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

Question 1. Write the correct alternative answer for each of the following questions.

(i) Which of the following data is not primary ?
(A) By visiting a certain class, gathering information about attendance of students.
(B) By actual visit to homes, to find number of family members.
(C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.
(D) Review the cleanliness status of canals by actually visiting them.
Answer: (C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.
In simple words: Primary data is collected directly by the researcher. Secondary data is collected from a source that has already compiled it, like a Talathi's records.

🎯 Exam Tip: Understanding the difference between primary and secondary data is crucial for data collection methods in statistics.

Question.
(ii) What is the upper class limit for the class 25 - 35?
(A) 25
(B) 35
(C) 60
(D) 30
Answer: (B) 35
In simple words: In a class interval, the upper class limit is the highest value in that specific interval. For '25 - 35', 35 is the upper limit.

🎯 Exam Tip: Accurately identifying upper and lower class limits is fundamental for constructing frequency distributions.

Question.
(iii) What is the class-mark of class 25 - 35?
(A) 25
(B) 35
(C) 60
(D) 30
Answer: (D) 30
In simple words: The class-mark is the midpoint of a class interval, calculated by adding the upper and lower limits and dividing by two: \((25 + 35) / 2 = 30\).

🎯 Exam Tip: The class-mark is vital for calculating the mean of grouped data and constructing frequency polygons.

Question.
(iv) If the classes are 0 - 10, 10 - 20, 20 - 30, ..., then in which class should the observation 10 be included?
(A) 0 - 10
(B) 10 - 20
(C) 0 - 10 and 10-20 in these 2 classes
(D) 20 - 30
Answer: (B) 10 - 20
In simple words: In exclusive class intervals, the upper limit of a class is not included in that class but in the next one. So, 10 belongs to the 10 - 20 class.

🎯 Exam Tip: Always remember the rule for exclusive class intervals: the upper limit is excluded from the current class and included in the subsequent class to avoid overlapping.

Question.
(v) If \(\bar{x}\) is the mean of \(x_1, x_2, \dots, x_n\) and \(\bar{y}\) is the mean of \(y_1, y_2, \dots y_n\) and \(\bar{z}\) is the mean of \(x_1,x_2, \dots, x_n, y_1, y_2, \dots y_n\), then \(\bar{z} = ?\)
(A) \(\frac{\bar{x}+\bar{y}}{2}\)
(B) \(\bar{x}+\bar{y}\)
(C) \(\frac{\bar{x}+\bar{y}}{n}\)
(D) \(\frac{\bar{x}+\bar{y}}{2n}\)
Answer: (A) \(\frac{\bar{x}+\bar{y}}{2}\)
Solution:
We know that \(\bar{x} = \frac{\Sigma x}{n}\)
\(\implies n\bar{x} = \Sigma x\)
Similarly, \(n\bar{y} = \Sigma y\)
Now, for \(\bar{z}\), the total sum of observations is \(\Sigma x + \Sigma y\), and the total number of observations is \(n + n = 2n\).
So, \(\bar{z} = \frac{\Sigma x + \Sigma y}{n + n}\)

\(\implies \bar{z} = \frac{n\bar{x} + n\bar{y}}{2n}\)

\(\implies \bar{z} = \frac{n(\bar{x} + \bar{y})}{2n}\)

\(\implies \bar{z} = \frac{\bar{x} + \bar{y}}{2}\)
In simple words: When combining two datasets of equal size, the combined mean is simply the average of their individual means.

🎯 Exam Tip: For combined means with an equal number of observations in each group, the formula simplifies significantly. Ensure you can derive this relationship.

Question.
(vi) The mean of five numbers is 50, out of which mean of 4 numbers is 46, find the 5th number.
(A) 4
(B) 20
(C) 434
(D) 66
Answer: (D) 66
Solution:
Sum of five numbers = \(5 \times 50 = 250\)
Sum of four numbers = \(4 \times 46 = 184\)
5th number = Sum of five numbers - Sum of four numbers
= \(250 - 184\)
= \(66\)
In simple words: To find a missing number when the mean of a set and a subset are known, subtract the sum of the subset from the total sum of the entire set.

🎯 Exam Tip: This type of problem tests your understanding of the definition of the mean: Sum = Mean \(\times\) Number of Observations.

