(A) The given observations are $7, 15, x-1, x+1, 24, 28$.
Since the number of observations $n = 6$ (which is even),the median is the average of the $(\frac{n}{2})^{th}$ and $(\frac{n}{2} + 1)^{th}$ terms.
Median $= \frac{(\text{3rd term} + \text{4th term})}{2}$.
Given median $= 21$,so $21 = \frac{(x-1) + (x+1)}{2}$.
$21 = \frac{2x}{2}$.
$21 = x$.
Therefore,the value of $x$ is $21$.

(B) The given observations are $56, 58, 63, x-5, x+1, 75, 81, 85$.
Total number of observations $n = 8$,which is an even number.
For an even number of observations,the median is the average of the $(\frac{n}{2})^{th}$ and $(\frac{n}{2} + 1)^{th}$ terms.
Here,$n = 8$,so the median is the average of the $4^{th}$ and $5^{th}$ terms.
$4^{th}$ term $= x-5$
$5^{th}$ term $= x+1$
Median $= \frac{(x-5) + (x+1)}{2} = 67$
$\frac{2x - 4}{2} = 67$
$x - 2 = 67$
$x = 69$.

(C) Let the $10$ observations be $x_1, x_2, \dots, x_{10}$.
The mean is given by $\bar{x} = \frac{\sum_{i=1}^{10} x_i}{10} = 68$.
This implies $\sum_{i=1}^{10} x_i = 680$.
According to the problem, each observation is divided by $2$ and then $6$ is added. Let the new observations be $y_i = \frac{x_i}{2} + 6$.
The mean of the new observations is $\bar{y} = \frac{\sum_{i=1}^{10} y_i}{10}$.
Substituting $y_i$, we get $\bar{y} = \frac{\sum_{i=1}^{10} (\frac{x_i}{2} + 6)}{10} = \frac{\frac{1}{2} \sum x_i + 10 \times 6}{10}$.
$\bar{y} = \frac{1}{2} \bar{x} + 6$.
Substituting $\bar{x} = 68$, we get $\bar{y} = \frac{68}{2} + 6 = 34 + 6 = 40$.

(A) $(1)$ True. The observations are $12, 13, 14, 15, 16$. Since the number of observations $n = 5$ (which is odd),the median is the value of the $\left(\frac{n+1}{2}\right)^{th}$ observation. Here,the $\left(\frac{5+1}{2}\right)^{th} = 3^{rd}$ observation is $14$. Thus,the statement is true.
$(2)$ False. The mode is the observation that occurs most frequently. In the set $1, 3, 5, 7, 9$,each observation occurs only once. Therefore,there is no mode for this data set. Thus,the statement is false.

(N/A) $(1)$ True: The first four odd natural numbers are $1, 3, 5, 7$. Their mean is $\frac{1+3+5+7}{4} = \frac{16}{4} = 4$. Thus,the statement is true.
$(2)$ False: The range of a data set is defined as the difference between the highest and lowest values (Range = Maximum value - Minimum value). The highest value itself is not the range.
$(3)$ False: Mean = $\frac{\text{Sum of observations}}{\text{Number of observations}}$. Given Mean = $30$ and Number of observations = $30$,the Sum = $30 \times 30 = 900$. Since $900 \neq 60$,the statement is false.

(B) To find the frequency of the class $0-5$,we count the number of observations that fall within this range.
In a continuous frequency distribution,the class $0-5$ includes values from $0$ up to,but not including,$5$ (i.e.,$0 \le x < 5$).
The given observations are: $0, 3, 2, 5, 7, 8, 10, 12, 5, 6, 6, 4, 14, 18, 20$.
Identifying the values that satisfy $0 \le x < 5$:
$0, 3, 2, 4$.
There are $4$ such observations.
Therefore,the frequency of the class $0-5$ is $4$.

(C) The range of a set of observations is defined as the difference between the maximum value and the minimum value in the data set.
Formula: $\text{Range} = \text{Maximum value} - \text{Minimum value}$
Given observations: $7, 9, 7, 5, 6, 6, 9, 18, 8, 8, 6$
Maximum value = $18$
Minimum value = $5$
$\text{Range} = 18 - 5 = 13$
Therefore,the range of the observations is $13$.

(D) The class mark of a class interval is calculated using the formula: $\text{Class mark} = \frac{\text{Upper limit} + \text{Lower limit}}{2}$.
Given the class interval is $70-90$,the lower limit is $70$ and the upper limit is $90$.
Substituting these values into the formula:
$\text{Class mark} = \frac{70 + 90}{2} = \frac{160}{2} = 80$.
Therefore,the correct option is $D$.

