Maharashtra Board Class 9 Maths Chapter 7 Set 7.3 Algebra Standard Part 1 Statistics Solutions

Question 1. For class interval 20 - 25 write the lower class limit and the upper class limit.
Answer:
Lower class limit = 20
Upper class limit = 25
In simple words: For a given class interval, the first number represents the lower limit and the second number represents the upper limit. This helps define the range of values included in that class.

🎯 Exam Tip: Accurately identifying lower and upper class limits is fundamental for constructing frequency distribution tables and calculating class marks.

Question 2. Find the class-mark of the class 35-40.
Answer:
Class-mark \( = \frac{\text{Lower class limit + Upper class limit}}{2} \)
\( = \frac{35+40}{2} \)
\( = \frac{75}{2} \)
\( = 37.5 \)
... Class-mark of the class 35 - 40 is 37.5
In simple words: The class-mark is the midpoint of a class interval, calculated by averaging its lower and upper limits. It represents the central value of the class.

🎯 Exam Tip: Class-mark is crucial for calculating mean from grouped data. Ensure accuracy in arithmetic operations.

Question 3. If class-mark is 10 and class width is 6, then find the class.
Answer:
Let the upper class limit be x and the lower class limit be y.
Class mark = 10 ...[Given]
Class-mark
Class-mark \( = \frac{\text{Lower class limit + Upper class limit}}{2} \)
..
\( 10 = \frac{x+y}{2} \)
.. \( x + y = 20 \)..(i)
Class width = 6 ... [Given]
Class width = Upper class limit - Lower class limit
.. \( x - y = 6 \)..(ii)
Adding equations (i) and (ii),
\( x + y = 20 \)
\( x - y = 6 \)
\( 2x = 26 \)
.. \( x = 13 \)
Substituting \( x = 13 \) in equation (i),
\( 13 + y = 20 \)
∴ \( y = 20 - 13 \)
∴ \( y = 7 \)
.. The required class is 7 - 13.
In simple words: Given the class-mark (midpoint) and class width, we can set up a system of two linear equations to find the lower and upper limits of the class.

🎯 Exam Tip: This question tests your ability to apply definitions of class-mark and class width to solve for unknown class limits. Clearly state your variables and equations.

Question 4. Complete the following table.

Solution:
Let frequency of the class 14 - 15 be x then, from table,
\( 5 + 14 + x + 4 = 35 \)
.. \( 23 + x = 35 \)
.. \( x = 35 - 23 \)
.. \( x = 12 \)

In simple words: To complete the frequency table, we sum the known frequencies and subtract this sum from the total number of students to find the missing frequency.

🎯 Exam Tip: Ensure all tally marks are correctly converted to frequencies and that the sum of frequencies matches the total N given or calculated.

Question 5. In a 'tree plantation' project of a certain school there are 45 students of 'Harit Sena.' The record of trees planted by each student is given below:
3, 5, 7, 6, 4, 3, 5, 4, 3, 5, 4, 7, 5, 3, 6, 6, 5, 3, 4, 5, 7, 3, 5, 6, 4, 4, 3, 5, 6, 6, 4, 3, 5,7, 3, 4, 5, 7, 6, 4, 3, 5, 4, 4, 7.
Prepare a frequency distribution table of the data.
Answer:

In simple words: To prepare a frequency distribution table for ungrouped data, list each unique value (number of trees) and count its occurrences using tally marks to determine its frequency.

🎯 Exam Tip: When dealing with ungrouped data, ensure every data point is counted accurately. A common mistake is miscounting tally marks.

Question 6. The value of n upto 50 decimal places is given below:
3.14159265358979323846264338327950288419716939937510
From this information prepare an ungrouped frequency distribution table of digits appearing after the decimal point.
Answer:

In simple words: This table categorizes each digit (0-9) that appears after the decimal point in the given number and records how many times each digit occurs.