Question.
(vii) Mean of 100 observations is 40. The 9th observation is 30. If this is replaced by 70 keeping all other observations same, find the new mean.
(A) 40.6
(B) 40.4
(C) 40.3
(D) 40.7
Answer: (B) 40.4
Solution:
Initial sum of 100 observations = \(100 \times 40 = 4000\)
New sum = Initial sum - Replaced value + New value
New sum = \(4000 - 30 + 70 = 4040\)
New mean = \(\frac{\text{New sum}}{\text{Total number of observations}} = \frac{4040}{100} = 40.4\)
In simple words: When a value is replaced in a dataset, adjust the total sum by subtracting the old value and adding the new one, then recalculate the mean.

🎯 Exam Tip: For replacement problems, directly adjust the total sum before dividing by the count of observations. This is more efficient than recalculating all values.

Question.
(viii) What is the mode of 19, 19, 15, 20, 25, 15, 20, 15?
(A) 15
(B) 20
(C) 19
(D) 25
Answer: (A) 15
In simple words: The mode is the number that appears most frequently in a given dataset. In this series, 15 appears three times, which is more than any other number.

🎯 Exam Tip: To find the mode, list all unique numbers and count their occurrences. The one with the highest count is the mode. A dataset can have multiple modes.

Question.
(ix) What is the median of 7, 10, 7, 5, 9, 10 ?
(A) 7
(B) 9
(C) 8
(D) 10
Answer: (C) 8
Solution:
First, arrange the data in ascending order: 5, 7, 7, 9, 10, 10.
There are 6 observations (an even number).
The median is the average of the two middle terms (3rd and 4th).
Median = \(\frac{7+9}{2} = \frac{16}{2} = 8\)
In simple words: The median is the middle value of an ordered dataset; for an even number of items, it's the average of the two middle values.

🎯 Exam Tip: Always arrange the data in ascending or descending order *before* finding the median. For an even number of observations, calculate the mean of the two central values.

Question.
(x) From following table, what is the cumulative frequency of less than type for the class 30-40?

(A) 13
(B) 15
(C) 35
(D) 22
Answer: (C) 35
Solution:
Cumulative frequency of less than type for the class 30 - 40 is the sum of frequencies of all classes up to and including 30 - 40.
Cumulative frequency = Frequency(0-10) + Frequency(10-20) + Frequency(20-30) + Frequency(30-40)
= \(7 + 3 + 12 + 13 = 35\)
In simple words: The 'less than' cumulative frequency for a class sums up the frequencies of all classes with values less than or equal to the upper limit of that class.

🎯 Exam Tip: Cumulative frequency (less than type) helps understand how many observations fall below a certain value and is critical for identifying median class in grouped data.

Question 2. The mean salary of 20 workers is Rs.10,250. If the salary of office superintendent is added, the mean will increase by Rs. 750. Find the salary of the office superintendent.
Answer:
Mean = \(\frac{\text{The sum of all observations}}{\text{Total number of observations}}\)
Therefore, The sum of all observations = Mean \(\times\) Total number of observations
The mean salary of 20 workers is Rs. 10,250.
Thus, Sum of the salaries of 20 workers = \(20 \times 10,250 = \text{Rs. } 2,05,000\) ...(i)
If the superintendent's salary is added, then mean increases by 750.
New mean = \(10,250 + 750 = 11,000\)
Total number of people after adding superintendent = \(20 + 1 = 21\)
Therefore, Sum of the salaries including the superintendent's salary = \(21 \times 11,000 = \text{Rs. } 2,31,000\) ...(ii)
Therefore, Superintendent salary = sum of the salaries including superintendent's salary - sum of salaries of 20 workers
= \(2,31,000 - 2,05,000\) ...[From (i) and (ii)]
= \(26,000\)
Therefore, The salary of the office superintendent is Rs. 26,000.
In simple words: Calculate the total sum of salaries before and after adding the superintendent, then find the difference to determine the superintendent's salary.

🎯 Exam Tip: This problem requires careful calculation of total sums based on the mean formula before and after an addition to the dataset.

