Class 9 Mathematics Chapter 12 Statistics

Key Concept: Histogram Adjustment

c) 3.0

[Solution Description]

Adjusted length formula: $\frac{\text{original length}}{w} \times \text{minimum class size}$. Original length = frequency = 6, $w$ = 30. Calculation: $\frac{6}{30} \times 15 = 3$.

Key Concept: Axes Representation

a) Horizontal

[Solution Description]

According to the syllabus, bars of uniform width are drawn on one axis (typically x-axis) representing the variable. The vertical axis (y-axis) shows values. Thus, categories are plotted on horizontal axis.

Key Concept: Advantages of Graphical Representation

b) Facilitates easier visual comparison

[Solution Description]

Graphical representation allows easier visual comparison of relative characteristics through visual elements like bars or lines. This makes patterns and differences more apparent compared to numerical tables.

Key Concept: Variables on Axes

b) Horizontal axis (\$x-axis)

[Solution Description]

Bar graphs place categorical variables (expenditure heads) on the horizontal axis ($x$-axis) and numerical values (expenditure amounts) on the vertical axis ($y$-axis).

Key Concept: Vertical Axis Scaling

a) 44

[Solution Description]

For classes with width equal to minimum class size (2), adjusted length = original frequency.

The 6-8 interval has frequency 44, which remains unchanged as its width matches the minimum class size.

Key Concept: Class-mark Calculation

b) \$45\$

[Solution Description]

Class-mark = \$(Lower limit + Upper limit)/2\$

For \$40 – 50\$:

Calculation: \$(40 + 50)/2 = 90/2 = 45\$

Key Concept: Imaginary Classes in Frequency Polygon

c) To ensure the area matches the histogram’s area

[Solution Description] Imaginary classes help complete the polygon by connecting it to the horizontal axis. This ensures the total area under the frequency polygon equals the area of the original histogram. When these classes (with midpoints A and H) are added, the triangles formed between the histogram’s bars and the polygon are congruent, keeping the total area unchanged.

Key Concept: Rectangle width for equal class intervals

c) The value of the class width itself

[Solution Description]

In histograms with equal class intervals, the width of rectangles is exactly equal to the class size (here 5 units). This ensures uniform spacing.

Key Concept: Comparison Using Frequency Polygons

c) Overlaying both polygons on the same axes

[Solution Description]

The syllabus states that multiple frequency polygons for different datasets can be overlaid on the same axes. This enables direct visual comparison of their distributions since their lines will show differences in peaks, spreads, and central tendencies.

Key Concept: Tally Marks Purpose

c) To count occurrences efficiently

[Solution Description]

Tally marks (||||

– represent 5 counts) are used during manual data collection to quickly record frequencies before finalizing the table. They simplify counting processes.

Key Concept: Adjusting Lengths in Varying Class Intervals

d) 30 units

[Solution Description]

Class width = 25 – 15 = 10. Base width for standardization = 10. Adjusted length = $\left(\frac{\text{Frequency}}{\text{Class Width}}\right) \times \text{Base Width}$. Substituting values: $\frac{30}{10} \times 10 = 30$. Check which option matches 30.

Key Concept: Bar Graphs

d) Number of students

[Solution Description]

In bar graphs, the variable being measured (e.g., number of students) is always represented on the y-axis. The categories (e.g., school names) are placed on the x-axis. Here, “number of students” is the measured quantity.

Key Concept: Primary Purpose

b) Enables visual comparison of individual items

[Solution Description]

Bar graphs enable immediate visual comparison of individual items at a glance, as stated in their key characteristics.

Key Concept: Imaginary Class Interval

a) (25, 0)

[Solution Description]

Step 1: First real class is 30-40. To add imaginary class before it: subtract class width (10) from lower limit.

Step 2: Imaginary class = 20-30. Its midpoint = (20+30)/2 = 25. Frequency = 0.

Step 3: Coordinate = (midpoint, 0) → (25, 0)

Key Concept: Axis Representation

a) Horizontal axis ($x$-axis)

[Solution Description] According to step 1 of constructing a bar graph, the variable (e.g., types of fruits, expenditure heads) is always represented on the horizontal axis ($x$-axis). The vertical axis ($y$-axis) represents the values corresponding to the variable.

Key Concept: Class-mark Calculation

b) 150-160

[Solution Description]

To find the class interval, use the class-mark formula:

$\text{Class-mark} = \frac{\text{Upper limit} + \text{Lower limit}}{2}$

Given class-mark = 155, solve for upper and lower limits:

$155 = \frac{\text{Upper limit} + \text{Lower limit}}{2} \\ \implies \text{Upper limit} + \text{Lower limit} = 310$

From the options, 150 + 160 = 310. Hence, the class interval is 150-160.

