CBSE Class 9 Mathematics Frequency Distribution Assignment Set 04

IX - Mathematics Assignment No-01 - Statistics

Question 1. Explain the meaning of the following terms:
(a) Primary Data:
Answer: Data collected directly by the investigator (or a person acting on their behalf) for a specific purpose or research study for the first time is called primary data. It is original in character.
In simple words: Data gathered firsthand by the person doing the study.

Exam Tip: Mention the word "original" and "firsthand" to score full marks in definition-based questions.

(b) Secondary Data:
Answer: Data that has already been collected, processed, and published by someone else (an agency, government department, or another researcher) and is subsequently used by the investigator is called secondary data.
In simple words: Data taken from existing sources like books, websites, or reports.

Exam Tip: Always distinguish between primary and secondary data based on who collected it initially.

(c) Raw data:
Answer: Data obtained in its original, unorganized form, just as it was collected without any arrangement, classification, or processing, is called raw data.
In simple words: Unorganized numbers straight from observation.

Exam Tip: Raw data is the starting point for constructing frequency distribution tables.

(d) Grouped data:
Answer: Data that has been organized and classified into groups, classes, or categories along with their respective frequencies is called grouped data.
In simple words: Data organized into classes or intervals.

Exam Tip: Grouped data makes large sets of information easy to analyze and interpret.

(e) Frequency:
Answer: The number of times a particular observation or value occurs in a given data set is called its frequency.
In simple words: How often a value repeats.

(f) variate:
Answer: Any characteristic that can take on different numerical values or quantities is called a variate or variable.
In simple words: A characteristic that changes and can be measured numerically.

Exam Tip: Variates can be discrete (taking specific isolated values) or continuous (taking any value within an interval).

(g) class interval:
Answer: Each individual group or sub-division into which the total raw data is divided in a grouped frequency distribution is called a class interval (e.g., \(10 - 20\)).
In simple words: The range of values forming a group in statistics.

Exam Tip: A class interval consists of a lower limit and an upper limit.

(h) class size:
Answer: The difference between the upper limit and the lower limit of a class interval is called the class size or class width. Formula: Class size = Upper limit - Lower limit.
In simple words: The width or length of a class interval.

Exam Tip: In standard continuous frequency tables, all class intervals usually have the same class size.

(i) class mark or mid value:
Answer: The central value of a class interval, obtained by taking the average of its upper limit and lower limit, is called the class mark. Formula: Class mark = \(\frac{\text{Lower limit} + \text{Upper limit}}{2}\).
In simple words: The exact middle point of a class interval.

Exam Tip: Class marks are extensively used in calculating the mean of grouped frequency distributions.

(j) Comulative frequency:
Answer: The cumulative frequency corresponding to a specific class interval is the running total of frequencies of all classes up to and including that class.
In simple words: The sum of frequencies added progressively row by row.

Exam Tip: The cumulative frequency of the last class always equals the total number of observations (\(N\)).

(k) class limits:
Answer: The lowest and highest values that can be included in a particular class interval are called its class limits (lower class limit and upper class limit).
In simple words: The boundary numbers defining a class.

Exam Tip: In inclusive classes, limits are part of the class; in exclusive classes, the upper limit is excluded.

(l) True class limits:
Answer: The adjusted class boundaries used in inclusive series to convert them into continuous exclusive series by subtracting \(0.5\) from lower limits and adding \(0.5\) to upper limits are called true class limits.
In simple words: Exact continuous boundaries with no gaps between classes.

Exam tip: True class limits are mandatory for drawing histograms.

(m) Range:
Answer: The difference between the highest observation and the lowest observation in a given set of raw data is called the range. Formula: Range = Highest value - Lowest value.
In simple words: The spread from the smallest to the largest value.

Exam Tip: Range helps in deciding the number of classes and class size when constructing a frequency table.

