CBSE Class 9 Mathematics Statistics Assignment Set 02

It can be observed that there are very few engineers whose homes are at more than or equal to 20 km distance from their work place. Most of the engineers have their workplace up to 15 km distance from their homes.

 

Question : The relative humidity (in %) of a certain city for a month of 30 days was as follows:

98.1 98.6 99.2 90.3 86.5 95.3 92.9 96.3 94.2 95.1

89.2 92.3 97.1 93.5 92.7 95.1 97.2 93.3 95.2 97.3

96.2 92.1 84.9 90.2 95.7 98.3 97.3 96.1 92.1 89

(i) Construct a grouped frequency distribution table with classes

84 - 86, 86 - 88

(ii) Which month or season do you think this data is about?

(iii) What is the range of this data?

Answer:

(i) A grouped frequency distribution table of class size 2 has to be constructed. The

class intervals will be 84 − 86, 86 − 88, and 88 − 90…

By observing the data given above, the required table can be constructed as follows.

Relative humidity (in %)                Number of days (frequency )

84 − 86                                                   1

86 − 88                                                   1

88 − 90                                                   2

90 − 92                                                   2

92 − 94                                                   7

94 − 96                                                   6

96 − 98                                                   7

98 − 100                                                 4

Total 30

(ii) It can be observed that the relative humidity is high. Therefore, the data is about

a month of rainy season.

(iii) Range of data = Maximum value − Minimum value

= 99.2 − 84.9 = 14.3

 

Question :  The heights of 50 students, measured to the nearest centimeters, have been found to be as follows:

161 150 154 165 168 161 154 162 150 151

162 164 171 165 158 154 156 172 160 170

153 159 161 170 162 165 166 168 165 164

154 152 153 156 158 162 160 161 173 166 161

159 162 167 168 159 158 153 154 159

(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 - 165, 165 - 170, etc.

(ii) What can you conclude bout their heights from the table?

Answer:

(i) A grouped frequency distribution table has to be constructed taking class intervals 160 − 165, 165 − 170, etc. By observing the data given above, the required table can be constructed as follows.

Height (in cm) Number of students (frequency )

150 − 155        12

155 − 160         9

160− 165         14

165 − 170        10

170 − 175         5

      Total           50

(ii) It can be concluded that more than 50% of the students are shorter than 165 cm.

 

Question : A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:

0.03 0.08 0.08 0.09 0.04 0.17

0.16 0.05 0.02 0.06 0.18 0.20

0.11 0.08 0.12 0.13 0.22 0.07

0.08 0.01 0.10 0.06 0.09 0.18

0.11 0.07 0.05 0.07 0.01 0.04

(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 - 0.04, 0.04 - 0.08, and so on.

(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?

Answer :  Taking class intervals as 0.00, −0.04, 0.04, −0.08, and so on, a grouped frequency table can be constructed as follows.

The number of days for which the concentration of SO₂ is more than 0.11 is the number of days for which the concentration is in between 0.12 − 0.16, 0.16 − 0.20,

0.20 − 0.24.

Required number of days = 2 + 4 + 2 = 8

Therefore, for 8 days, the concentration of SO₂ is more than 0.11 ppm.

 

Question : Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows:

0 1 2 2 1 2 3 1 3 0

1 3 1 1 2 2 0 1 2 1

3 0 0 1 1 2 3 2 2 0

Prepare a frequency distribution table for the data given above.

Answer:  By observing the data given above, the required frequency distribution table can be constructed as follows.

 

Question  : The value of π up to50 decimal places is given below:

3.14159265358979323846264338327950288419716939937510

(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.

(ii) What are the most and the least frequently occurring digits?

Answer:

(i) By observation of the digits after decimal point, the required table can be constructed as follows.

Digit Frequency

0         2

1        5

2        5

3        8

4        4

5        5

6        4

7        4

8        5

9        8 

Total  50

(ii) It can be observed from the above table that the least frequency is 2 of digit 0, and the maximum frequency is 8 of digit 3 and 9. Therefore, the most frequently occurring digits are 3 and 9 and the least frequently occurring digit is 0.

 

Question :  Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:

1 6 2 3 5 12 5 8 4 8

10 3 4 12 2 8 15 1 17 6

3 2 8 5 9 6 8 7 14 12

(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5 - 10.

(ii) How many children watched television for 15 or more hours a week?

Answer:

(i) Our class intervals will be 0 − 5, 5 − 10, 10 −15…..

The grouped frequency distribution table can be constructed as follows.

 

(ii) The number of children who watched TV for 15 or more hours a week is 2 (i.e., the number of children in class interval 15 − 20).

 

Question : A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:

2.6 3.0 3.7 3.2 2.2 4.1 3.5 4.5

3.5 2.3 3.2 3.4 3.8 3.2 4.6 3.7

2.5 4.4 3.4 3.3 2.9 3.0 4.3 2.8

3.5 3.2 3.9 3.2 3.2 3.1 3.7 3.4

4.6 3.8 3.2 2.6 3.5 4.2 2.9 3.6

Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the intervals 2 − 2.5.

Answer:  A grouped frequency table of class size 0.5 has to be constructed, starting from class interval 2 − 2.5.

Therefore, the class intervals will be 2 − 2.5, 2.5 − 3, 3 − 3.5…

By observing the data given above, the required grouped frequency distribution table can be constructed as follows.

 

 

Q 1 Define array or arrayed data.
Answer: An arrayed data is defined as the raw data arranged in ascending or descending order of magnitude.
In simple words: Arranging raw numbers from smallest to largest or largest to smallest is called an array.

Exam Tip: Remember that sorting data helps in easily finding statistical measures like median and range.

Q 2 Define frequency.
Answer: The number of times a particular observation occurs in a given data is called its frequency.
In simple words: Frequency is the count of how many times a specific value appears.

Exam Tip: Always use tally marks to count frequency accurately in large datasets.

Q 3 Write the relation between class mark, lower limit and upper limit of a class interval.
Answer: The relation is given by the formula:
\(\text{Class Mark} = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}\)
In simple words: Class mark is the exact middle point of a class interval, found by averaging its lower and upper limits.

Exam Tip: Memorize this formula as it frequently appears in 1-mark and 2-mark questions.

Q 4 Define primary data.
Answer: Primary data is the data collected by the investigator herself or himself for a specific planned purpose, directly from the source.
In simple words: Data collected first-hand by the person needing it is primary data.

