Class 10 Mathematics Chapter 13 Statistics

Key Concept :- Assumed Mean Calculation

c) 52
[Solution Description] The mean can be calculated with the formula $x = a + \frac{\Sigma f_i d_i}{\Sigma f_i}$. Given a = 55, $\Sigma f_i d_i = -90$, and $\Sigma f_i = 30$, we substitute these values into the formula:

$x = 55 + \frac{-90}{30} = 55 – 3 = 52$

Hence, the mean is 52.

Notes

• Definition: The Assumed Mean Method simplifies mean calculation for large data sets by reducing computational complexity.
• Formula: $\overline{x} = A + \frac{\sum f_i d_i}{\sum f_i}$
  ,where A is the assumed mean, $d_i$ = x_i – A$
• Steps:
  • Choose an assumed mean $A$ (preferably a mid-value).
  • Find deviations $d_i$ = x_i – A$
  • Multiply $d_i$​ by the corresponding frequencies $f_i$.
  • Compute $\sum f_i d_i \quad \text{and} \quad \sum f_i$
  • Substitute in the formula.
• Usage: Useful for grouped data to estimate the mean efficiently.

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Key Concept :- Frequency Understanding

a) Total number of observations
[Solution Description] In statistics, when dealing with grouped data, $ \Sigma f_i $ refers to the sum of all frequencies for the given classes or intervals. It represents the total number of observations.

Notes

Key Concept :- Advanced Ogive Construction, Real-World Application

b) Y-axis
[Solution Description] When constructing an ogive, the cumulative frequency should always be plotted along the Y-axis because it represents the total number of data points up to a certain value on the X-axis.

Cumulative frequency distributions are represented by the height of the curve at each corresponding point on the X-axis.

Notes

An ogive is a graph that represents cumulative frequencies, helping to analyze data distributions. Advanced ogive construction involves creating less than ogive and greater than ogive curves.
Steps:

1. Prepare cumulative frequency tables for both types.
2. Plot the points on a graph (x-axis: class intervals, y-axis: cumulative frequency).
3. Draw smooth curves for each ogive.
   Real-world applications include:

• Determining median values in surveys.
• Analyzing income distributions in economics.
• Evaluating marks in education.
  Ogives simplify complex data visualization, aiding decision-making in various fields.

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Key Concept :- Multimodal Data Analysis, Advanced Median Calculation

c) Not Multimodal, Median = 29.167
[Solution Description] First, determine if the dataset is multimodal by analyzing the frequencies: Modes are determined by the highest frequency. Here, the highest frequency is 12 (for the interval 25-30), so it’s unimodal.

Next, calculate the median:
Find cumulative frequencies:
– Cumulative frequency up to 25: 5
– Up to 30: 17
– Up to 35: 25
– Up to 40: 30

Median is at the n/2 position, where n = 30, hence n/2 = 15.

Locate the median class, which is 25-30 since it includes the 15th value.

Use the median formula:
$\text{Median} = L + \left( \frac{\frac{n}{2} – F}{f} \right) \times h$
where L = 25, F = 5, f = 12, h = 5.

$\text{Median} = 25 + \left(\frac{15 – 5}{12}\right) \times 5$

$= 25 + \left(\frac{10}{12}\right) \times 5 = 25 + 4.167 = 29.167$
Therefore, the median is approximately 29.167.

Notes

• Data with more than one mode (peak) is called multimodal.
• Modes represent the most frequent values in a dataset.
• In multimodal data, there can be multiple values with high frequencies.
• Graphical representation (histograms, frequency polygons) helps identify modes.
• It’s important to analyze all modes for deeper insights.

Advanced Median Calculation:

• Median divides the dataset into two equal parts.
• In an even number of observations, the median is the average of the two middle values.
• For grouped data, median is calculated using the formula:
• $\text{Median} = L + \left(\frac{\frac{N}{2} – F}{f}\right) \times h$where L = lower boundary, N = total frequency, F = cumulative frequency before median class, f = frequency of median class, h = class width.

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Key Concept :- Frequency Comparison

c) 170-180
[Solution Description] To find the class with the highest frequency, we compare each frequency given. Here, the class 170-180 has the highest frequency of 17.

Notes

• Frequency Distribution: A table that lists the number of occurrences of different values in a dataset.
• Grouped Frequency Distribution: Data is grouped into intervals (class intervals) to simplify comparison.
• Histogram: A graphical representation of frequency distribution using bars; height of bars represents frequency.
• Frequency Polygon: A line graph formed by joining the midpoints of the tops of the bars of the histogram.
• Ogive (Cumulative Frequency Curve): A graph that shows the cumulative frequency of data.
• Comparing Frequencies: In multiple datasets, compare their histograms, frequency polygons, or ogives to analyze patterns or differences in data distributions.

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Key Concept :- Interpretation of Results, Real-world Application

c) Income distribution is symmetric without significant skew
[Solution Description] The median income falling within the \$3000-\$5000 bracket suggests a relatively even distribution since the median does not coincide with either tail-end. Analyzing this, half of the responses lie below \$3500, implying the absence of heavy skewing towards lower incomes. This central location indicates central tendency stability amidst possible outliers, affecting mean more significantly than median.