(A) The width (or size) of a class interval is calculated by subtracting the lower limit from the upper limit.
Class interval $= 50.5 - 55.5$
Upper limit $= 55.5$
Lower limit $= 50.5$
Class width $= \text{Upper limit} - \text{Lower limit} = 55.5 - 50.5 = 5$.

(B) The class mark of a class interval is defined as the midpoint of the class.
Let the width of the class be $h$.
For two successive classes with class marks $x_1 = 72$ and $x_2 = 77$,the difference between the class marks is equal to the width of the class.
Therefore,the width of the class $h = x_2 - x_1$.
$h = 77 - 72 = 5$.
Thus,the width of the class is $5$.

(C) To find the mode,we identify the value that appears most frequently in the data set.
Let us count the frequency of each mark:
$3$ appears $2$ times.
$4$ appears $3$ times.
$5$ appears $2$ times.
$6$ appears $3$ times.
$7$ appears $3$ times.
$9$ appears $5$ times.
$10$ appears $2$ times.
Since the value $9$ has the highest frequency ($5$ times),the mode of the data is $9$.

(D) The given observations are $5, 6, 8, 9, 6, 7, 8, 9, 4, 5, 8, 9, x, 10$.
First,let us count the frequency of each observation:
$4$ appears $1$ time.
$5$ appears $2$ times.
$6$ appears $2$ times.
$7$ appears $1$ time.
$8$ appears $3$ times.
$9$ appears $3$ times.
$10$ appears $1$ time.
For the mode of the data to be $9$,the frequency of $9$ must be greater than the frequency of any other observation.
Currently,$8$ and $9$ both appear $3$ times.
Therefore,for $9$ to be the mode,$x$ must be equal to $9$,which makes the frequency of $9$ equal to $4$.

(A) The formula for the mean of a set of observations is given by $\bar{x} = \frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}$.
Given that the number of observations $\Sigma f_{i} = 20$ and the mean $\bar{x} = 27.2$.
Substituting these values into the formula: $27.2 = \frac{\Sigma f_{i} x_{i}}{20}$.
Therefore,$\Sigma f_{i} x_{i} = 27.2 \times 20 = 544$.

(B) The first ten natural numbers are $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$.
Here,the total number of observations $n = 10$,which is an even number.
The formula for the median of an even number of observations is $\text{Median} = \frac{(\frac{n}{2})^{th} \text{ observation} + (\frac{n}{2} + 1)^{th} \text{ observation}}{2}$.
Substituting $n = 10$,we get $\text{Median} = \frac{5^{th} \text{ observation} + 6^{th} \text{ observation}}{2}$.
The $5^{th}$ observation is $5$ and the $6^{th}$ observation is $6$.
Therefore,$\text{Median} = \frac{5 + 6}{2} = \frac{11}{2} = 5.5$.

(A) The mean of a set of observations is calculated by dividing the sum of all observations by the total number of observations.
Sum of observations = $(x+5) + (x+3) + (x+11) + (x+1)$
Sum = $4x + (5+3+11+1) = 4x + 20$
Total number of observations = $4$
Mean = $\frac{\text{Sum of observations}}{\text{Total number of observations}}$
Mean = $\frac{4x + 20}{4}$
Mean = $\frac{4(x + 5)}{4} = x+5$
Therefore,the mean is $x+5$.

(D) Let the $10$ observations be $x_1, x_2, ..., x_{10}$.
The mean of these observations is given by $\bar{x} = \frac{\sum_{i=1}^{10} x_i}{10} = 52$.
Therefore,the sum of the observations is $\sum_{i=1}^{10} x_i = 52 \times 10 = 520$.
When $12$ is subtracted from each of the $10$ observations,the new sum becomes $\sum_{i=1}^{10} (x_i - 12) = \sum_{i=1}^{10} x_i - (10 \times 12) = 520 - 120 = 400$.
The new mean is $\frac{400}{10} = 40$.
Alternatively,if a constant $k$ is subtracted from each observation,the new mean is the original mean minus $k$. Thus,$52 - 12 = 40$.