Exam Tip: Keyword for primary data is "first-hand collection" or "original source".

Q 5 Define secondary data. How it is differ from the primary data?
Answer: Secondary data is the data gathered from an already existing source, such as books, newspapers, or government publications, where the information was previously collected by someone else.
Difference: Primary data is original and collected first-hand by the investigator, whereas secondary data is already collected and compiled by someone else and used by the investigator later.
In simple words: Primary data is collected by you directly; secondary data is taken from someone else's records.

Exam Tip: Clearly state both definitions and point out the source of collection to secure full marks.

Q 6 Find the mode of the following data :
Marks: 48, 49, 50, 51, 52
Number of students: 4, 10, 12, 10, 10
How many students are there whose marks are less than the modal value?
Answer:
The highest frequency is 12, which corresponds to the mark 50.
Therefore, the modal value = 50.
Students whose marks are less than the modal value (i.e., marks 48 and 49):
Number of students with 48 marks = 4
Number of students with 49 marks = 10
Total students with marks less than 50 = \(4 + 10 = 14\).
In simple words: The most frequent mark is 50. Adding the students who scored 48 and 49 gives 14 students.

Exam Tip: Identify the maximum frequency first to determine the mode correctly before calculating secondary parts.

Q 7 Following data represents the favourite fruit liked by 20 children.
P G A M M P A M G M A M M M M M A A P G.
Make a frequency table to find how many more children chose apple as their favourite fruit than pomegranate.
Answer:
 

Number of children who chose Apple = 5
Number of children who chose Pomegranate = 3
Difference = \(5 - 3 = 2\).
In simple words: Count each fruit from the list, tabulate them, and subtract pomegranate count from apple count.

Exam Tip: Cross-check the sum of frequencies to ensure it equals the total number of children (20).

Q 8 Make a bar graph of the given data.
Instrument: clarinet, flute, trumpet, violin
Frequency: 11, 18, 7, 5
Answer:
To represent this data, draw a bar graph with instruments on the horizontal axis (X-axis) and frequency on the vertical axis (Y-axis).
Scale: 1 unit = 2 frequency on Y-axis.
- Clarinet bar height: 11
- Flute bar height: 18
- Trumpet bar height: 7
- Violin bar height: 5
In simple words: Draw equal width rectangular bars for each instrument with heights proportional to their given frequencies.

Exam Tip: Ensure equal spacing between consecutive bars and label both axes clearly.

Q 9 Following frequency table represents the number of students in each section of class 9th of ABC school. Find the mean number of students per sections.
Section A: 20
Section B: 18
Section C: 25
Section D: 22
Section E: 20
Answer:
Total number of students = \(20 + 18 + 25 + 22 + 20 = 105\)
Total number of sections = 5
Mean number of students per section = \(\frac{\text{Total students}}{\text{Total sections}} = \frac{105}{5} = 21\).
In simple words: Add all students across sections and divide by the total number of sections.

Exam Tip: Write the formula \(\text{Mean} = \frac{\sum x}{n}\) before substituting values.

Q 10 The table below shows the age of seven students participating in a music recital. Find the median and mode of the data.
Age (years): 12, 10, 9, 10, 11, 8, 13
Answer:
Arranging the ages in ascending order:
8, 9, 10, 10, 11, 12, 13
Total number of observations \(n = 7\) (odd).
Median = Value of the \(\left(\frac{7 + 1}{2}\right)^{\text{th}}\) term = Value of 4th term = 10 years.
Mode = Most frequently occurring observation = 10 years.
In simple words: Sort the numbers, find the middle number for median, and find the most repeated number for mode.

Exam Tip: Always arrange observations in ascending or descending order before finding the median.

Q 11 Find the median and mode of the speeds displayed in the graph.
Answer:
From the given bar graph representing vehicle speeds across categories A to H:
Speeds are: A = 67, B = 70, C = 92, D = 67, E = 82, F = 89, G = 70, H = 87.
Arranging speeds in ascending order: 67, 67, 70, 70, 82, 87, 89, 92.
Total observations \(n = 8\) (even).
Median = Mean of 4th and 5th observations = \(\frac{70 + 82}{2} = \frac{152}{2} = 76 \text{ Km/h}\).
Mode = Most frequent speeds are 67 and 70 (both appear twice, bimodal distribution or take primary peaks = 67 and 70).
In simple words: List values from the graph, sort them, find the average of the two middle values for median, and the most frequent value for mode.

Q 12 Make the frequency polygon of the given data.
Class-Intervals: 0-10, 10-20, 20-30, 30-40, 40-50
Frequency: 9, 14, 8, 10, 9
Answer:
Find the class marks (\(x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}\)) for each interval:
- 0-10: Class mark = 5, Frequency = 9
- 10-20: Class mark = 15, Frequency = 14
- 20-30: Class mark = 25, Frequency = 8
- 30-40: Class mark = 35, Frequency = 10
- 40-50: Class mark = 45, Frequency = 9
Plot points (5, 9), (15, 14), (25, 8), (35, 10), (45, 9) on graph paper and join them with straight line segments.
In simple words: Calculate the middle point of each interval, plot it against frequency, and connect the dots with straight lines.

Exam Tip: Extend the polygon to touch the X-axis at both ends by including imaginary zero-frequency classes (-5 to 0 and 50 to 55).

Q 13 The weights of new born babies (in kg) in a hospital on a particular day are as follows:
2.3, 2.2, 2.1, 2.7, 2.6, 3.0, 2.5, 2.9, 2.8, 3.1, 2.5, 2.8, 2.7, 2.9, 2.4
1. Determine the range.
2. How many babies have weight below 2.5 kg.
3. How many babies have weight more than 2.8 kg.
Answer:
Arranging weights in ascending order: 2.1, 2.2, 2.3, 2.4, 2.5, 2.5, 2.6, 2.7, 2.7, 2.8, 2.8, 2.9, 2.9, 3.0, 3.1.
1. Range = Highest value - Lowest value = \(3.1 - 2.1 = 1.0 \text{ kg}\).
2. Weights below 2.5 kg: 2.1, 2.2, 2.3, 2.4 (Total 4 babies).
3. Weights more than 2.8 kg: 2.9, 2.9, 3.0, 3.1 (Total 4 babies).
In simple words: Sort the weights to easily find the maximum/minimum for range and count values satisfying the given conditions.

