Given: A frequency distribution table representing the number of cars passing through a spot in 100 periods of 3 minutes each.

To find: The mode of the given grouped frequency distribution.

Step 1: Identify the Modal Class

The modal class is the class interval with the highest frequency. Observing the table, the highest frequency is $20$, which corresponds to the class interval $40 - 50$.

Therefore, the Modal Class is $40 - 50$.

Step 2: State the Formula for Mode

The formula for calculating the mode of grouped data is:

$\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$

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

Step 3: Extract the Variables

Based on the modal class $40 - 50$:

• $l = 40$
• $h = 50 - 40 = 10$
• $f_1 = 20$
• $f_0 = 12$ (frequency of the class $30 - 40$)
• $f_2 = 11$ (frequency of the class $50 - 60$)

Step 4: Substitute and Calculate

Substituting the values into the formula:

$\text{Mode} = 40 + \left( \frac{20 - 12}{2(20) - 12 - 11} \right) \times 10$

$\text{Mode} = 40 + \left( \frac{8}{40 - 23} \right) \times 10$

$\text{Mode} = 40 + \left( \frac{8}{17} \right) \times 10$

$\text{Mode} = 40 + \frac{80}{17}$

Performing the division: $80 \div 17 \approx 4.7058...$

$\text{Mode} = 40 + 4.71$ (rounded to two decimal places)

$\text{Mode} = 44.71$

Final Answer: The mode of the data is 44.71 cars.

Given: The frequency distribution of ages of patients admitted to a hospital:

To Find: The Mode and the Mean of the given data, and to compare and interpret the results.

Part 1: Calculation of Mode

Formula: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$

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

Step 1.1: Identify the Modal Class. The maximum frequency is $23$, which corresponds to the class interval $35-45$. Thus, the modal class is $35-45$.

Step 1.2: Assign values. $l = 35$, $f_1 = 23$, $f_0 = 21$, $f_2 = 14$, $h = 10$.

Step 1.3: Substitute into the formula.

$\text{Mode} = 35 + \left( \frac{23 - 21}{2(23) - 21 - 14} \right) \times 10$

$\text{Mode} = 35 + \left( \frac{2}{46 - 35} \right) \times 10 = 35 + \left( \frac{2}{11} \right) \times 10 = 35 + \frac{20}{11} \approx 35 + 1.818 = 36.82$

Part 2: Calculation of Mean

Step 2.1: Prepare the table for Assumed Mean Method ($A = 40$).

Step 2.2: Apply the Mean formula.

$\text{Mean} (\bar{x}) = A + \frac{\sum f_i d_i}{\sum f_i} = 40 + \left( \frac{-370}{80} \right) = 40 - 4.625 = 35.375$

Part 3: Comparison and Interpretation

The modal age is approximately $36.82$ years, representing the age group with the highest number of patients admitted. The mean age is $35.38$ years, representing the average age of all patients admitted. Since the mean is slightly less than the mode, it indicates that the distribution is slightly negatively skewed, but both measures suggest that the majority of patients admitted are in the mid-30s age range.

Final Answer: The Mode is 36.82 years and the Mean is 35.38 years.

Given: The frequency distribution of the lifetimes (in hours) of 225 electrical components is as follows:

To Find: The modal lifetime of the electrical components.

Step 1: Identifying the Modal Class
The modal class is the class interval with the highest frequency. Looking at the frequency distribution table, the highest frequency is $61$, which corresponds to the class interval $60-80$.
Therefore, the Modal Class is $60-80$.

Step 2: Stating the Formula for Mode
The formula for calculating the mode of a grouped frequency distribution is:
$\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
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

Step 3: Extracting the Values
From the given data:
$l = 60$
$h = 80 - 60 = 20$
$f_1 = 61$
$f_0 = 52$
$f_2 = 38$

Step 4: Substituting Values into the Formula
$\text{Mode} = 60 + \left( \frac{61 - 52}{2(61) - 52 - 38} \right) \times 20$
[Substituting the identified values into the formula]

Step 5: Performing the Arithmetic Calculations
$\text{Mode} = 60 + \left( \frac{9}{122 - 90} \right) \times 20$
$\text{Mode} = 60 + \left( \frac{9}{32} \right) \times 20$
$\text{Mode} = 60 + \left( \frac{180}{32} \right)$
$\text{Mode} = 60 + 5.625$
$\text{Mode} = 65.625$

Final Answer: The modal lifetime of the electrical components is 65.625 hours.

Given: The frequency distribution of monthly household expenditure of 200 families.