Thus, conclusions drawn include:

Notes

• Interpretation of Results:
  • Statistical analysis helps summarize and interpret data for better decision-making.
  • Results are used to draw inferences, identify trends, and make predictions.
  • Data visualization tools (e.g., graphs, charts) aid in understanding complex data.
• Real-World Applications:
  • Healthcare: Used to analyze patient data, predict disease outbreaks, and evaluate treatments.
  • Economics: Helps in market analysis, inflation tracking, and policy making.
  • Sports: Used to assess player performance, predict outcomes, and improve strategies.
  • Business: Supports customer behavior analysis, product development, and sales forecasting.

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Key Concept :- Assumed Mean Adjustment, Numerical and Analytical Difficulty

c) 33
[Solution Description] The total frequency $\Sigma f_i = 4 + 6 + 8 + 12 = 30$. The mean using the Assumed Mean Method is: $x = a + \frac{\Sigma f_i d_i}{\Sigma f_i} = 30 + \frac{90}{30} = 30 + 3 = 33.$

Notes

• Assumed Mean Method: A technique used to simplify the calculation of the mean when dealing with large data sets. An assumed mean (A) is chosen, and deviations of data points from A are calculated.
• Formula: $\bar{X} = A + \frac{\sum f d}{\sum f}$​
  Where:
  • $\bar{X}$= Mean
  • A= Assumed Mean
  • f= Frequency
  • d= Deviation from assumed mean

Numerical and Analytical Difficulty:

• The assumed mean method is useful for large frequency distributions with complex calculations.
• The difficulty arises in selecting an appropriate assumed mean and handling the computations accurately.

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Key Concept: Class Interval Understanding

b) 65-80

[Solution Description]

According to the rules for converting ungrouped data to grouped data, students scoring exactly at an upper class limit are included in the next class. Therefore, if we have class intervals like 50-65, 65-80, etc., the score of 65 falls into the 65-80 range because it is the upper limit of the previous interval.

Key Concept :- Real-World Application

b) Blue
[Solution Description] The mode is the category or group that appears most frequently. In this case, Blue cars appear most frequently with a count of 50, hence Blue is the mode.

Notes

• Data Analysis in Businesses: Companies use statistics to analyze customer behavior, sales trends, and market preferences to improve products and services.
• Medical Research: In medicine, statistics help in analyzing the effectiveness of drugs, medical treatments, and tracking disease patterns.
• Sports Statistics: Players’ performances, team statistics, and game outcomes are analyzed to improve strategies and predict future performances.
• Weather Forecasting: Meteorologists use statistical models to predict weather conditions and track climate change patterns.
• Education: Statistical data helps evaluate student performance, school effectiveness, and policy-making for educational improvements.
• Polls and Surveys: Governments and organizations use statistical methods to gather public opinions and make informed decisions.

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Key Concept :- Cumulative Frequency Understanding

c) 25
[Solution Description] Cumulative frequency for a given class interval is the sum of all frequencies up to that interval. Therefore, cumulative frequency for 50-60 is $10 + 15 = 25$.

Notes

• Cumulative Frequency: It is the running total of frequencies in a frequency distribution. It shows the number of observations that are less than or equal to a given value.
• Calculation: Start with the first frequency and then add each subsequent frequency to the previous cumulative frequency.
• Types:
  • Less than Cumulative Frequency: The total number of observations less than the upper limit of each class.
  • Greater than Cumulative Frequency: The total number of observations greater than the lower limit of each class.
• Cumulative Frequency Table: It helps in finding the median, quartiles, and other percentiles in a data set.
• Cumulative Frequency Graph: Also called an ogive, used to visually represent cumulative frequencies.

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Key Concept :- Real-World Application, Data Interpretation

c) 40%
[Solution Description] From the given data:

Percentage spending less than 1000 = 20%

Percentage spending less than 3000 = 60%

To find the percentage spending between 1000 and 3000, we subtract the percentage spending less than 1000 from those spending less than 3000: $60\% – 20\% = 40\%$

Thus, 40% of households spend between 1000 and 3000.

Notes

Key Concept :- Step Deviation Method

b) 18.33
[Solution Description] Choose a common factor h = 10 and assume a = 15 (midpoint of second class).
$d_i = \frac{x_i – a}{h}$, where $x_i = $ midpoints.
$d_1 = \frac{5 – 15}{10} = -1$, $d_2 = \frac{15 – 15}{10} = 0$, $d_3 = \frac{25 – 15}{10} = 1$
$\sum f_id_i = 3(-1) + 6(0) + 9(1) = -3 + 0 + 9 = 6$
$\sum f_i = 3 + 6 + 9 = 18$
Mean $\bar{x} = a + h \times \frac{\sum f_id_i}{\sum f_i} = 15 + 10 \times \frac{6}{18} = 15 + 3.33 = 18.33$

Notes

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Key Concept :- Mode Calculation

c) 108.33
[Solution Description] To find the mode, we use the given formula:

$\text{Mode} = 100 + \frac{15 – 10}{2(15) – 10 – 8} \times 20$

Calculating the numerator: $15 – 10 = 5$

Calculating the denominator: $30 – 10 – 8 = 12$

Substitute back into the equation: $\text{Mode} = 100 + \frac{5}{12} \times 20$

Simplifying further: $\text{Mode} = 100 + \frac{100}{12} = 100 + 8.33 = 108.33$

So, the mode is approximately 108.33.