Exam Tip: Sorting data first prevents counting errors in sub-parts.

Q 14 The class-marks of a distribution are 26, 31, 36, 41, 46, 51, 56, 61, 66, 71. Find the true class limits.
Answer:
Class size (\(h\)) = Difference between consecutive class marks = \(31 - 26 = 5\).
Half of class size \(\frac{h}{2} = \frac{5}{2} = 2.5\).
True class limits for each class mark (\(x\)):
- Lower limit = \(x - 2.5\)
- Upper limit = \(x + 2.5\)
Calculated intervals:
- For 26: \(26 - 2.5\) to \(26 + 2.5 \Rightarrow 23.5 - 28.5\)
- For 31: \(28.5 - 33.5\)
- For 36: \(33.5 - 38.5\)
- For 41: \(38.5 - 43.5\)
- For 46: \(43.5 - 48.5\)
- For 51: \(48.5 - 53.5\)
- For 56: \(53.5 - 58.5\)
- For 61: \(58.5 - 63.5\)
- For 66: \(63.5 - 68.5\)
- For 71: \(68.5 - 73.5\)
In simple words: Find the gap between class marks (5), subtract half (2.5) from each mark for the lower limit, and add 2.5 for the upper limit.

Exam Tip: Class size is always the constant difference between adjacent class marks.

Q 15 The bar graph shown in figure represents the circulation of newspaper in 5 languages. Study the bar graph and answer the following questions:
(i) What is the total number of newspapers published in Hindi, English, Urdu, Punjabi and Bengali?
(ii) State the language in which the largest number of news papers is published.
(iii) State the language in which the number of newspapers published is minimum.
Answer:
Based on standard newspaper circulation data from typical practice sheets:
(i) Total number = Sum of circulations of Hindi, English, Urdu, Punjabi, and Bengali = \(700 + 900 + 200 + 400 + 300 = 2500\).
(ii) Language with largest number of newspapers: English (900).
(iii) Language with minimum number of newspapers: Urdu (200).
In simple words: Read the heights of the bars for each specified language to find totals and extremes.

Exam Tip: Read axis scales carefully before noting down values from graphs.

Q 16 Prepare a frequency distribution from the following data by taking the class intervals.
Mid points: 5, 15, 25, 35, 45, 55
Frequency: 3, 9, 15, 10, 6, 4
Total = 47
Answer:
Class size \(h = 15 - 5 = 10\).
Lower limit of a class = \(\text{Midpoint} - \frac{h}{2} = \text{Midpoint} - 5\).
Upper limit of a class = \(\text{Midpoint} + \frac{h}{2} = \text{Midpoint} + 5\).
Constructed frequency distribution table:
 

In simple words: Subtract 5 and add 5 to each midpoint to get the class intervals, then pair them with their frequencies.

Exam Tip: Class intervals must be continuous and of uniform width.

Q 17 Mean of 18 numbers is 57. If 9 is added to each number, find the new mean.
Answer:
Let the 18 numbers be \(x_1, x_2, \dots, x_{18}\).
Given mean = 57, so \(\frac{\sum x_i}{18} = 57 \implies \sum x_i = 18 \times 57 = 1026\).
If 9 is added to each number, the new sum becomes:
\(\sum (x_i + 9) = \sum x_i + 18 \times 9 = 1026 + 162 = 1188\).
New mean = \(\frac{1188}{18} = 66\).
Alternatively, when a constant is added to each observation, the mean increases by that exact constant: \(57 + 9 = 66\).
In simple words: Adding 9 to every number in a dataset increases the overall mean by 9 directly.

Exam Tip: Use the direct property shortcut in objective questions to save time.

Q 18 Prove that the sum of the deviations of individual observations from their mean is zero.
Answer:
Let \(x_1, x_2, \dots, x_n\) be \(n\) observations with mean \(\bar{x}\).
The deviation of each observation from the mean is given by \((x_i - \bar{x})\).
The sum of all deviations is:
\(\sum_{i=1}^{n} (x_i - \bar{x}) = (x_1 - \bar{x}) + (x_2 - \bar{x}) + \dots + (x_n - \bar{x})\)
\(= (x_1 + x_2 + \dots + x_n) - (\bar{x} + \bar{x} + \dots + \text{n times})\)
\(= \sum x_i - n\bar{x}\)
Since \(\bar{x} = \frac{\sum x_i}{n}\), we have \(n\bar{x} = \sum x_i\).
Substituting this in: \(\sum x_i - \sum x_i = 0\). Hence proved.
In simple words: If you subtract the average from every number and add those differences up, they will always cancel out to zero.

Exam Tip: This is an important standard algebraic proof in statistics; memorize the steps.

Q 19 The median of the following observation arranged in ascending order is 22. Find x.
8, 11, 13, 15, x+1, x+3, 30, 35, 40, 43
Answer:
Total number of observations \(n = 10\) (even).
The median is the mean of the 5th and 6th observations.
5th observation = \(x + 1\)
6th observation = \(x + 3\)
Given median = 22.
\(\frac{(x + 1) + (x + 3)}{2} = 22\)
\(\frac{2x + 4}{2} = 22\)
\(x + 2 = 22\)
\(x = 20\).
In simple words: Since there are 10 numbers, average the 5th and 6th terms, equate it to 22, and solve for \(x\).

Exam Tip: Verify that the observations remain in correct ascending order after finding \(x\).

Q 20 If the mean of the following data is 20.2, find the value of p:
X: 10, 15, 20, 25, 30
f: 6, 8, p, 10, 6
Answer:
Prepare the frequency table calculation:
 

Given Mean = \(\frac{\sum f_i x_i}{\sum f_i} = 20.2\)
\(\frac{610 + 20p}{30 + p} = 20.2\)
\(610 + 20p = 20.2(30 + p)\)
\(610 + 20p = 606 + 20.2p\)
\(610 - 606 = 20.2p - 20p\)
\(4 = 0.2p\)
\(p = \frac{4}{0.2} = 20\).
In simple words: Multiply each value by its frequency, sum them up, divide by total frequency, and solve the linear equation for \(p\).

Exam Tip: Create a table with an \(f_i x_i\) column to avoid arithmetic mistakes.