Question 3. The mean of nine numbers is 77. If one more number is added to it, then the mean increases by 5. Find the number added in the data.
Answer: :
Mean = \(\frac{\text{The sum of all observations}}{\text{Total number of observations}}\)
Therefore, The sum of all observations = Mean \(\times\) Total number of observations
Mean of nine numbers is 77.
Therefore, sum of 9 numbers = \(9 \times 77 = 693\) ...(i)
If one more number is added, then mean increases by 5.
Mean of 10 numbers = \(77 + 5 = 82\)
Therefore, sum of the 10 numbers = \(10 \times 82 = 820\) ...(ii)
Therefore, Number added = sum of the 10 numbers - sum of the 9 numbers = \(820 - 693\) ...[From (i) and (ii)]
= \(127\)
Therefore, The number added in the data is 127.
In simple words: To find the new number added, calculate the total sum before and after the addition, then subtract the initial sum from the new total sum.

🎯 Exam Tip: Similar to Question 2, this problem emphasizes using the total sum derived from the mean to find a specific value added to the dataset.

Question 4. The monthly maximum temperature of a city is given in degree Celsius in the following data. By taking suitable classes, prepare the grouped frequency distribution table
29.2, 29.0, 28.1, 28.5, 32.9, 29.2, 34.2, 36.8, 32.0, 31.0, 30.5, 30.0, 33, 32.5, 35.5, 34.0, 32.9, 31.5, 30.3, 31.4, 30.3, 34.7, 35.0, 32.5, 33.5.29.0. 29.5.29.9.33.2.30.2
From the table, answer the following questions.
i. For how many days the maximum temperature was less than 34°C?
ii. For how many days the maximum temperature was 34°C or more than 34°C?
Answer:

i. Number of days for which the maximum temperature was less than 34°C
= \(8 + 8 + 8 = 24\)
ii. Number of days for which the maximum temperature was 34°C or more than 34°C
= \(5 + 1 = 6\)
In simple words: First, organize the raw temperature data into a frequency table using given class intervals, then use the frequencies to answer questions about specific temperature ranges.

🎯 Exam Tip: When constructing a frequency distribution table, ensure that class intervals are consistent and observations are accurately tallied. Pay attention to inclusive/exclusive limits if specified.

Question 5. If the mean of the following data is 20.2, then find the value of p.

Answer:

Mean \((\bar{x}) = \frac{\Sigma f_i x_i}{\Sigma f_i}\)
Therefore, \(20.2 = \frac{610 + 20p}{30 + p}\)

\(\implies 20.2 (30 + p) = 610 + 20p\)

\(\implies 606 + 20.2p = 610 + 20p\)

\(\implies 20.2p - 20p = 610 - 606\)

\(\implies 0.2p = 4\)
Therefore, \(p = \frac{4}{0.2} = \frac{40}{2} = 20\)
Therefore, \(p = 20\)
In simple words: Use the formula for the mean of grouped data, substitute the given mean and frequency sum, then solve the resulting linear equation for the unknown frequency 'p'.

🎯 Exam Tip: Remember the formula for the mean of grouped data \(\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}\). Algebraic manipulation is key when solving for an unknown frequency.

Question 6. There are 68 students of 9th standard from Model Highschool, Nandpur. They have scored following marks out of 80, in written exam of mathematics.
70, 50, 60, 66, 45, 46, 38, 30, 40, 47, 56, 68,
80, 79, 39, 43, 57, 61, 51, 32, 42, 43, 75, 43,
36, 37, 61, 71, 32, 40, 45, 32, 36, 42, 43, 55,
56, 62, 66, 72, 73, 78, 36, 46, 47, 52, 68, 78,
80, 49, 59, 69, 65, 35, 46, 56, 57, 60, 36, 37,
45, 42, 70, 37,45, 66, 56, 47
By taking classes 30 - 40, 40 - 50, .... prepare the less than type cumulative frequency table. Using the table, answer the following questions:
i. How many students have scored marks less than 80?
ii. How many students have scored marks less than 40?
iii. How many students have scored marks less than 60?
Answer:
Class

i. 66 students have scored marks less than 80.
ii. 14 students have scored marks less than 40.
iii. 45 students have scored marks less than 60.
In simple words: Organize the raw data into class intervals, count frequencies using tally marks, then calculate the cumulative frequency (less than type) to answer questions about specific score ranges.