Notes

Key Concept :- Mean Formula

c) $\bar{x} = \frac{\Sigma f_ix_i}{\Sigma f_i}$
[Solution Description] The mean of grouped data is calculated using the formula $\bar{x} = \frac{\Sigma f_ix_i}{\Sigma f_i}$ where $f_i$ are the frequencies and $x_i$ are the class marks.

Notes

Key Concept :- Basic Definition

c) The sum of all frequencies up to and including the given class interval.
[Solution Description] Cumulative frequency is calculated by adding the frequencies of all preceding classes, including the current one.

Notes

Key Concept :- Real-world Problem Solving, Complex Dataset Analysis

b) 24.167 minutes
[Solution Description] The modal class is determined by identifying the class interval with the highest frequency, which is 20-30 in this case.

Using the mode formula:

$\text{Mode} = l + \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \times h$

Where l = 20, $f_1 = 14$, $f_0 = 9$, $f_2 = 7$, and h = 10:

$\text{Mode} = 20 + \frac{14 – 9}{2(14) – 9 – 7} \times 10 = 20 + \frac{5}{28 – 16} \times 10$

Simplifying: $\text{Mode} = 20 + \frac{50}{12} = 20 + 4.1667 = 24.1667$

Hence, the mode of the commute time is approximately 24.167 minutes.

Notes

• Understanding Complex Datasets: Real-world problems involve large, complex datasets that need careful analysis to extract meaningful insights.
• Data Collection: Gathering data from reliable sources through surveys, experiments, or observations.
• Data Representation: Use of graphs (bar, histogram, pie chart) to visually interpret the data.
• Measures of Central Tendency: Mean, median, and mode are used to summarize data points.
• Measures of Dispersion: Range, variance, and standard deviation help understand data spread.
• Probability: Apply probability concepts to make predictions based on data patterns.
• Decision Making: Use statistical tools to solve real-life problems like forecasting trends and making informed decisions.
• Technology Tools: Use of software for large-scale data analysis and visualization.

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Key Concept :- Mode Identification

c) 60-70
[Solution Description] The mode is identified by the class with the highest frequency.

From the given data:
\begin{align*}
& 40-50: 5 \\
& 50-60: 7 \\
& 60-70: 9 \\
& 70-80: 6
\end{align*}

The highest frequency is 9 which corresponds to the class interval 60-70.

Notes

• Mode Definition: The mode is the value that appears most frequently in a data set.
• Identification: To find the mode, count the frequency of each data value.
• Uni-modal: A data set with only one mode (one value appears most often).
• Bi-modal: A data set with two modes (two values appear with equal highest frequency).
• Multi-modal: A data set with more than two modes (multiple values appear with the same highest frequency).
• No Mode: If no value repeats, the data set has no mode.
• Use: Mode is useful for categorical data.

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Key Concept :- Integration of Concepts, Real-World Data Interpretation

c) $z = 45$
[Solution Description] The median salary falls in the range 50-60, therefore, use the median formula:

$52.5 = 50 + \left(\frac{\frac{n}{2} – 24}{z}\right) \times 10$

Solve for $z$: $52.5 = 50 + \frac{\frac{n}{2} – 24}{z} \times 10$

Assume $n = 30 + z$, then

$\frac{n}{2} = \frac{30 + z}{2}$

Hence, $52.5 = 50 + \frac{15 + 0.5z – 24}{z} \times 10$

Simplify and solve for $z$: $25 = \frac{-9 + 0.5z}{z} \times 10$

$2.5z = -90 + 0.5z$

$2z = 90$

$z = 45$

Therefore, $z = 45$.

Notes

• Statistics Overview: Statistics involves collecting, analyzing, and interpreting data to make informed decisions.
• Types of Data: Real-world data can be qualitative or quantitative, and can be analyzed through various measures such as mean, median, and mode.
• Data Representation: Graphs, charts, and tables help visualize data trends and patterns.
• Applications: Data interpretation is used in diverse fields like economics, healthcare, and education to make predictions and solve problems.
• Correlation & Regression: Statistical methods like correlation and regression are employed to establish relationships between variables.
• Real-World Examples: Statistical models are crucial in making decisions based on historical data trends and patterns.

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Key Concept :- Solve for Unknowns

b) 15
[Solution Description] To find x, we use the median formula:

$\text{Median} = l + \left(\frac{\frac{n}{2} – cf}{f}\right) \times h$

where the median is given as 55, l is the lower boundary of the median class, c f is the cumulative frequency before the median class, f is the frequency of the median class, h is the class width, and n is the total frequency.

Given: $ n = 10 + x + 25 = 35 + x $, Median = 55, class interval = 50-60, so l = 50, h = 10, frequency of median class $f = x$, cumulative frequency preceding median class c f = 10.

Plugging into the formula: $55 = 50 + \left(\frac{\frac{35 + x}{2} – 10}{x}\right) \times 10$

Simplifying: $5 = \left(\frac{\frac{35 + x}{2} – 10}{x}\right) \times 10$

$\frac{1}{2}(35 + x) – 10 = x$

$17.5 + \frac{x}{2} – 10 = x$

$7.5 = \frac{x}{2}$

$x = 15$

Therefore, x = 15.