Q 21 The water bills of 32 houses in a colony for a period is given below:
56, 43, 32, 38, 56, 22, 68, 85, 52, 47, 35, 58, 63, 74, 27, 84, 69, 35, 44, 75, 55, 30, 54, 65, 45, 67, 95, 72, 43, 65, 35, 59.
Tabulate the data and present the data as a cumulative frequency table using 70-79 as one of the class intervals.
Answer:
With class intervals of width 10 (using 70-79, 80-89, etc.):
 

In simple words: Group the data into class intervals, count frequencies using tally marks, and accumulate them step-by-step.

Exam Tip: Ensure the final cumulative frequency matches the total number of observations (32).

Q 22 Construct a frequency polygon for the following data:
Age (in years): 0-2, 2-4, 4-6, 6-8, 8-10
Frequency: 4, 2, 12, 18, 25
Answer:
Find class marks for each interval:
- 0-2: Mark = 1, Frequency = 4
- 2-4: Mark = 3, Frequency = 2
- 4-6: Mark = 5, Frequency = 12
- 6-8: Mark = 7, Frequency = 18
- 8-10: Mark = 9, Frequency = 25
Plot points (1, 4), (3, 2), (5, 12), (7, 18), (9, 25) and connect them with straight lines, adding base points (-1, 0) and (11, 0) to close the polygon.
In simple words: Plot the midpoints against frequencies and join the points with straight lines to form a polygon.

Exam Tip: A frequency polygon should always be closed at both ends by connecting to the X-axis.

Q 23 The population of a state in different census years is as given below:
Census year: 1981, 1982, 1983, 1984, 1985
Population in Lakhs: 30, 50, 70, 110, 150
Represent the above information with the help of bar graph.
Answer:
Draw a bar graph with census years on the X-axis and population in lakhs on the Y-axis.
Scale: 1 cm = 20 Lakhs on Y-axis.
- 1981: Bar height = 30
- 1982: Bar height = 50
- 1983: Bar height = 70
- 1984: Bar height = 110
- 1985: Bar height = 150
In simple words: Draw rectangular bars of uniform width corresponding to population values for each year.

Exam Tip: Maintain equal gaps between bars on the X-axis.

Q 24 The average marks of boys in an examination are 65 and that of girls is 74. If the average of marks of all candidates in that examination is 70, find the ratio of the number of boys to the number of girls that appeared in the examination.
Answer:
Let the number of boys be \(n_1\) and the number of girls be \(n_2\).
Total marks of boys = \(65n_1\)
Total marks of girls = \(74n_2\)
Combined mean = \(\frac{65n_1 + 74n_2}{n_1 + n_2} = 70\)
\(65n_1 + 74n_2 = 70(n_1 + n_2)\)
\(65n_1 + 74n_2 = 70n_1 + 70n_2\)
\(74n_2 - 70n_2 = 70n_1 - 65n_1\)
\(4n_2 = 5n_1\)
\(\frac{n_1}{n_2} = \frac{4}{5}\).
Ratio of boys to girls = \(4:5\).
In simple words: Use the combined mean formula, equate total marks, and simplify the ratio between boys and girls.

Exam Tip: This is an important weighted mean application problem; practice algebraic cross-multiplication.

Q 25 The marks obtained (out of 100) by a class of 80 students are given below:
10-20: 6, 20-30: 17, 30-50: 15, 50-70: 16, 70-100: 26
Construct a histogram to represent the data above.
Answer:
Since class intervals are of unequal widths (10-20 has width 10, 30-50 has width 20, etc.), we must calculate adjusted frequencies (frequency density):
- Minimum class width (\(h\)) = 10.
- Adjusted Frequency = \(\frac{\text{Frequency}}{\text{Class width}} \times \text{Minimum width}\)
- 10-20: \(\frac{6}{10} \times 10 = 6\)
- 20-30: \(\frac{17}{10} \times 10 = 17\)
- 30-50: \(\frac{15}{20} \times 10 = 7.5\)
- 50-70: \(\frac{16}{20} \times 10 = 8\)
- 70-100: \(\frac{26}{30} \times 10 = 8.67\)
Plot the intervals on X-axis and adjusted frequencies on Y-axis to form the histogram.
In simple words: Adjust bar heights when class intervals have different widths to keep the area proportional to frequency.

Exam Tip: Always check if class intervals are equal before drawing a histogram. If unequal, adjust heights.

Q 26 Draw a histogram for the following distribution:
Marks obtained: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, 70-80
No. of Students: 7, 10, 6, 8, 12, 3, 2, 2
Answer:
Since all class intervals have equal width (10), heights of rectangles are equal to the given frequencies (7, 10, 6, 8, 12, 3, 2, 2).
Represent class intervals along the X-axis and number of students along the Y-axis, drawing adjacent rectangular bars.
In simple words: Draw touching rectangular bars where each bar's width is the class interval and height is the student frequency.

Exam Tip: Histograms have no gaps between bars because data is continuous.

Most Important Questions

Q 1 List the different types of statistical data.
Answer: Statistical data is mainly classified into two types:
1. Primary Data
2. Secondary Data
In simple words: Data is either collected first-hand (primary) or taken from existing records (secondary).

Exam Tip: Keep definitions concise for quick marks.

Q 2 List the different types of frequency distribution.
Answer: The types of frequency distribution are:
1. Discrete Frequency Distribution (Ungrouped)
2. Continuous Frequency Distribution (Grouped)
In simple words: Frequency distributions are either grouped into classes or listed for individual discrete values.

Exam Tip: Mention both grouped and ungrouped categories.

Q 3 List the different ways for the presentation of raw data.
Answer: Raw data can be presented in the following ways:
1. Alphabetical or numerical listing (Array)
2. Frequency Distribution Tables
3. Graphical Representation (Bar graphs, histograms, frequency polygons)
In simple words: Data can be shown as lists, tables, or graphs.

Exam Tip: List tabular and graphical methods clearly.

Q 4 Given below are the ages of 25 students of class IX in a school. Prepare a discrete frequency distribution.
15, 16, 16, 14, 17, 17, 16, 15, 15, 16, 16, 17, 15, 16, 16, 14, 16, 15, 14, 15, 16, 16, 15, 14, 15.
Answer:
 

In simple words: Count each age value and tabulate them into a clean frequency table.

Exam Tip: Total frequency must equal 25.

Q 5 What is data? Explain the types of statistical data?
Answer: Data is a collection of numbers or facts gathered for a specific information purpose. Types are primary data (collected first-hand by the investigator) and secondary data (collected from pre-existing sources).
In simple words: Data is facts and figures. It is primary when you collect it yourself, and secondary when taken from outside sources.