(n) Array:
Answer: Arranging the raw data observations in an ascending or descending order of magnitude is called an array or arrayed data.
In simple words: Data sorted in order from smallest to largest (or vice versa).

Exam Tip: Making an array prevents missing numbers while counting frequencies.

Question 2. The weights (in grams) of 40 apples are given below. Arrange the data in increasing order (an array) and construct a frequency table.
90, 75, 85, 70, 60, 70, 70, 75, 100, 75, 110, 30, 40, 50, 45, 84, 60, 55, 65, 100, 30, 85, 35, 80, 65, 30, 40, 70, 75, 110, 70, 45, 45, 40, 60, 55, 100, 80, 95, 85.
Answer:
Step 1: Ascending Order Array (40 values):
30, 30, 30, 35, 40, 40, 40, 45, 45, 45, 50, 55, 55, 60, 60, 60, 65, 65, 70, 70, 70, 70, 70, 75, 75, 75, 75, 80, 80, 84, 85, 85, 85, 90, 95, 100, 100, 100, 110, 110.

Step 2: Frequency Table (Exclusive Form, Class Size = 10):

In simple words: Sorted the weights from smallest to largest and counted how many weights fall into each 10-unit interval.

Exam Tip: Always cross-check that the sum of frequencies matches the total number of items given in the question (\(40\)).

Question 3. The temperature of a certain city in any month is given below. Construct f.d.t. (both inclusive and exclusive form) having 6 C.I.
31.4, 34.4, 35.6, 36.3, 36.7, 36.9, 34.7, 31.6, 32.5, 33.3, 32.6, 33.5, 37.8, 37.5, 34.9, 33.8, 30.4, 37.3, 37.0, 32.5, 36.4, 35.2, 31.0, 32.4, 37.0, 36.9, 34.4, 35.3, 30.3, 33.9.
Answer:
Step 1: Find Range and Class Size
Lowest value = 30.3, Highest value = 37.8
Range = \(37.8 - 30.3 = 7.5\)
Number of classes = 6
Approximate class size = \(\frac{7.5}{6} = 1.25\). Let us choose a convenient class size of \(1.3\) or \(1.5\). Let class size = \(1.2\) or range covers \(30.0\) to \(38.0\) with 6 classes of size \(1.33\) (or simply use ranges \(30 - 32, 32 - 34\)... if size = 2, giving 4 classes). Let us use uniform class size = \(1.3\):
Classes: \(30.0 - 31.3, 31.3 - 32.6, 32.6 - 33.9, 33.9 - 35.2, 35.2 - 36.5, 36.5 - 37.8\) (Exclusive form).
For Inclusive form: \(30.0 - 31.2, 31.3 - 32.5, 32.6 - 33.8, 33.9 - 35.1, 35.2 - 36.4, 36.5 - 37.7\) (Adjusted slightly to cover all values up to \(37.8\)).

Frequency Distribution Table (Exclusive Form):

In simple words: Divided the temperature data into 6 continuous intervals and counted the frequencies.

Exam Tip: In exclusive form, a value equal to the upper limit goes into the next higher class.

Question 3-B. The class marks of a distribution are: 93, 163, 123, 103, 113, 133, 143, 153 in mixed form. Find class size, class lts. and C.I.
Answer:
Class marks given: 93, 103, 113, 123, 133, 143, 153, 163
Class size (\(h\)) = Difference between consecutive class marks = \(103 - 93 = 10\)
Class Limits for each class mark (\(x\)) are given by \(x - \frac{h}{2}\) to \(x + \frac{h}{2}\):
- For class mark 93: \(93 - 5\) to \(93 + 5\) = \(88 - 98\)
- For class mark 103: \(98 - 108\)
- For class mark 113: \(108 - 118\)
- For class mark 123: \(118 - 128\)
- For class mark 133: \(128 - 138\)
- For class mark 143: \(138 - 148\)
- For class mark 153: \(148 - 158\)
- For class mark 163: \(158 - 168\)
In simple words: Subtract and add half the class size (\(5\)) from each class mark to find the class intervals.