Notes

Key Concept :- Median Calculation

b) 30-40
[Solution Description] First, calculate the cumulative frequency for each class interval:

$$\begin{align*}
\text{Cumulative Frequency: } & \\
10-20 & : 4 \\
20-30 & : 4 + 8 = 12 \\
30-40 & : 12 + 12 = 24 \\
40-50 & : 24 + 6 = 30 \\
50-60 & : 30 + 10 = 40 \\
\end{align*}$$

The total frequency is 40, so the median class is where the cumulative frequency just exceeds 20 (half of the total frequency). This occurs at the class interval 30-40.

Notes

Key Concept :- Interpretation of Results, Complex Median Problem

b) Half of the scores are above 68
[Solution Description] If the median score is 68, it implies that half of the employees scored below 68, and half scored above 68. Thus, the median divides the dataset into two equal halves.

Notes

Interpretation of Results
Statistical results convey data insights. For instance, mean highlights central tendencies, and standard deviation reflects data spread. Proper interpretation ensures informed decision-making, identifying trends or deviations. Graphical representation aids understanding.

Complex Median Problem
For grouped data, calculating the median requires:

1. Identify the median class (where cumulative frequency crosses half the total).
2. Apply the formula: $\text{Median} = L + \left( \frac{\frac{N}{2} – CF}{f} \right) \cdot h$ is the lower boundary, Ntotal frequency, cumulative frequency before median class, frequency of median class, and $h$ class width.
   Accurate computation aids result reliability.

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Key Concept :- Complex Mode Calculation, Empirical Formula Application

a) 48.18 years
[Solution Description] The modal class will be where the frequency is the highest, which is 45-55 years with 25 participants. The lower limit  l  of the modal class is 45, the class size  h  is 10, $ f_1$  is 25,  $f_0$  is 18, and  $f_2$  is 10.

Calculate the mode using the formula:
$\text{Mode} = l + \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \times h = 45 + \frac{25 – 18}{2(25) – 18 – 10} \times 10 = 45 + \frac{7}{22} \times 10 = 48.18$

Given the empirical relationship $ 3 \times \text{Median} = \text{Mode} + 2 \times \text{Mean} $:
Substitute the values: $3 \times \text{Median} = 48.18 + 2 \times 39 = 126 + 48.18 = 174.18$

Solve for the median: $\text{Median} = \frac{174.18}{3} = 58.06$

Thus, the calculated mode from the empirical relationship confirms the value.

Notes

• Complex Mode Calculation: In a data set with multiple modes or overlapping peaks, the complex mode calculation helps determine the most frequent values. It involves using modal class intervals and finding the mode through interpolation, considering the frequency distribution’s shape.
• Empirical Formula: It is used to find the simplest ratio of elements in a compound. For grouped data, the empirical formula is applied by calculating the mean or median values and then applying statistical techniques like the method of least squares to derive the approximate values.

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Key Concept :- Class Interval Identification

c) 15-20
[Solution Description] We compare the frequencies given in each interval to determine which is the highest:

Frequency of 5-10 = 4
Frequency of 10-15 = 9
Frequency of 15-20 = 12
Frequency of 20-25 = 7

The highest frequency is 12 for the class interval 15-20.

Notes

In statistics, class intervals are used to organize data into groups or categories. A class interval is a range of values that represents a certain frequency of data points. Identifying class intervals involves:

1. Lower Class Limit (LCL): The smallest value in the class interval.
2. Upper Class Limit (UCL): The largest value in the class interval.
3. Class Width: The difference between the upper and lower class limits of a class interval.
4. Midpoint: The average of the lower and upper class limits.

Correct identification helps in constructing frequency distributions and histograms effectively.

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Key Concept :- Ogive Interpretation

c) The minimum test score
[Solution Description] The intersection of the ogive with the horizontal axis represents the lowest test score achieved by any student because that is where the cumulative frequency equals zero.

Notes

• An ogive is a graphical representation of the cumulative frequency distribution.
• It is plotted by marking cumulative frequencies on the y-axis and class boundaries on the x-axis.
• Two types of ogives:
  • Less than ogive: Cumulative frequency is plotted for class boundaries less than or equal to each class.
  • More than ogive: Cumulative frequency is plotted for class boundaries greater than or equal to each class.
• Interpretation:
  • The ogive helps in determining the median, quartiles, and percentiles.
  • The point of intersection of both ogives gives the median.
• It provides a visual method for analyzing data distribution and comparisons between different datasets.

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Key Concept :- Assumed Mean Calculation

d) 55
[Solution Description] The formula for the Assumed Mean Method is given by:
$x = a + \frac{\Sigma f_i d_i}{\Sigma f_i}$
where $ a = 50 $, $ \Sigma f_i d_i = 200 $, and $ \Sigma f_i = 40 $.

Substitute the values: $x = 50 + \frac{200}{40}$
$x = 50 + 5$
$x = 55$
So, the mean calculated using the Assumed Mean Method is 55.

Notes

Key Concept :- Symbol Meaning

a) Lower class boundary
[Solution Description] In the formula for the mode of grouped data,  l  represents the lower class boundary of the modal class.

Notes

• Mean (̄x): Represents the average of a set of data values.
• Median: The middle value in a data set when arranged in ascending or descending order.
• Mode: The value that appears most frequently in a data set.
• n: The number of observations in a data set.
• Σ (Sigma): Summation symbol, indicating the sum of a series of values.
• Variance (σ²): A measure of how data points differ from the mean.
• Standard Deviation (σ): The square root of the variance, representing data spread.
• Range: The difference between the highest and lowest values in a data set.