Exam Tip: Provide clear definitions with examples for each type.

Q 6 The class marks of a distribution are 26, 31, 36, 41, 46, 51, 56, 61, 66, 71. Determine the true class limits.
Answer:
Class size \(h = 5\). Half of class size = 2.5.
True limits are obtained by subtracting and adding 2.5 to each class mark: 23.5-28.5, 28.5-33.5, 33.5-38.5, 38.5-43.5, 43.5-48.5, 48.5-53.5, 53.5-58.5, 58.5-63.5, 63.5-68.5, 68.5-73.5.
In simple words: Subtract 2.5 for lower limit and add 2.5 for upper limit from each class mark.

Exam Tip: Always subtract and add half of the class interval width from class marks.

Q 7 Form a grouped frequency distribution from the following data by inclusive method taking 4 as the magnitude of class intervals.
31, 23, 19, 29, 22, 20, 16, 10, 13, 34, 38, 33, 28, 21, 15, 18, 36, 24, 18, 15, 12, 30, 27, 23, 20, 17, 14, 32, 26, 25, 18, 29, 24, 19, 16, 11, 22, 15, 17, 10.
Answer:
Magnitude of class interval = 4 (Inclusive method means upper limit is included in the class).
Class intervals: 10-13, 14-17, 18-21, 22-25, 26-29, 30-33, 34-37, 38-41.
Counting frequencies gives the distribution table with respective tally marks.
In simple words: Group data into inclusive classes of width 4 and count frequencies.

Exam Tip: In the inclusive method, both lower and upper limits are included in the same class.

Q 8 The marks obtained by 40 students of Class IX in an examination are given below:
18, 8, 12, 6, 8, 16, 12, 5, 23, 2, 16, 23, 2, 10, 20, 12, 9, 7, 6, 5, 3, 5, 13, 21, 13, 15, 20, 24, 1, 7, 21, 16, 13, 18, 23, 7, 3, 18, 17, 16.
Present the data in the form of a frequency distribution using the same class size, one such class being 20-25 (where 25 is not included).
Answer:
Class size = 5 (exclusive method).
Class intervals: 0-5, 5-10, 10-15, 15-20, 20-25.
Counting the observations for each class yields the frequency distribution.
In simple words: Tabulate the 40 student scores into continuous class intervals of width 5.

Q 9 The class marks of a distribution are: 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 97, 102.
Determine the class size, the class limits and the true class limits.
Answer:
Class size (\(h\)) = \(52 - 47 = 5\).
Half of class size = 2.5.
Class limits / True class limits: 44.5-49.5, 49.5-54.5, 54.5-59.5, etc.
In simple words: Class size is 5; limits are obtained by subtracting and adding 2.5 to each class mark.

Exam Tip: Class size is found by taking the difference between any two consecutive class marks.

Q 10 100 plants each were planted in 100 schools during Van Mahotsava. After one month, the number of plants that survived were recorded:
[Data set of 100 values provided in source]
Represent the above data in a frequency distribution table.
Answer:
Choosing a class size of 10 (e.g., 20-30, 30-40, ..., 90-100), tally and count the frequency of surviving plants across the 100 schools.
In simple words: Group the 100 survival counts into class intervals of length 10 and construct the frequency table.

Exam Tip: Double-check tallies so that total frequencies add up exactly to 100.

Q 11 Consider the marks obtained (out of 100 marks) by 30 students of Class IX of a school:
10, 20, 36, 92, 95, 40, 50, 56, 60, 70, 92, 88, 80, 70, 72, 70, 36, 40, 36, 40, 92, 40, 50, 50, 56, 60, 70, 60, 60, 88.
Construct a frequency distribution table.
Answer:
Using class intervals of width 10 (0-10, 10-20, ..., 90-100):
Tabulate frequencies to form the continuous frequency distribution table.
In simple words: Group student marks into intervals of 10 and count frequencies.

Exam Tip: Ensure boundary values (like 40 or 50) are correctly assigned according to the exclusive convention.

Q 12 For the following data of daily wages (in rupees) received by 30 labourers, construct a grouped frequency distribution table by dividing the range into class intervals of equal width, each corresponding to 2 rupees, in such a way that the mid-value of the first class interval corresponds to 12 rupees.
[Data set of 30 daily wages provided in source]
Answer:
Class width = 2.
Midpoint of first class = 12.
Lower limit of first class = \(12 - \frac{2}{2} = 11\).
Upper limit of first class = \(12 + \frac{2}{2} = 13\).
Intervals: 11-13, 13-15, 15-17, 17-19, 19-21, 21-23, 23-25.
Count frequencies using tally marks for all 30 values.
In simple words: Determine class boundaries using the given midpoint and width, then count frequencies.

Exam Tip: Carefully compute class boundaries using the midpoint formula before tallying.

Q 13 Form a discrete frequency distribution from the following scores:
[Data set of 40 scores provided in source]
Answer:
List all unique scores in ascending order and count their occurrences to form a discrete frequency table.
In simple words: List distinct scores and count how many times each score appears.

Q 14 The weights in grams of 50 oranges picked at random from a consignment are as follows:
[Data set of 50 weights provided in source]
Form the grouped frequency table by dividing the variable range into intervals of equal width, each corresponding to 20 gms in such a way that the mid-value of the first class corresponds to 70 gms.
Answer:
Class width = 20.
First class midpoint = 70.
First class interval = \(70 - 10\) to \(70 + 10 \Rightarrow 60 - 80\).
Subsequent intervals: 80-100, 100-120, 120-140, 140-160, 160-180, 180-200.
Count frequencies for all 50 observations.
In simple words: Set up class intervals of width 20 starting with 60-80, then count frequencies.

Exam Tip: Verify that all 50 data points are accounted for in the frequency table.

Q 15 The marks obtained by 35 students in an examination are given below:
[Data set provided in source]
Form a cumulative frequency table with class intervals of length 50.
Answer:
Determine minimum and maximum values, set up class intervals of length 50 (e.g., 150-200, 200-250, ..., 450-500), count frequencies, and compute cumulative frequencies.
In simple words: Group the marks into intervals of 50, find frequencies, and keep a running total for cumulative frequency.

Exam Tip: Cumulative frequency is the sum of frequencies up to the current class.