🎯 Exam Tip: For digit frequency, systematically count each digit's occurrence. Double-check the total frequency against the total number of digits provided (50 in this case).

Question 7. In the tables given below, class-mark and frequencies is given. Construct the frequency tables taking inclusive and exclusive classes.
i.

Answer:
Solution:
i. Let the Lower class limit and upper class limit of the class mark 5 be x and y respectively.
Class mark \( = \frac{\text{Lower class limit + Upper class limit}}{2} \)
..
\( 5 = \frac{x+y}{2} \)
.. \( x + y = 10 \)
Here, class width = 15 - 5 = 10
But, Class width = Upper class limit - Lower class limit
∴ \( y - x = 10 \)
.. \( -x + y = 10 \)..(ii)
Adding equations (i) and (ii),
\( x + y = 10 \)
\( -x + y = 10 \)
.. \( 2y = 20 \)
∴ \( y = 10 \)
Substituting \( y = 10 \) in equation (i),
.. \( x + 10 = 10 \)
.. \( x = 0 \)
.. class with class-mark 5 is 0 - 10
Similarly, we can find the remaining classes.
.. frequency table taking inclusive and exclusive classes.

In simple words: Given class marks and frequencies, we first determine the class width and then use the class-mark formula to calculate the lower and upper limits for the first class, and subsequently for all other classes, distinguishing between inclusive and exclusive formats.

🎯 Exam Tip: Pay close attention to the definition of exclusive (upper limit not included) versus inclusive (both limits included) classes. Class boundaries often shift by 0.5 for inclusive classes.

ii.

Answer:
ii. Let the lower class limit and upper class limit of the class mark 22 be x andy respectively.
Class mark \( = \frac{\text{Lower class limit+Upper class limit}}{2} \)
.:
\( 22 = \frac{x+y}{2} \)
.. \( x + y = 44 \)..(i)
Here, class width = 24 - 22 = 2
But, Class width = Upper class limit - Lower class limit
∴\( y - x = 2 \)
.. \( -x + y = 2 \) .... (ii)
Adding equations (i) and (ii),
\( x + y = 44 \)
\( -x + y = 2 \)
\( 2y = 46 \)
∴ \( y = 23 \)
Substituting \( y = 23 \) in equation (i),
.. \( x + 23 = 44 \)
.. \( x = 21 \)
... class with class-mark 22 is 21 - 23
Similarly, we can find the remaining classes
.. frequency table taking inclusive and exclusive classes.

In simple words: Similar to part (i), we use the class-mark and the derived class width to calculate the lower and upper limits for both exclusive and inclusive class intervals. The class width is determined by the difference between consecutive class marks.

🎯 Exam Tip: Be methodical in calculating class limits. A slight error in the first class can propagate throughout the table. Check that inclusive and exclusive classes correctly represent the same data range.

Question 8. In a school, 46 students of 9th standard, were told to measure the lengths of the pencils in their compass-boxes in Centimetres. The data collected was as follows:
16, 15, 7, 4.5, 8.5, 5.5, 5, 6.5, 6, 10, 12, 13, 4.5, 4.9, 16, 11, 9.2, 7.3, 11.4, 12.7, 13.9, 16, 5.5, 9.9, 8.4, 11.4, 13.1, 15, 4.8, 10, 7.5, 8.5, 6.5, 7.2, 4.5, 5.7, 16, 5.7, 6.9, 8.9, 9.2, 10.2, 12.3, 13.7, 14.5, 10
By taking exclusive classes 0-5, 5-10, 10-15,.... prepare a grouped frequency distribution table.
Answer:

In simple words: A grouped frequency distribution table organizes raw data into specified class intervals (like 0-5, 5-10) and then counts how many data points fall into each interval using tally marks. For exclusive classes, the upper limit of an interval is not included in that class but in the next.

🎯 Exam Tip: When forming grouped frequency tables with exclusive classes, be careful with values that fall exactly on the upper limit of a class. They belong to the next class.