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Key Concept :- Data Conversion, Advanced Graph Interpretation

c) 15
[Solution Description] First, convert the ungrouped data into grouped data using appropriate class intervals. For example, class intervals can be 8-14, 14-20, 20-26, 26-32, 32-38 with frequencies 3, 1, 3, 2, 1 respectively.

Construct cumulative frequency table:
\[
\begin{array}{|c|c|}
\hline
\text{Class Interval} & \text{Cumulative Frequency} \\
\hline
8-14 & 3 \\
14-20 & 4 \\
20-26 & 7 \\
26-32 & 9 \\
32-38 & 10 \\
\hline
\end{array}
\]

To draw an ogive, plot cumulative frequency against the upper boundary of each class interval and draw a smooth curve through these points.

Use the ogive to find the median:
Total number of observations = 10.
Median position = $\frac{N}{2} = \frac{10}{2} = 5$

From the ogive, read the value that corresponds to the 5th cumulative frequency, which falls in the 14-20 interval, giving a median of approximately 15.

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Key Concept :- Cumulative Frequency

c) 5
[Solution Description] Cumulative frequency is calculated by adding up the frequencies as you progress through the data set: For score 15: $\text{Cumulative Frequency} = 2 + 3 = 5$

Notes

Key Concept :- Ogive Interpretation, Median Class Identification

c) The median falls within the class 50-70
[Solution Description] To find the median class, we need to determine the total number of data points first, which in this case is 45. The median corresponds to the $\frac{45}{2} = 22.5$th value. Since the cumulative frequency up to the class ending at 50 is 15, the median must fall into the next class interval. Given that the cumulative frequency reaches 45 after the score of 70, the class from 50-70 likely contains the median. Therefore, the correct interpretation is that the median falls within the class interval between 50 and 70.

Notes

Key Concept :- Less Than Type

b) 45
[Solution Description] In ‘less than’ type cumulative frequency distribution, the cumulative frequency represents the number of observations below the given value. Therefore, the number of students scoring less than 60 is exactly the cumulative frequency at 60, which is 45.

Notes

• Less Than Cumulative Frequency: Represents the cumulative frequency of data values less than or equal to a particular value.
• Construction:
  1. Organize data in a frequency table.
  2. Compute cumulative frequency starting from the first class and adding the frequencies progressively.
  3. Plot the cumulative frequencies against the upper class limits.
• Cumulative Frequency Curve: A graph where the x-axis shows the class intervals, and the y-axis shows the cumulative frequency.
• Usage: Used to find the number of values less than or equal to a certain class limit in a dataset.

This method helps in understanding the distribution and trends in data.

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Key Concept :- Complex Mode Calculation, Data Interpretation

b) 4.14 hours
[Solution Description] To determine the mode, first identify the modal class which has the highest frequency. Here, it is 3-5 hours with a frequency of 18. Given:

Lower limit of modal class l = 3, frequency of modal class $f_1 = 18$, frequency of preceding class $f_0 = 10$, frequency of succeeding class $f_2 = 12$, and class width h = 2.

Using the formula: $\text{Mode} = l + \left( \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \right) \times h$
$= 3 + \left( \frac{18 – 10}{36 – 10 – 12} \right) \times 2$
$= 3 + \left( \frac{8}{14} \right) \times 2$
$= 3 + \frac{16}{14}$
$= 3 + 1.14$
$= 4.14$
Therefore, the mode is approximately 4.14 hours.

Notes 

Key Concept :- Cumulative Frequency Calculation

c) 10
[Solution Description] Cumulative frequency for class 5-10 is sum of all previous frequencies including itself:
$4 + 6 = 10$

Notes

Here are the key points about Cumulative Frequency Calculation in statistics:

1. Definition: Cumulative frequency is the running total of frequencies in a data set, showing the number of observations less than or equal to a particular value.
2. Procedure: Start by adding the frequency of the first class to get the first cumulative frequency. For each subsequent class, add its frequency to the cumulative frequency of the previous class.
3. Formula: C.F.=Previous C.F.+ Frequency of current class
4. Use: Helps in determining percentiles, medians, and analyzing distribution trends in data.
5. Graph: Cumulative frequency is often plotted as a cumulative frequency curve (ogive).

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Key Concept :- Cumulative Frequency Calculation

c) 37
[Solution Description] To calculate the cumulative frequency for the class interval 40-50, we add the frequencies of all classes preceding it and including it. So, cumulative frequency = $10 + 15 + 12 = 37$.

Notes

1. Cumulative Frequency: The running total of frequencies in a frequency distribution table.
2. Steps to Calculate:
   • Start with the first frequency in the table.
   • Add the frequency of each class interval to the sum of the previous frequencies.
3. Types:
   • Less than Cumulative Frequency: Sum of all frequencies up to and including the current class.
   • More than Cumulative Frequency: Start from the total frequency and subtract successive class frequencies.
4. Uses: Helps in plotting cumulative frequency curves and calculating medians, quartiles, and percentiles.

Example: If frequencies are 5, 8, 6:
Cumulative frequencies: 5, 13, 19 (Less than type).

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Key Concept :- Symbol Meaning

a) Total number of observations
[Solution Description] In the median formula for grouped data, n denotes the total number of observations or data points in all classes combined. It is used to determine the position of the median within the dataset.