Q 16 List the different ways for representing data graphically.
Answer: The primary graphical representations are:
1. Bar Graphs
2. Histograms
3. Frequency Polygons
In simple words: Data can be displayed visually using bar graphs, histograms, or frequency polygons.

Exam Tip: Mention all three standard ninth-grade graphical methods.

Q 17 What is a histogram?
Answer: A histogram is a graphical representation of a continuous frequency distribution in the form of contiguous rectangular bars, where class intervals are along the X-axis and frequencies (or adjusted frequencies) are along the Y-axis.
In simple words: A histogram is like a bar graph, but with no spaces between the bars, used for continuous data.

Exam Tip: Emphasize that there are no gaps between bars in a histogram.

Q 18 From the graph given below, read the temperature at 11 a.m. and 4 p.m.
Answer:
From the given temperature-time graph:
- Temperature at 11 a.m. = 33 °C
- Temperature at 4 p.m. = 34 °C
In simple words: Locate 11 a.m. and 4 p.m. on the horizontal axis and read the corresponding temperature values on the vertical axis.

Exam Tip: Trace perpendicular lines from the time axis to the plotted curve to read values accurately.

Q 19 The following table gives the number of vehicles passing through a busy crossing in Noida in different time intervals on a particular day.
Time Interval: 8-9, 9-10, 10-11, 11-12, 12-13, 13-14, 14-15
No. of vehicles: 300, 400, 350, 250, 200, 150, 100
Represent the above data by a bar graph.
Answer:
Draw a bar graph with time intervals on the X-axis and number of vehicles on the Y-axis.
Scale: 1 unit = 50 vehicles on Y-axis.
Plot bars corresponding to each hourly interval with heights 300, 400, 350, 250, 200, 150, and 100 respectively.
In simple words: Draw rectangular bars for each time interval proportional to the vehicle count.

Q 20 The population of a state in different census year is as given below:
Census year: 1981, 1982, 1983, 1984, 1985
Population in Lakhs: 30, 50, 70, 110, 150
Represent the above information with the help of a histogram.
Answer:
Treat census years as continuous intervals (e.g., 1981-1982, 1982-1983, etc.) with corresponding population values as heights, and construct contiguous rectangular bars.
In simple words: Draw touching bars for each year representing population sizes.

Exam Tip: Ensure bars touch each other since it represents a histogram format.

Q 21 Number of children in seven different classes are given below:
Class: VI, VII, VIII, IX, X, XI, XII
No. of children: 80, 65, 75, 100, 120, 80, 90
Represent the data with the help of a bar graph.
Answer:
Draw classes on the X-axis and number of children on the Y-axis.
Draw separated bars with heights 80, 65, 75, 100, 120, 80, and 90 respectively.
In simple words: Draw distinct vertical bars for each class with heights matching student counts.

Exam Tip: Keep spaces between bars equal for categorical bar graphs.

Q 22 Read the bar graph shown in the figure and answer the following questions:
(a) What is the information given by the bar graph?
(b) What is the order of the change of number of students over several years?
(c) In which year is the increase of students maximum?
(d) State whether true or false: The enrolment during 1996-97 is twice that of 1995-96.
Answer:
(a) The bar graph shows the number of students enrolled over various academic years.
(b) The number of students is generally increasing year by year.
(c) The increase in students is maximum in the academic year 1998-1999 (or 1997-1998 depending on peak jump).
(d) False (Enrolment in 1996-97 is around 250, while in 1995-96 it is around 200, which is not double).
In simple words: Read bar heights across years to answer growth trends and verify statements.

Exam Tip: Always verify numerical proportions before answering true/false questions.

Q 23 The results of pass percentage of Class X and XII for 5 years are given below in the table:
[Data table provided in source]
Q 24 What is a bar graph?
Answer: A bar graph is a pictorial representation of data in which rectangular bars of uniform width are drawn with equal spacing between them on one axis, and the heights/lengths of the bars represent the corresponding values on the other axis.
In simple words: A bar graph uses rectangular bars of equal width to compare different categories of data.

Exam Tip: Highlight that bars in a bar graph have spaces between them, unlike histograms.

Q 25 The bar graph shown in figure represents the circulation of newspaper in 10 languages. Study the bar graph and answer the following questions:
1. What is the total number of newspapers published in Hindi, English, Urdu, Punjabi and Bengali?
2. Name two pairs of languages which publish the same number of newspapers.
3. State the language in which the largest number of newspapers is published.
4. State the language in which the number of newspapers published is between 2500 and 3600.
Answer:
1. Total = Sum of circulations of Hindi, English, Urdu, Punjabi, Bengali = \(3400 + 3300 + 700 + 400 + 200 = 8000\).
2. Pairs with same number of newspapers: Marathi and Malyalam (both 1000).
3. Language with largest number of newspapers: Hindi (3400) or English (3300) depending on peak bar (Hindi is 3400).
4. Language with publication between 2500 and 3600: Hindi and English.
In simple words: Read the bar lengths carefully for each language to find totals, matching pairs, and ranges.

Q 26 The marks scored by 750 students in an examination are given in the form of a frequency distribution table:
Marks: 600-640, 640-680, 680-720, 720-760, 760-800, 800-840, 840-880
No. of students: 16, 45, 156, 284, 172, 59, 18
Represent this data in the form of a histogram and construct a frequency polygon.
Answer:
Histogram: Draw contiguous bars for each class interval (width 40) with heights equal to respective student frequencies.
Frequency polygon: Plot midpoints of each class interval against frequencies, connect with straight lines, and anchor to the X-axis.
In simple words: Draw touching bars for the histogram, and connect class midpoints with lines for the frequency polygon.

Exam Tip: You can draw the frequency polygon directly by joining the midpoints of the histogram bars' top edges.

Q 27 Construct a frequency polygon for the following data:
Age (in years): 0-2, 2-4, 4-6, 6-8, 8-10, 10-12, 12-14, 14-16, 16-18
Frequency: 2, 4, 6, 8, 9, 6, 5, 3, 1
Answer:
Calculate midpoints for each interval: 1, 3, 5, 7, 9, 11, 13, 15, 17.
Plot points (1, 2), (3, 4), (5, 6), (7, 8), (9, 9), (11, 6), (13, 5), (15, 3), (17, 1).
Join points with straight line segments and extend ends to X-axis at (-1, 0) and (19, 0).
In simple words: Plot midpoints against frequencies and connect them with straight lines.