(A) Given that the number of observations $n = 10$ and the incorrect mean $\bar{x}_{incorrect} = 22.7$.
Sum of observations (incorrect) $= n \times \bar{x}_{incorrect} = 10 \times 22.7 = 227$.
The correct sum is obtained by subtracting the incorrect value and adding the correct value: $\text{Correct Sum} = 227 - 15 + 51$.
$\text{Correct Sum} = 227 + 36 = 263$.
The correct mean is $\bar{x}_{correct} = \frac{\text{Correct Sum}}{n} = \frac{263}{10} = 26.3$.
Therefore,the correct mean is $26.3$.

(B) The natural numbers between $1$ and $20$ that are multiples of $3$ are: $3, 6, 9, 12, 15, 18$.
To find the mean, we calculate the sum of these numbers and divide by the count of numbers.
Sum $= 3 + 6 + 9 + 12 + 15 + 18 = 63$.
Number of terms $= 6$.
Mean $= \frac{\text{Sum}}{\text{Number of terms}} = \frac{63}{6} = 10.5$.

(A) The mean of a set of data is calculated by the formula: $\text{Mean} = \frac{\text{Sum of observations}}{\text{Number of observations}}$.
Given data: $\frac{4}{7}, \frac{4}{9}, \frac{3}{7}, \frac{5}{9}$.
Sum of observations = $\frac{4}{7} + \frac{4}{9} + \frac{3}{7} + \frac{5}{9}$.
Grouping terms with the same denominators: $(\frac{4}{7} + \frac{3}{7}) + (\frac{4}{9} + \frac{5}{9}) = \frac{7}{7} + \frac{9}{9} = 1 + 1 = 2$.
Number of observations = $4$.
Mean = $\frac{2}{4} = \frac{1}{2}$.

(D) The mean of $10$ observations is $11$,so the sum of $10$ observations = $10 \times 11 = 110$.
After including one new observation,the total number of observations becomes $10 + 1 = 11$.
The new mean is $12$,so the sum of $11$ observations = $11 \times 12 = 132$.
The value of the new observation = (Sum of $11$ observations) - (Sum of $10$ observations).
Value = $132 - 110 = 22$.

(A) The facts or figures,which are numerical or otherwise,collected with a definite purpose are called data. Data is the plural form of the Latin word 'datum'.

(B) Statistics is the branch of Mathematics that deals with the collection,organization,analysis,interpretation,and presentation of data. It is specifically concerned with the extraction of meaningful information from raw data.

(C) The mean (average) of a set of numbers is calculated by dividing the sum of the numbers by the total count of the numbers.
Sum of the given numbers = $3 + (-3) + (-3) + 3 + 3 + (-3) = 3 - 3 - 3 + 3 + 3 - 3 = 0$.
Total count of numbers = $6$.
Mean = $\frac{\text{Sum of numbers}}{\text{Total count of numbers}} = \frac{0}{6} = 0$.
Therefore,the mean is $0$.

(D) When data is collected by an investigator from a source that already exists (such as government publications,newspapers,or research reports),it is known as secondary data. Primary data,on the other hand,is data collected directly by the investigator for a specific purpose.

(A) The difference between the maximum (highest) value and the minimum (lowest) value of a given dataset is defined as the range of that data. Therefore,the correct answer is range.

(B) When dealing with a large set of data,it is difficult to interpret individual values. To make the data more manageable and understandable,we group the values into intervals or ranges. These groups are known as $classes$ or $class$ $intervals$.

(C) When an investigator collects data personally for a specific purpose,it is known as $primary$ $data$. This type of data is original in character and is collected for the first time by the investigator.

(D) In a frequency polygon,the $x$-axis represents the class marks (mid-points) of the class intervals,and the $y$-axis represents the corresponding frequencies. The class mark is calculated as $\frac{\text{Upper limit} + \text{Lower limit}}{2}$.

(A) In statistics,for a given set of ungrouped data,the observation that occurs with the highest frequency is defined as the mode. Therefore,the observation which is repeated most often is called the mode of the data.

(B) When a set of data is arranged in ascending or descending order,the middle observation is defined as the median of the data. It represents the central value of the distribution.

(C) The mean of a set of observations is given by the sum of observations divided by the total number of observations.
Given observations: $18, 2x, 27, 12, 3x, 5x, 21$.
Total number of observations $(n)$ = $7$.
Mean = $\frac{\text{Sum of observations}}{n} = 24$.
Sum of observations = $18 + 2x + 27 + 12 + 3x + 5x + 21 = 10x + 78$.
Setting up the equation: $\frac{10x + 78}{7} = 24$.
Multiply both sides by $7$: $10x + 78 = 168$.
Subtract $78$ from both sides: $10x = 90$.
Divide by $10$: $x = 9$.