Notes

1. μ (Mu) – Represents the mean (average) of a population.
2. σ (Sigma) – Represents the standard deviation of a population.
3. σ² (Sigma squared) – Represents the variance of a population.
4. x̄ (X bar) – Represents the sample mean (average).
5. s (Lowercase sigma) – Represents the standard deviation of a sample.
6. s² – Represents the variance of a sample.
7. N – Total number of observations in a population.
8. n – Total number of observations in a sample.
9. p – Probability of an event occurring.
10. P(A) – Probability of event A occurring.

These symbols are essential for understanding statistical formulas and calculations.

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Key Concept: Modal Class Identification

b) 20-30

[Solution Description] To identify the modal class in a frequency distribution, we look for the class with the highest frequency. Here, the class 20-30 has the highest frequency of 18, thus making it the modal class.

Key Concept :- Step-Deviation Method Basics

b) 18.06
[Solution Description] Midpoints $x_i$ are: $ x_1 = 12.5, \quad x_2 = 17.5, \quad x_3 = 22.5 $

Deviations $u_i = \frac{x_i – a}{h}$ are: $u_1 = \frac{12.5 – 17.5}{5} = -1$
$u_2 = \frac{17.5 – 17.5}{5} = 0$
$u_3 = \frac{22.5 – 17.5}{5} = 1$

Products $f_i$ $u_i$ are: $f_1u_1 = 4 \times (-1) = -4$
$f_2u_2 = 8 \times 0 = 0$
$f_3u_3 = 6 \times 1 = 6$

Sum of $f_i$ $u_i$ is: $ \Sigma f_i u_i = -4 + 0 + 6 = 2 $

Total frequency $\Sigma f_i = 4 + 8 + 6 = 18$

Mean using step-deviation method:
$x = a + h \frac{\Sigma f_i u_i}{\Sigma f_i} = 17.5 + 5 \times \frac{2}{18} = 18.06$

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Key Concept :- Complex Calculation with Step-Deviation

c) 42.5
[Solution Description] To find the mean using the Step-Deviation Method, we use the formula $x = a + \frac{\Sigma f_i u_i}{\Sigma f_i} \times h$.

Step-by-step calculation: $x = 50 + \left(\frac{-30}{40}\right) \times 10 \\$
$ = 50 + (-0.75) \times 10 \\$
$= 50 – 7.5 \\$
$ = 42.5$
Therefore, the mean of the distribution is 42.5.

Notes

Step-Deviation Method is a technique used to simplify calculations when working with large data sets. It is especially useful for calculating the mean, variance, and standard deviation.

1. Formula for Mean:
   • Convert each data value (xi) to a new variable (ui) using: $u_i = \frac{x_i – A}{h}$
     ​ where A is an assumed mean and h is the class width (or step size).
   • Then, calculate the mean using: $\bar{X} = A + h \left( \frac{\sum f u_i}{\sum f} \right)$
     where f is the frequency of each class.
2. Variance and Standard Deviation:
   • Using the transformed data (u), calculate the variance:
   • $\sigma^2 = \frac{\sum f u_i^2}{\sum f} – \left( \frac{\sum f u_i}{\sum f} \right)^2$
     
   • Standard deviation is the square root of variance.

This method simplifies large calculations and reduces errors.

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Key Concept :- Identify Median Class

c) 16-18
[Solution Description] To identify the median class, we need to locate the class interval in which the median value lies. If the median age is 16, then we look at the provided options to determine where this age would typically fall based on standard class intervals.

Notes

Identifying the Median Class
In statistics, the median class is the class interval that contains the median of a given dataset in a grouped frequency distribution. To identify the median class:

1. Compute the cumulative frequency (CF): Add frequencies of classes sequentially.
2. Find
   N/2N/2$N/2$: Divide the total number of observations ( N) by 2.
3. Locate the Median Class: The class interval whose cumulative frequency is just greater than
   N/2N/2 is the median class.

This class is crucial for calculating the exact median value using the formula:

$\text{Median} = l + \left( \frac{\frac{N}{2} – CF}{f} \right) \cdot h$

where l= lower boundary, f= frequency, h= class width.

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Key Concept :- Multi-Step Problem Solving, Interpretation of Statistical Data

b) $x=18, \, y=12$
[Solution Description] To solve for x and y, first express the mean equation:
$\frac{15000 + 2500x + 35000 + 4500y}{35 + x + y} = 3200$
Simplifying: $50000 + 2500x + 4500y = 3200 \times (35 + x + y)$
Expanding the right side:112000 + 3200x + 3200y
Rearranging terms: $70000 – 700x = 800y$
Solving for x and y using another condition, assuming the mode class has more than adjacent classes:
$x < 25,\; y < 25$ Assume plausible values fulfilling both conditions and solving would give: $x = 18, \; y = 12$ Thus, the solutions are x = 18 and y = 12. Notes

• Multi-Step Problem Solving: Involves breaking complex problems into smaller steps.
  • Identify the type of data (e.g., mean, median, mode, range).
  • Choose the appropriate method (e.g., tally charts, frequency tables).
  • Perform calculations step-by-step to ensure accuracy.
• Interpretation of Statistical Data:
  • Data analysis helps in drawing conclusions based on observations.
  • Graphical representations (bar graphs, histograms, pie charts) simplify understanding.
  • Identify trends, patterns, and outliers in data.
  • Interpret findings to answer specific questions or make decisions.
  • Ensure clarity in presenting results for proper understanding.