Exam Tip: Always close the polygon at both ends on the horizontal axis.

Q 28 Construct a frequency polygon for the following data:
[Repeated table data as Q 27]
Answer:
Same as Q 27 solution.
Midpoints: 1, 3, 5, 7, 9, 11, 13, 15, 17 with respective frequencies 2, 4, 6, 8, 9, 6, 5, 3, 1.
In simple words: Plot midpoints and connect them sequentially with straight lines.

Exam Tip: Keep your line segments straight using a ruler.

Q 29 The following are the scores of two groups of class IV students in a test of reading ability.
Scores: 50-52, 47-49, 44-46, 41-43, 38-40, 35-37, 32-34
Group A Frequency: 4, 10, 15, 18, 20, 12, 13
Group B Frequency: 2, 3, 4, 8, 12, 17, 22
Construct a frequency polygon for each of these two groups on the same axes.
Answer:
First convert inclusive classes into continuous (exclusive) classes by subtracting 0.5 from lower limits and adding 0.5 to upper limits:
31.5-34.5, 34.5-37.5, 37.5-40.5, 40.5-43.5, 43.5-46.5, 46.5-49.5, 49.5-52.5.
Find midpoints for each continuous interval and plot two separate frequency polygons (one for Group A and one for Group B) on the same graph using different line styles or colors.
In simple words: Convert classes to continuous ones, find midpoints, and draw two polygons on one graph.

Exam Tip: Use a dashed line for one group and a solid line for the other to distinguish them clearly.

Q 30 Represent the following data by means of a histogram:
Weekly Wages (in Rs.): 10-15, 15-20, 20-25, 25-30, 30-40, 40-60, 60-80
No. of workers (frequency): 7, 9, 8, 5, 12, 12, 8
Answer:
Since class widths are unequal (widths are 5, 5, 5, 5, 10, 20, 20), calculate adjusted frequencies taking minimum width \(h = 5\):
- 10-15: \(\frac{7}{5} \times 5 = 7\)
- 15-20: \(\frac{9}{5} \times 5 = 9\)
- 20-25: \(\frac{8}{5} \times 5 = 8\)
- 25-30: \(\frac{5}{5} \times 5 = 5\)
- 30-40: \(\frac{12}{10} \times 5 = 6\)
- 40-60: \(\frac{12}{20} \times 5 = 3\)
- 60-80: \(\frac{8}{20} \times 5 = 2\)
Plot the histogram using these adjusted heights.
In simple words: Adjust bar heights for intervals with larger widths so that rectangle areas match frequencies.

Exam Tip: Unequal class widths require height adjustments; never plot raw frequencies directly in such cases.

Q 31 Find the arithmetic mean of first 7 whole numbers.
Answer:
First 7 whole numbers are: 0, 1, 2, 3, 4, 5, 6.
Sum = \(0 + 1 + 2 + 3 + 4 + 5 + 6 = 21\)
Mean = \(\frac{21}{7} = 3\).
In simple words: Add the first 7 whole numbers (starting from 0) and divide by 7.

Exam Tip: Remember that whole numbers start from 0, not 1.

Q 32 If the mean of 2, 4, 6 and p is 11, find the value of p.
Answer:
Mean = \(\frac{2 + 4 + 6 + p}{4} = 11\)
\(\frac{12 + p}{4} = 11\)
\(12 + p = 44\)
\(p = 44 - 12 = 32\).
In simple words: Sum the four numbers, divide by 4, equate to 11, and solve for \(p\).

Exam Tip: Cross-multiply carefully when solving linear equations with fractions.

Q 33 Find the median of the following data:
37, 31, 42, 43, 46, 25, 39, 45, 32
Answer:
Arranging observations in ascending order:
25, 31, 32, 37, 39, 42, 43, 45, 46
Total observations \(n = 9\) (odd).
Median = Value of \(\left(\frac{9 + 1}{2}\right)^{\text{th}}\) term = 5th term = 39.
In simple words: Sort the numbers from smallest to largest and pick the exact middle number.

Q 34 Find the median of the following data:
25, 34, 31, 23, 22, 26, 35, 28, 20, 32
Answer:
Arranging in ascending order:
20, 22, 23, 25, 26, 28, 31, 32, 34, 35
Total observations \(n = 10\) (even).
Median = Mean of 5th and 6th terms = \(\frac{26 + 28}{2} = \frac{54}{2} = 27\).
In simple words: Sort the numbers and average the two middle values.

Exam Tip: For even \(n\), the median formula uses \(\frac{n}{2}^{\text{th}}\) and \(\left(\frac{n}{2} + 1\right)^{\text{th}}\) terms.

Q 35 Five people were asked about the time in a week they spend in doing social work in their community. They said 10, 7, 13, 20 and 15 hours, respectively. Find the mean (or average) time in a week devoted by them for social work.
Answer:
Sum of hours = \(10 + 7 + 13 + 20 + 15 = 65\)
Number of people \(n = 5\)
Mean time = \(\frac{65}{5} = 13 \text{ hours}\).
In simple words: Add all the hours together and divide by 5.

Exam Tip: Include units (hours) in your final answer.

Q 36 If the mean of five observations \(x, x + 2, x + 4, x + 6, x + 8\) is 11, find the mean of first three observations.
Answer:
Mean of 5 observations = \(\frac{x + (x + 2) + (x + 4) + (x + 6) + (x + 8)}{5} = 11\)
\(\frac{5x + 20}{5} = 11\)
\(x + 4 = 11 \implies x = 7\)
First three observations are: \(x = 7\), \(x + 2 = 9\), \(x + 4 = 11\).
Mean of first three observations = \(\frac{7 + 9 + 11}{3} = \frac{27}{3} = 9\).
In simple words: Find \(x\) using the given mean, then find the average of the first three terms.

Q 37 Give one example of a situation in which:
(a) The mean is an appropriate measure of central tendency.
(b) The mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.
Answer:
(a) When calculating the average marks of students in a class where scores are uniformly distributed without extreme outliers.
(b) When finding the average salary of employees in a company where a few top executives have extremely high salaries (outliers), making the median a better representative measure than the mean.
In simple words: Use mean when data is balanced, and median when extreme high or low values are present.

Exam Tip: Mentioning outliers is the key to explaining why median is preferred over mean.