(D) The formula for the mean $(\bar{x})$ of $n$ observations is given by $\bar{x} = \frac{\Sigma f_{i} x_{i}}{n}$.
Given that the number of observations $n = 35$ and the mean $\bar{x} = 42$.
Substituting these values into the formula: $42 = \frac{\Sigma f_{i} x_{i}}{35}$.
Therefore,the sum of the observations $\Sigma f_{i} x_{i} = 42 \times 35$.
Calculating the product: $42 \times 35 = 1470$.
Thus,the sum of the observations is $1470$.

(A) The mean of a set of observations is calculated by dividing the sum of the observations by the total number of observations.
Formula: $\text{Mean} = \frac{\text{Sum of observations}}{\text{Number of observations}}$
Given:
Sum of observations = $900$
Number of observations = $25$
Calculation:
$\text{Mean} = \frac{900}{25} = 36$
Therefore, the mean of the observations is $36$.

(B) The first seven natural numbers are $1, 2, 3, 4, 5, 6, 7$.
The formula for the mean is: $\text{Mean} = \frac{\text{Sum of observations}}{\text{Number of observations}}$.
Sum of the first seven natural numbers = $1 + 2 + 3 + 4 + 5 + 6 + 7 = 28$.
Number of observations = $7$.
Mean = $\frac{28}{7} = 4$.
Therefore,the mean of the first seven natural numbers is $4$.

(C) The given observations are $41, 48, x+1, x+5, 62, 72$.
Total number of observations $n = 6$,which is an even number.
The median of an even number of observations is the average of the $(\frac{n}{2})^{th}$ and $(\frac{n}{2} + 1)^{th}$ terms.
Here,the $3^{rd}$ term is $x+1$ and the $4^{th}$ term is $x+5$.
Median $= \frac{(x+1) + (x+5)}{2} = 53$.
$\frac{2x + 6}{2} = 53$.
$x + 3 = 53$.
$x = 53 - 3 = 50$.

(B) composite number is a positive integer greater than $1$ that has at least one divisor other than $1$ and itself.
The first five composite numbers are $4, 6, 8, 9, 10$.
To find the median,we arrange the data in ascending order (which is already done: $4, 6, 8, 9, 10$).
Since the number of observations $(n = 5)$ is odd,the median is the middle term,which is the $\left(\frac{n+1}{2}\right)^{th}$ term.
Median $= \left(\frac{5+1}{2}\right)^{th} = 3^{rd}$ term.
The $3^{rd}$ term in the sequence is $8$.
Therefore,the median is $8$.

(A) The range of a data set is defined as the difference between the maximum value and the minimum value.
Given data: $28, 16, 9, 17, 82, 71, 41, 18, 25, 15$.
Maximum value $(Max)$ = $82$.
Minimum value $(Min)$ = $9$.
Range = $Max - Min = 82 - 9 = 73$.
Therefore,the range of the data is $73$.

(B) The formula for the mean of a set of observations is given by: $\text{Mean} = \frac{\text{Sum of observations}}{\text{Number of observations}}$.
Given that the sum of observations is $975$ and the mean is $32.5$.
Let the number of observations be $n$.
Substituting the values into the formula: $32.5 = \frac{975}{n}$.
Rearranging the equation to solve for $n$: $n = \frac{975}{32.5}$.
$n = \frac{9750}{325} = 30$.
Therefore,the number of observations is $30$.

(C) The first five prime numbers are $2, 3, 5, 7,$ and $11$.
To find the mean, we calculate the sum of these numbers and divide by the count of numbers.
Sum $= 2 + 3 + 5 + 7 + 11 = 28$.
Number of terms $= 5$.
Mean $= \frac{\text{Sum of terms}}{\text{Number of terms}} = \frac{28}{5} = 5.6$.

(D) To find the median,first arrange the given observations in ascending order:
$3, 4, 5, 8, 10, 12, 18, 20$.
The total number of observations $n = 8$,which is an even number.
For an even number of observations,the median is the average of the $(\frac{n}{2})^{th}$ and $(\frac{n}{2} + 1)^{th}$ terms.
Here,$\frac{n}{2} = \frac{8}{2} = 4^{th}$ term and $(\frac{n}{2} + 1) = 5^{th}$ term.
The $4^{th}$ term is $8$ and the $5^{th}$ term is $10$.
Median = $\frac{8 + 10}{2} = \frac{18}{2} = 9$.