Key Concept: Characteristics of Primary Data

c) It is customized to address specific research objectives

[Solution Description]

Primary data is original, collected specifically for the research problem at hand. Secondary data is repurposed from existing sources. Primary data collection is costlier and time-consuming compared to secondary data.

Key Concept: Adjusted Length Calculation

b) 8

[Solution Description]

Use formula: Adjusted length = \$\left(\frac{\text{Frequency}}{\text{Class width}}\right) \times \text{Minimum class size}\$.

Substitute values:

Frequency = 24, Class width = 15, Min class size = 5

Calculation: $\frac{24}{15} = 1.6$, then $1.6 \times 5 = 8$

Key Concept: Type of Question: Class Mark Calculation

b) 35

[Solution Description]

Class-mark is calculated using:

$\frac{\text{Upper limit} + \text{Lower limit}}{2} = \frac{40 + 30}{2} = 35$. All arithmetic operations are performed step-by-step.

Key Concept: Budget Planning

d) Mean of the expenses

[Solution Description]

To create a realistic budget, one would calculate the average expenditure. The mean is calculated by adding all values and dividing by the number of months.

Calculation:

Mean = $(1200 + 1350 + 1100 + 1450 + 1300 + 1250) \div 6$

Sum = $7650$

Mean = $7650 \div 6 = 1275$.

This average gives a balanced estimate for budgeting.

Key Concept: Constructing Frequency Polygon

b) To equalize area between histogram and frequency polygon

[Solution Description]

To create closed shape ABCDEFGHI, we add imaginary class intervals before first and after last real intervals. This ensures the total area under frequency polygon equals area of original histogram through congruent triangle properties (as per hint in syllabus).

Key Concept: Misleading Histogram Detection

c) Class intervals 0-50 (width 50), 50-60 (width 10)

[Solution Description]

Misleading histograms occur when class widths vary but frequencies aren’t adjusted. A bar spanning 0-50 (width 50) vs another spanning 50-60 (width 10) with same frequency distorts area proportionality.

Key Concept: Frequency Calculation

b) 105

[Solution Description]

Customers after 10 AM exclude the 8-10 AM slot. Add frequencies of 10-12 PM, 12-2 PM, and 2-4 PM: $35 + 45 + 25 = 105$.

Key Concept: Minimum Width Identification

d) 5

[Solution Description]

Minimum class width is the smallest interval difference. Calculate widths: 10-0=10, 15-10=5, 25-15=10. The smallest value is 5. Always compare actual numerical differences, not frequency values.

Key Concept: Application Comparison

c) Two continuous datasets of similar type

[Solution Description]

The syllabus states that frequency polygons enable visual comparison between two continuous datasets of the same nature, such as academic performance across different classes. Their graphical overlap allows identifying patterns or disparities easily.

Key Concept: Definition

c) Sales figures from a retail store’s internal database

[Solution Description]

Secondary data refers to information collected by someone other than the original user. Examples include census reports, published journal articles, or company annual reports. Primary data is collected directly by the researcher (e.g., surveys conducted personally).

Key Concept: Histogram Application

b) Continuous class intervals

[Solution Description]

A histogram is specifically designed for continuous grouped data with adjacent rectangles. It cannot be used for discrete categories or ungrouped data which require bar graphs.

Key Concept: Histogram Properties

c) Total area of rectangles represents total frequency

[Solution Description] Correct property: Areas of rectangles are proportional to frequencies. False statements claim heights directly equal frequencies in unequal intervals or presence of gaps.

Key Concept: Bar Graph Interpretation

d) March had the maximum sales.

[Solution Description]

The tallest bar indicates the highest value. The question asks for a definite truth. Option d states March has maximum sales (directly inferred from tallest bar). Other options involve assumptions beyond graphical representation (like growth rate or comparison with previous year).

Key Concept: Discrete Data Representation

b) Gaps between bars clarify separation of distinct categories

[Solution Description] Syllabus states bar graphs use gaps between bars to emphasize categorical comparisons for discrete data. Histograms (no gaps) are for continuous data. Correct answer must refer to gaps enabling categorical distinction.

Key Concept: Bar Graphs

b) Horizontal axis (\$x-axis)

[Solution Description]

According to the syllabus, in bar graphs, the categories (or “Heads”) are represented on the horizontal axis (x-axis). The example specifies placing Heads on the horizontal axis and values on the vertical axis.