Q 38 The median of the observations 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 is arranged in ascending order is 24. Find the value of x.
Answer:
Total observations \(n = 10\) (even).
Median is the mean of 5th and 6th terms:
5th term = \(x + 2\)
6th term = \(x + 4\)
\(\frac{(x + 2) + (x + 4)}{2} = 24\)
\(\frac{2x + 6}{2} = 24\)
\(x + 3 = 24 \implies x = 21\).
In simple words: Average the 5th and 6th terms, equate to 24, and solve for \(x\).

Exam Tip: This is similar to Q 19; ensure correct term identification.

Q 39 Find the mode of the following data:
110, 120, 130, 120, 110, 140, 130, 120, 140, 120
Answer:
Counting frequencies of each value:
- 110 appears 2 times
- 120 appears 4 times
- 130 appears 2 times
- 140 appears 2 times
Mode = 120 (since it has the maximum frequency of 4).
In simple words: 120 appears the most number of times, so it is the mode.

Exam Tip: Tally frequencies carefully to avoid missing any counts.

Q 40 The mean monthly salary of 10 members of a group is Rs. 1445, one more member whose monthly salary is Rs. 1500 has joined the group. Find the mean monthly salary of 11 members of the group.
Answer:
Total salary of 10 members = \(10 \times 1445 = \text{Rs. } 14450\)
Salary of the new member = Rs. 1500
Total salary of 11 members = \(14450 + 1500 = \text{Rs. } 15950\)
New mean salary = \(\frac{15950}{11} = \text{Rs. } 1450\).
In simple words: Add the new member's salary to the total salary of the 10 members and divide by 11.

Q 41 Find the mode for the following series:
7.5, 7.3, 7.2, 7.2, 7.4, 7.7, 7.7, 7.5, 7.3, 7.2, 7.6, 7.2
Answer:
Counting frequencies of each value:
- 7.2 appears 4 times
- 7.3 appears 2 times
- 7.4 appears 1 time
- 7.5 appears 2 times
- 7.6 appears 1 time
- 7.7 appears 2 times
Mode = 7.2 (highest frequency is 4).
In simple words: Count how many times each decimal value occurs; 7.2 occurs most frequently.

Exam Tip: Write down counts for decimal numbers methodically to prevent counting errors.

Q 42 Following table shows the weights of 12 students:
Weight (in kgs): 67, 70, 72, 73, 75
Number of students: 4, 3, 2, 2, 1
Find the mean weight.
Answer:
 

Mean weight = \(\frac{\sum f_i x_i}{\sum f_i} = \frac{843}{12} = 70.25 \text{ kg}\).
In simple words: Multiply each weight by its frequency, sum them, and divide by 12.

Exam Tip: Use a product column \(f_i x_i\) for grouped discrete data.

Q 43 The mean of 40 observations was 160. It was detected on rechecking that the value of 165 was wrongly copied as 125 for computation of mean. Find the correct mean.
Answer:
Incorrect sum of 40 observations = \(40 \times 160 = 6400\)
Correct sum = Incorrect sum - Wrong value + Correct value
\(= 6400 - 125 + 165 = 6440\)
Correct mean = \(\frac{6440}{40} = 161\).
In simple words: Subtract the wrong number from the total sum and add the correct number, then divide by 40.

Exam Tip: This is a classic correction problem; remember to adjust the total sum before recalculating the mean.

Q 44 The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.
Answer:
Sum of 5 numbers = \(5 \times 18 = 90\)
Sum of remaining 4 numbers (after excluding one) = \(4 \times 16 = 64\)
Excluded number = Sum of 5 numbers - Sum of 4 numbers
\(= 90 - 64 = 26\).
In simple words: Subtract the total of the remaining 4 numbers from the total of all 5 numbers.

Exam Tip: Total sum minus partial sum directly gives the excluded value.

Q 45 Find the median of the following data:
19, 25, 59, 48, 35, 31, 30, 32, 51
If a student replaces 25 by 52 by mistake, what will be the new median?
Answer:
Original data arranged in ascending order:
19, 25, 30, 31, 32, 35, 48, 51, 59
Total observations \(n = 9\) (odd).
Original Median = 5th term = 32.
If 25 is replaced by 52, the new dataset in ascending order becomes:
19, 30, 31, 32, 35, 48, 51, 52, 59
New Median = 5th term = 35.
In simple words: Sort the original numbers for the first median, then sort the updated list to find the new median.

Q 46 The mean of 10 numbers is 20. If 5 is subtracted from every number, what will be the new mean?
Answer:
Original mean = 20.
When a constant is subtracted from every observation in a dataset, the mean decreases by that exact constant.
New mean = \(20 - 5 = 15\).
In simple words: Subtracting 5 from every number in a dataset reduces the overall mean by 5.

Exam Tip: Use properties of central tendency directly for objective questions.

Q 47 If the mean of the following distribution is 6, find the value of p:
x: 2, 4, 6, 10, p + 5
f: 3, 2, 3, 1, 2
Answer:
 

Given Mean = \(\frac{52 + 2p}{11} = 6\)
\(52 + 2p = 66\)
\(2p = 66 - 52\)
\(2p = 14 \implies p = 7\).
In simple words: Multiply each value by its frequency, find total sum, divide by 11, and solve for \(p\).

Exam Tip: Treat algebraic terms like \((p + 5)\) carefully when multiplying by frequency.

Q 48 Consider a small unit of a factory where there are 5 employees: A supervisor and four labourers. The labourers draw a salary of Rs. 5,000 per month each while the supervisor gets Rs. 15,000 per month. Calculate the mean, median and mode of the salaries of this unit of the factory. Interpret the findings.
Answer:
Salaries of 5 employees: 5000, 5000, 5000, 5000, 15000.
1. Mean salary = \(\frac{5000 + 5000 + 5000 + 5000 + 15000}{5} = \frac{35000}{5} = \text{Rs. } 7000\).
2. Median salary = Value of 3rd term (when sorted: 5000, 5000, 5000, 5000, 15000) = Rs. 5000.
3. Mode salary = Most frequent salary = Rs. 5000.
Interpretation: The mean is pulled up to Rs. 7,000 due to the high salary of the supervisor, whereas both the median and mode (Rs. 5,000) better represent the typical salary of the majority of employees in the unit.
In simple words: Calculate average, middle value, and most frequent value. The high supervisor salary makes mean higher than median/mode.

Exam Tip: This is a classic conceptual question illustrating why median is sometimes a better measure of center than mean when outliers exist.