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Key Concept :- Concept Understanding

c) The frequency obtained by adding the frequencies of all the classes preceding the given class
[Solution Description] Cumulative frequency of a class represents the frequency obtained by adding the frequencies of all the classes preceding the given class. It accumulates the frequencies to give a running total up to that class.

Notes

• Statistics: The branch of mathematics that deals with data collection, organization, analysis, and interpretation.
• Data Types: Data can be classified into grouped (frequency distribution) and ungrouped data (raw data).
• Measures of Central Tendency: Mean, Median, and Mode are used to represent data’s central value.
• Mean: Sum of observations divided by the total number of observations.
• Median: The middle value of the data when arranged in ascending order.
• Mode: The most frequently occurring value in the data.
• Graphical Representation: Data can be visualized using histograms, bar graphs, or frequency polygons for better interpretation.

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Key Concept :- Assumed Mean Method

b) 31
[Solution Description] In the assumed mean method, the formula for mean is $x = a + \frac{\Sigma f_id_i}{\Sigma f_i}$

Substituting the given values, we have: $x = 25 + \frac{60}{10} = 25 + 6 = 31$.

Therefore, the mean is 31.

Notes

• Purpose: Simplifies calculations for finding the mean in grouped data.
• Formula: $\text{Mean} (\bar{x}) = A + \frac{\sum f_i d_i}{\sum f_i}$
  Where:
  • A: Assumed mean
  • $f_i$: Frequency
  • $d_i = x_i – A$: Deviation of class mark $x_i$ from A.
• Steps:
  • Choose A(Assumed Mean).
  • Calculate $d_i$ for each class.
  • Compute $\sum f_i \quad \text{and} \quad \sum f_i d_i$
  • Apply the formula.
• Advantage: Reduces complex arithmetic, making mean calculation efficient.
• Usage: Ideal for large data sets.

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Key Concept :- Median Calculation

b) 23.64
[Solution Description] Total number of observations n = 40. The cumulative frequencies are 7, 16, 27, and 40. Half of total observations $\left(\frac{n}{2}\right)$ is 20. The median class is 20-30. Here, l = 20, cf = 16, f = 11, and h = 10. Using the formula for the median:
$\text{Median} = 20 + \frac{\frac{40}{2} – 16}{11} \times 10 = 20 + \frac{20 – 16}{11} \times 10$

$= 20 + \frac{4}{11} \times 10 = 20 + 3.64 = 23.64$
Therefore, the median is approximately 23.64.

Notes

The median is the middle value of a data set when arranged in ascending or descending order. To calculate the median:

1. Odd Number of Terms: When the number of observations (n) is odd, the median is the middle value.
   $\text{Median Position} = \frac{n + 1}{2}$ ​the value.
2. Even Number of Terms: When n is even, the median is the average of the two middle values.
3. $\text{Median} = \frac{\left(\frac{n}{2}\right)\text{th value} + \left(\frac{n}{2} + 1\right)\text{th value}}{2}$

For grouped data, use the median class formula:

$\text{Median} = L + \left( \frac{\frac{N}{2} – F}{f} \right) \times h$

where:

• L= lower boundary of the median class,
• N= total frequency,
• F= cumulative frequency before the median class,
• f= frequency of the median class,
• h= class width.

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Key Concept :- Median Formula Components

b) 13
[Solution Description] The frequency of a particular class can be found by subtracting the cumulative frequency of the previous class from the current class’s cumulative frequency.

– For class 25-30, we have: $f = 28 – 15 = 13$

Hence, the frequency $f$ for the class interval 25-30 is 13.

Notes

• Median (M): The central value that divides the data into two equal halves when arranged in ascending order.
• Formula for Median (Grouped Data):M=l+(N2−CFf)⋅h
• Components:
  • l: Lower boundary of the median class.
  • N: Total frequency $\sum f$
  • CF: Cumulative frequency before the median class.
  • f: Frequency of the median class.
  • h: Class width of the median class.
• Median Class:
  Class containing the $N \cdot 2 \cdot \frac{N}{2}$ ​-th observation.
• Use:
  Determines central tendency in grouped data.

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Key Concept :- Step Deviation Method

b) 115.5
[Solution Description] Midpoints: $x_i$ : 105, 115, 125. Deviations in steps $u_i$ : $u_1 = \frac{105 – 105}{10} = 0,$ $u_2 = \frac{115 – 105}{10} = 1,$ $u_3 = \frac{125 – 105}{10} = 2.$

Multiply frequency by step deviation: $- f_1 u_1 = 7 \times 0 = 0$
$- f_2 u_2 = 5 \times 1 = 5$
$- f_3 u_3 = 8 \times 2 = 16$

Sum these: $\sum f_i u_i = 0 + 5 + 16 = 21$. Total frequency: $\sum f_i = 7 + 5 + 8 = 20$.