Exam Tip: Class size is always constant for equally spaced class marks.

Question 4. Form a frequency table for the following variables.
18, 27, 27, 20, 20, 16, 22, 17, 27, 20, 21, 16, 25, 24, 16, 19, 21, 25, 20, 18, 27, 21, 16, 26, 16, 18, 27, 25, 23, 26, 27, 16, 25, 27, 21, 22, 18, 24, 20, 20, 20, 27, 24, 26, 16, 22, 19, 22, 18, 19.
Answer:
Since the data consists of discrete variable values from 16 to 27, we can form an ungrouped (or discrete) frequency table:

In simple words: Counted the exact occurrences of each distinct number in the dataset.

Exam Tip: For discrete variables with a small range, a discrete frequency table is preferred over grouped classes.

Question 5. Form a c.f.d.t with C.I. of 50.
628, 668, 650, 665, 620, 480, 560, 300, 328, 525, 420, 360, 440, 480, 360, 305, 326, 370, 470, 328, 306, 325, 520, 360, 326, 440, 406, 664, 421, 450, 480, 426, 405, 421, 565.
Answer:
Lowest value = 300, Highest value = 668. Class size = 50.
Classes: \(300 - 350, 350 - 400, 400 - 450, 450 - 500, 500 - 550, 550 - 600, 600 - 650, 650 - 700\).

Cumulative Frequency Distribution Table (c.f.d.t):

In simple words: Grouped data into classes of size 50 and calculated cumulative frequencies by adding frequencies progressively.

Exam Tip: Ensure that the final cumulative frequency matches the total count of observations (\(35\)).

Question 6. Following is the record of ages of 50 students in a certain school. Form a f.d.t and c.f.d.t keeping one class as 12-15, 15 not included.
11, 15, 07, 20, 15, 18, 06, 19, 19, 16, 8, 17, 12, 19, 18, 16, 06, 20, 13, 11, 10, 08, 14, 6, 19, 18, 09, 08, 12, 15, 12, 14, 08, 12, 16, 15, 17, 09, 11, 07, 17, 11, 11, 18, 14, 12, 07, 13, 17, 13.
Answer:
Class interval size = \(15 - 12 = 3\). Classes: \(6 - 9, 9 - 12, 12 - 15, 15 - 18, 18 - 21\).

Frequency and Cumulative Frequency Distribution Table:

In simple words: Grouped student ages into intervals of width 3 and calculated running cumulative totals.

Exam Tip: "15 not included" confirms that the intervals are in exclusive form (\(12 - 15\) means 12 is included, 15 is excluded).

Question 7. Following is the distribution of ages (in years) of teachers in a certain school:

Answer the following:
(a) Write the lower limit of the 2nd C.I.
Answer: 21
In simple words: The starting value of the second class interval (\(21 - 25\)) is 21.

Exam Tip: Be careful to identify whether the class interval is inclusive or exclusive when reading limits.

(b) Find the class-limits of the 4th C.I.
Answer: Lower limit = 31, Upper limit = 35 (Class interval: \(31 - 35\))
In simple words: The 4th interval is \(31 - 35\), so its limits are 31 and 35.

Exam Tip: Count class intervals carefully from left to right.

(c) Find the class-mark of the 6th C.I.
Answer: 6th C.I. is \(41 - 45\).
Class mark = \(\frac{41 + 45}{2} = \frac{86}{2} = 43\)
In simple words: Average of 41 and 45 is 43.

Exam Tip: Class mark formula is always \(\frac{\text{lower limit} + \text{upper limit}}{2}\).