🎯 Exam Tip: Accuracy in tallying and calculating cumulative frequencies is paramount. Double-check your sums to ensure the final cumulative frequency matches the total number of observations.

Question 7. By using data in example (6), and taking classes 30 - 40, 40 - 50,... prepare equal to or more than type cumulative frequency table and answer the following questions based on it.
i. How many students have scored marks 70 or more than 70?
ii. How many students have scored marks 30 or more than 30?
Answer:

i. 11 students have scored marks 70 or more than 70.
ii. 68 students have scored marks 30 or more than 30.
In simple words: To create a 'more than or equal to' cumulative frequency table, start with the total observations and subtract the frequencies of classes from the bottom up.

🎯 Exam Tip: For 'more than or equal to' cumulative frequency, start from the total frequency at the lowest class interval and subtract frequencies as you move up the classes.

Question 8. There are 10 observations arranged in ascending order as given below.
45, 47, 50, 52, JC, JC + 2, 60, 62, 63, 74. The median of these observations is 53. Find the value of JC. Also find the mean and the mode of the data.
Answer:
i. Given data in ascending order:
45, 47, 50, 52, x, x+2, 60, 62, 63, 74. (Using 'x' instead of 'JC' for clarity in equations)
Therefore, Number of observations (n) = 10 (i.e., even)
Therefore, Median is the average of middle two observations.
Here, the 5th and 6th numbers are in the middle position.
Median = \(\frac{\text{(5th observation) + (6th observation)}}{2}\)
\(53 = \frac{x + (x+2)}{2}\)

\(\implies 53 = \frac{2x+2}{2}\)

\(\implies 106 = 2x + 2\)

\(\implies 106 - 2 = 2x\)

\(\implies 104 = 2x\)

\(\implies x = 52\)
Therefore, The given data becomes:
45, 47, 50, 52, 52, 54, 60, 62, 63, 74.

ii. Mean = \(\frac{\text{The sum of all observations in the data}}{\text{Total number of observations}}\)
Mean = \(\frac{45+47+50+52+52+54+60+62+63+74}{10}\)
Mean = \(\frac{559}{10} = 55.9\)
Therefore, The mean of the given data is 55.9.

iii. Given data in ascending order:
45, 47, 50, 52, 52, 54, 60, 62, 63, 74.
Therefore, The observation repeated maximum number of times = 52
Therefore, The mode of the given data is 52.
In simple words: Use the median definition for an even dataset to find the unknown 'JC', then calculate the mean by summing all values and dividing by the count, and identify the mode by finding the most frequent number.

🎯 Exam Tip: For problems involving median with an unknown value, correctly identify the middle terms and set up the equation. Remember to re-list the complete dataset to find the mean and mode accurately.

Maharashtra Board Class 9 Maths Chapter 7 Statistics Problem Set 7 Intext Questions and Activities

Question 1. To show following information diagrammatically, which type of bar- diagram is suitable?
i. Literacy percentage of four villages.
ii. The expenses of a family on various items.
iii. The numbers of girls and boys in each of five divisions.
iv. The number of people visiting a science exhibition on each of three days.
v. The maximum and minimum temperature of your town during the months from January to June.
vi. While driving a two-wheeler, number of people wearing helmets and not wearing helmet in 100 families.
Answer:
i. Percentage bar diagram
ii. Sub-divided bar diagram
iii. Sub-divided bar diagram
iv. Sub-divided bar diagram
v. Sub-divided bar diagram
vi. Sub-divided bar diagram
In simple words: Different types of bar diagrams (percentage, sub-divided) are chosen based on whether you're showing parts of a whole, comparisons of totals, or ranges for different categories.

🎯 Exam Tip: Understand the purpose of each type of bar diagram (simple, sub-divided, percentage) to choose the correct one for representing given data effectively.

Question 2. You gather information for several reasons. Take a few examples and discuss whether the data is primary or secondary.
Answer:
[Students should attempt the above activity on their own.]
In simple words: Data is primary if you collect it yourself directly from the source; it's secondary if you obtain it from existing records or sources.

🎯 Exam Tip: Differentiating between primary and secondary data is a foundational concept in statistics, important for understanding data reliability and collection methods.