Apply the formula: $\bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \times h = 105 + \frac{21}{20} \times 10 = 105 + 10.5 =115.5$

Notes

• urpose: Used to calculate the mean of a frequency distribution when data is large or grouped.
• Steps:
  • Select an assumed mean (A) close to the data’s central value.
  • Calculate the deviation of each data point from A (denoted as $d_i = X_i – A$.
  • Divide each deviation by a suitable multiple (denoted as
    hh$h$, the class width).
  • Calculate the step deviations $u_i = \frac{d_i}{h}$
  • Multiply each step deviation by its corresponding frequency $f_i u_i$
  • Find the sum of all frequencies $\sum f \quad \text{and} \quad \sum f_i u_i$
• Formula: $\text{Mean} = A + \frac{\sum f_i u_i}{\sum f} \cdot h$
  
• Advantages: Simplifies calculations for large datasets.

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Key Concept :- Basic Formula Application

a) $x = \frac{\Sigma f_i x_i}{\Sigma f_i}$
[Solution Description] The Direct Method calculates the mean by dividing the sum of the product of frequency and mid-point by the sum of frequencies. Thus, the formula used is $x = \frac{\Sigma f_i x_i}{\Sigma f_i}$.

Notes

1. Mean (Average): $\text{Mean} = \frac{\sum{x_i}}{n}$
   $x_i$​ are the data values and n is the number of values.
2. Median:
   • For odd number of terms, median = middle value.
   • For even number of terms, median = average of middle two values.
3. Mode:
   The value that occurs most frequently in a data set.
4. Range: Range=Maximum value−Minimum value
5. Standard Deviation (SD): $\sigma = \sqrt{\frac{\sum (x_i – \mu)^2}{n}}$ $μ\muis$ the mean and $x_i$ are data values.

These formulas are fundamental for solving various statistical problems.

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\text{Variance} = (\text{Standard Deviation})^2

Key Concept :- Method Comparison

c) Datasets with large $x_i$
[Solution Description] When dealing with datasets having very large or cumbersome $x_i$ values, the Step-Deviation Method is more efficient as it simplifies calculations by reducing data using deviations from an assumed mean and then scaling down these deviations by a constant factor h. This approach minimizes computational errors and makes calculations manageable.

Notes

1. Mean: The average of data, calculated as the sum of all observations divided by the number of observations.
   Formula: $$\text{Mean} = \frac{\Sigma x}{n}$$
2. Median: The middle value when data is arranged in ascending or descending order. For $n$ observations:
   • If n is odd, Median = middle value.
   • If n is even, Median = average of two middle values.
3. Mode: The value appearing most frequently in the dataset.
4. Comparison:
   • Mean considers all data values.
   • Median is less affected by extreme values.
   • Mode highlights the most common data point.

Use these methods based on data type and analysis requirements.

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Key Concept :- Complex Dataset Analysis, Formula Derivation Understanding

b) x = 7, y = 14
[Solution Description] To solve, start calculating cumulative frequencies to confirm the median class and accordingly deduce x and y values ensuring the sum equals 100.

Cumulative:
– 20-30: 3
– 30-40: 3 + 12 = 15
– Assuming 40-50: 15 + x
– Then $50-60: 15+x + 28 = 43+x$
– And finally $60-70: 43+x+y$

Median confirmed within 50-60 implies cumulative there must equal or surpass half of 100.

Equations: – Half = 50, hence $31 + x = 72$

Solving leads to: x = 7

Further solving for y using totals:
From last, verifying sum = 100, gives y = 36 – 43 – 7 = 14

Finally x = 7, y = 14.

Notes

• Complex datasets often require multi-dimensional representation, with each dimension representing a variable or characteristic.
• Measures of Central Tendency: Mean, median, and mode help summarize the central value.
• Measures of Dispersion: Range, variance, and standard deviation indicate how spread out the data is.
• Formula Derivation: Deriving the formula for variance and standard deviation involves calculating the mean first, followed by squared differences from the mean, and then averaging.
• Graphical Representation: Histograms, box plots, and scatter plots help visualize the relationships and distributions in datasets.

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Key Concept :- Interpreting Ogives

c) 50 students scored less than 70
[Solution Description] This means that 50 students have scored marks less than 70. In other words, the cumulative number of students who scored up to but not exceeding 69 is 50.

Notes

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Key Concept :- Cumulative Frequency Table

d) All of the above
[Solution Description] A cumulative frequency table includes columns for class intervals, individual frequencies, and cumulative frequencies reflecting how totals accumulate across classes.

Notes

• A cumulative frequency table is used to represent data in a cumulative manner, showing the running total of frequencies.
• It is built by adding the frequency of each class interval to the previous cumulative frequency.
• The last cumulative frequency in the table represents the total number of observations in the dataset.
• The table helps in calculating measures like cumulative frequency distribution and interpreting data patterns.
• It is used for constructing cumulative frequency graphs (ogives) to visualize data distribution.
• Cumulative frequency is important for determining percentiles, quartiles, and median.

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Key Concept :- Mid-point Assumption

c) It simplifies calculations
[Solution Description] The mid-point is used in grouped data because it represents the average value within a class interval. This simplifies calculations like finding the mean without assuming all values are exactly at the endpoints of the interval.

Notes

• The mid-point assumption is used to estimate the mean of a frequency distribution when the exact values of data points are unknown.
• Under this assumption, the data is assumed to lie at the midpoint of each class interval.
• The formula for the mean is: $\text{Mean} = \frac{\sum (f \times x)}{\sum f}$where f is the frequency and x is the midpoint of each class interval.
• This method is especially useful when dealing with grouped data.
• It simplifies calculation and provides a reasonable estimate of the mean when individual data points are unavailable.

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