(d) Find the true class-limits of 5th C.I.
Answer: 5th C.I. is \(36 - 40\).
Since the given table is inclusive (\(20\) to \(21\) has a gap of \(1\)), subtract \(0.5\) from lower limit and add \(0.5\) to upper limit.
True class limits = \(35.5 - 40.5\)
In simple words: Expand the boundaries by \(0.5\) on both sides to make the classes continuous.

Exam Tip: Always check for gaps between classes to determine if true limits are needed.

(e) Find the class-size
Answer: Using true class limits for \(16 - 20\) (\(15.5 - 20.5\)), Class size = \(20.5 - 15.5 = 5\). (Alternatively, Upper limit - Lower limit + 1 for inclusive = \(20 - 16 + 1 = 5\)).
In simple words: Each class covers 5 years of age.

Exam Tip: For inclusive series, class size is \(\text{Upper Limit} - \text{Lower Limit} + 1\).

(f) Construct a c.f.d.t.
Answer:

In simple words: Added frequencies row by row to find the cumulative table.

Exam Tip: Verify addition at each step to avoid calculation errors.

(g) How many teachers are of age: (i) less than 31 years (ii) 36 years or more.
Answer:
(i) Less than 31 years: Includes classes \(16 - 20, 21 - 25, 26 - 30\) (true upper limit \(30.5\)).
Number of teachers = \(11 + 32 + 51 = 94\).
(ii) 36 years or more: Includes classes \(36 - 40, 41 - 45, 46 - 50\).
Number of teachers = \(27 + 6 + 4 = 37\).
In simple words: (i) Sum teachers up to age 30. (ii) Sum teachers from age 36 onwards.

Exam Tip: Use true boundaries when dealing with inclusive classes for "less than" or "more than" questions.

Question 8. Following is the table of class marks of various classes. Construct c.f.d.t.

Answer:
Class size \(h = 152 - 146 = 6\).
Class intervals are formed using \(x - 3\) to \(x + 3\):
- For mark 146: \(143 - 149\)
- For mark 152: \(149 - 155\)
- For mark 158: \(155 - 161\)
- For mark 164: \(161 - 167\)
- For mark 170: \(167 - 173\)
- For mark 176: \(173 - 179\)
- For mark 182: \(179 - 185\)
- For mark 188: \(185 - 191\)

Cumulative Frequency Distribution Table:

In simple words: Derived class intervals from class marks and computed cumulative frequencies.

Exam Tip: Subtract and add half of the class interval difference (\(6/2 = 3\)) to each class mark to find class boundaries.

Question 9. Extract a f.d.t from the following:

Answer:
(a) Frequency table from "Below" cumulative data (Class size = 10):

(b) Frequency table from "More than" cumulative data (Class size = 100):

In simple words: Subtracted consecutive cumulative frequencies to find individual class frequencies.

Exam Tip: For "Below" data, subtract downward; for "More than" data, subtract upward.

Question 10. Find the unknown entries \(a, b, c, d, e, f\) from the following c.f.d.t. shows the heights of 50 students in a class.

Answer:
- For first class, \(a = 12\)
- For second class, c.f. = \(a + b \implies 25 = 12 + b \implies b = 25 - 12 = 13\)
- For third class, c.f. (\(c\)) = Previous c.f. + frequency = \(25 + 10 = 35 \implies c = 35\)
- For fourth class, c.f. = \(c + d \implies 43 = 35 + d \implies d = 43 - 35 = 8\)
- For fifth class, c.f. = \(48 = \text{previous c.f. } (43) + e \implies e = 48 - 43 = 5\)
- For sixth class, c.f. (\(f\)) = Previous c.f. (\(48\)) + frequency (\(2\)) = \(48 + 2 = 50 \implies f = 50\) (Total students = 50).
Summary: \(a = 12, b = 13, c = 35, d = 8, e = 5, f = 50\).
In simple words: Found missing frequencies and cumulative frequencies step by step using addition and subtraction.

Exam Tip: The last cumulative frequency must always equal the total number of observations given in the problem statement (\(50\)).