To Find: The Modal monthly expenditure and the Mean monthly expenditure.

Part 1: Finding the Mode

Step 1: Identify the Modal Class. The modal class is the class interval with the highest frequency. Here, the highest frequency is $40$, which corresponds to the class interval $1500 - 2000$.

Step 2: Apply the Mode Formula. The formula for mode is: $Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$

Where:
$l$ (lower limit of modal class) = $1500$
$f_1$ (frequency of modal class) = $40$
$f_0$ (frequency of class preceding modal class) = $24$
$f_2$ (frequency of class succeeding modal class) = $33$
$h$ (class size) = $500$

Step 3: Calculation.
$Mode = 1500 + \left( \frac{40 - 24}{2(40) - 24 - 33} \right) \times 500$
$Mode = 1500 + \left( \frac{16}{80 - 57} \right) \times 500$
$Mode = 1500 + \left( \frac{16}{23} \right) \times 500$
$Mode = 1500 + \frac{8000}{23} \approx 1500 + 347.83 = 1847.83$

Part 2: Finding the Mean

Step 1: Calculate Class Marks ($x_i$). $x_i = \frac{\text{Upper Limit} + \text{Lower Limit}}{2}$. Let assumed mean $a = 2750$.

Step 2: Apply the Step-Deviation Formula. $\bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h$
$\bar{x} = 2750 + \left( \frac{-35}{200} \right) \times 500$
$\bar{x} = 2750 + (-0.175 \times 500)$
$\bar{x} = 2750 - 87.5 = 2662.5$

Final Answer: The modal monthly expenditure is ₹1847.83 and the mean monthly expenditure is ₹2662.50.

Given: The frequency distribution of the number of students per teacher in various states/UTs of India:

To Find: The Mode and the Mean of the given data and provide an interpretation.

Part 1: Calculation of Mode

Formula: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$

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

Step 1: Identify the Modal Class. The highest frequency is $10$, which corresponds to the class interval $30 - 35$. Thus, the modal class is $30 - 35$.

Step 2: Assign values. $l = 30$, $f_1 = 10$, $f_0 = 9$, $f_2 = 3$, $h = 5$.

Step 3: Substitute into the formula.

$\text{Mode} = 30 + \left( \frac{10 - 9}{2(10) - 9 - 3} \right) \times 5$

$\text{Mode} = 30 + \left( \frac{1}{20 - 12} \right) \times 5 = 30 + \frac{5}{8} = 30 + 0.625 = 30.625$

Part 2: Calculation of Mean

We use the Assumed Mean Method: $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$, where $d_i = x_i - a$. Let assumed mean $a = 32.5$.

Step 4: Calculate Mean.

$\bar{x} = 32.5 + \left( \frac{-115}{35} \right) = 32.5 - 3.2857 \approx 29.21$

Interpretation

The mode of $30.6$ indicates that the most common teacher-student ratio in the majority of states is approximately $30.6$. The mean of $29.2$ indicates that, on average, there are about $29$ students per teacher across the states.

Final Answer: Mode = 30.625, Mean = 29.21

Given: The frequency distribution of runs scored by top batsmen in one-day international cricket matches is as follows:

To Find: The mode of the given grouped frequency distribution.

Step 1: Identify the Modal Class
The modal class is the class interval with the highest frequency. Looking at the table, the highest frequency is $18$, which corresponds to the class interval $4000 - 5000$.
Therefore, the Modal Class is $4000 - 5000$.

Step 2: State the Formula for Mode
The formula for calculating the mode of grouped data is:
$\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
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

Step 3: Assign Values to Variables
From the distribution:
$l = 4000$
$h = 5000 - 4000 = 1000$
$f_1 = 18$
$f_0 = 4$
$f_2 = 9$

Step 4: Substitute Values into the Formula
$\text{Mode} = 4000 + \left( \frac{18 - 4}{2(18) - 4 - 9} \right) \times 1000$
$\text{Mode} = 4000 + \left( \frac{14}{36 - 13} \right) \times 1000$
$\text{Mode} = 4000 + \left( \frac{14}{23} \right) \times 1000$
$\text{Mode} = 4000 + \frac{14000}{23}$

Step 5: Perform the Arithmetic Calculation
Dividing $14000$ by $23$:
$14000 \div 23 \approx 608.6956...$
Rounding to two decimal places, we get $608.70$.
$\text{Mode} = 4000 + 608.70$
$\text{Mode} = 4608.70$

Final Answer: The mode of the given data is 4608.7 runs.