Question 9. In a village, the milk was collected from 50 milkmen at a collection center in litres as given below:
27, 75, 5, 99, 70, 12, 15, 20, 30, 35, 45, 80, 77, 90, 92, 72, 4, 33, 22, 15, 20, 28, 29, 14, 16, 20, 72, 81, 85, 10, 16, 9, 25, 23, 26, 46, 55, 56, 66, 67, 51, 57, 44, 43, 6, 65, 42, 36, 7, 35
By taking suitable classes, prepare grouped frequency distribution table.
Answer:

In simple words: To create a grouped frequency table from raw data, define appropriate class intervals, then tally how many data points fall into each interval, ensuring all data is covered and no overlaps occur.

🎯 Exam Tip: When choosing "suitable classes," aim for 5-10 classes of equal width. Ensure class boundaries are clearly defined and handle boundary values consistently (e.g., using exclusive classes).

Question 10. 38 people donated to an organisation working for differently abled persons. The amount in rupees were as follows:
101, 500, 401, 201, 301, 160, 210, 125, 175, 190, 450, 151, 101, 351, 251, 451, 151, 260, 360, 410, 150, 125, 161, 195, 351, 170, 225, 260, 290, 310, 360, 425, 420, 100, 105, 170, 250, 100
i. By taking classes 100 - 149, 150 - 199, 200 - 249... prepare grouped frequency distribution table.
ii. From the table, find the number of people who donated Rs.350 or more.
Answer:
i.

In simple words: This table groups the donation amounts into specific inclusive class intervals (e.g., 100-149 Rs.), then counts how many donations fall within each interval, providing a summarized view of the data.

🎯 Exam Tip: For inclusive classes, ensure that both the lower and upper limits are included in the count for that class. Be meticulous when tallying values near class boundaries.

ii. Number of people who donated Rs.350 or more = 4 + 4 + 2 + 1 = 11
In simple words: To find the number of people who donated Rs.350 or more, simply add the frequencies of all class intervals where the lower limit is Rs.350 or greater.

🎯 Exam Tip: When answering "more than" or "less than" questions from a frequency table, carefully identify all relevant class intervals and sum their frequencies.

Maharashtra Board Class 9 Maths Chapter 7 Statistics Practice Set 7.3 Intext Questions And Activities

Question 1. The record of marks out of 20 in Mathematics in the first unit test is as follows:
20,6, 14, 10, 13, 15, 12, 14, 17. 17, 18, 1119, 9, 16. 18, 14, 7, 17, 20, 8, 15, 16, 10, 15, 12. 18, 17, 12, 11, 11, 10, 16, 14, 16, 18, 10, 7, 17, 14, 20, 17, 13, 15, 18, 20, 12, 12, 15, 10
Answer the following questions, from the above information.
a. How many students scored 15 marks?
b. How many students scored more than 15 marks?
c. How many students scored less than 15 marks?
d. What is the lowest score of the group?
e. What is the highest score of the group? (Textbook pg. no. 114)
Answer:
a. 5 students scored 15 marks.
b. 20 students scored more than 15 marks.
c. 25 students scored less than 15 marks.
d. 6 is the lowest score of the group.
e. 20 is the highest score of the group.
In simple words: By carefully reviewing the raw data of student scores, we can directly count occurrences to answer questions about specific scores, ranges of scores, and extreme values.

🎯 Exam Tip: For raw data questions, it's helpful to sort the data first or use tally marks for accurate counting. Double-check counts for "more than" and "less than" conditions.

Question 2. For the above Question prepare Frequency Distribution Table. (Textbook pg. no. 115)
Answer:

In simple words: An ungrouped frequency distribution table lists each distinct score from the data and shows how often it occurs using tally marks and numerical frequency.

🎯 Exam Tip: When constructing an ungrouped frequency table, ensure every unique score is listed once and its frequency accurately reflects its count in the raw data. The sum of frequencies must match the total number of observations.