Maharashtra Board Class 9 Maths Chapter 7 Set 7.5 Algebra Standard Part 1 Statistics Solutions

Question 1. Yield of soyabean per acre in quintal in Mukund's field for 7 years was 10, 7, 5, 3, 9, 6, 9. Find the mean of yield per acre.
Solution: \[ \text{Mean} = \frac{\text{The sum of all observations in the data}}{\text{Total number of observations}} \] \[ = \frac{10+7+5+3+9+6+9}{7} \] \[ = \frac{49}{7} \] \[ \text{Mean} = 7 \] The mean of yield per acre is 7 quintals.
In simple words: To find the mean, add up all the given soybean yields and divide by the total number of years (observations). This gives the average yield per acre.

🎯 Exam Tip: Always show the formula for mean and clearly list all observations before calculating the sum and division to avoid errors.

Question 2. Find the median of the observations, 59, 75, 68, 70, 74, 75, 80.
Solution: Given data in ascending order: 59, 68, 70, 74, 75, 75, 80
\( \therefore \) Number of observations(n) = 7 (i.e., odd)
\( \therefore \) Median is the middle most observation Here, 4th number is at the middle position, which is = 74
\( \therefore \) The median of the given data is 74.
In simple words: The median is the middle value in a dataset when the numbers are arranged in order. If there's an odd number of observations, it's the exact middle one.

🎯 Exam Tip: Remember to always arrange the data in ascending or descending order first before identifying the median. For an odd number of observations, the median is the \(\left(\frac{n+1}{2}\right)^{\text{th}}\) term.

Question 3. The marks (out of 100) obtained by 7 students in Mathematics examination are given below. Find the mode for these marks. 99, 100, 95, 100, 100, 60, 90
Solution: Given data in ascending order: 60, 90, 95, 99, 100, 100, 100 Here, the observation repeated maximum number of times = 100
\( \therefore \) The mode of the given data is 100.
In simple words: The mode is the number that appears most frequently in a set of data.

🎯 Exam Tip: Arranging the data in ascending order helps easily spot repeated values and identify the mode. If multiple numbers appear with the same highest frequency, all of them are modes.

Question 4. The monthly salaries in rupees of 30 workers in a factory are given below. 5000, 7000, 3000, 4000, 4000, 3000, 3000, 3000, 8000, 4000, 4000, 9000, 3000, 5000, 5000, 4000, 4000, 3000, 5000, 5000, 6000, 8000, 3000, 3000, 6000, 7000, 7000, 6000, 6000, 4000 From the above data find the mean of monthly salary.
Solution:

\[ \text{Mean} (\bar{x}) = \frac{\Sigma f_i x_i}{\Sigma f_i} \] \[ = \frac{1,47,000}{30} \] \[ = 4900 \]
\( \therefore \) The mean of monthly salary is Rs. 4900.
In simple words: To find the mean salary when you have a large number of observations, it's easier to group them by frequency (how many workers earn each salary), multiply each salary by its frequency, sum these products, and then divide by the total number of workers.

🎯 Exam Tip: For large datasets with repeated values, using a frequency distribution table simplifies the calculation of the mean, reducing the chances of computational errors. Ensure your summations are accurate.

Question 5. In a basket there are 10 tomatoes. The weight of each of these tomatoes in grams is as follows: 60, 70, 90, 95, 50, 65, 70, 80, 85, 95. Find the median of the weights of tomatoes.
Solution: Given data in ascending order: 50, 60, 65, 70, 70, 80, 85, 90, 95, 95
\( \therefore \) Number of observations (n) = 10 (i.e., even)
\( \therefore \) Median is the average of middle two observations Here, 5th and 6th numbers are in the middle position
\( \therefore \) Median = \( \frac{70+80}{2} \)
\( \therefore \) Median = \( \frac{150}{2} \)
\( \therefore \) The median of the weights of tomatoes is 75 grams.
In simple words: When there's an even number of observations, arrange them in order, find the two middle values, add them together, and then divide by two to get the median.

🎯 Exam Tip: For an even number of observations (n), the median is the average of the \(\left(\frac{n}{2}\right)^{\text{th}}\) and \(\left(\frac{n}{2}+1\right)^{\text{th}}\) terms. Always ensure the data is ordered first.

Question 6. A hockey player has scored following number of goals in 9 matches: 5, 4, 0, 2, 2, 4, 4, 3, 3. Find the mean, median and mode of the data.
Solution: i. Given data: 5, 4, 0, 2, 2, 4, 4, 3, 3. Total number of observations = 9 \[ \text{Mean} = \frac{\text{The sum of all observations in the data}}{\text{Total number of observations}} \] \[ = \frac{5+4+0+2+2+4+4+3+3}{9} \] \[ = \frac{27}{9} \]
\( \therefore \) The mean of the given data is 3. ii. Given data in ascending order: 0, 2, 2, 3, 3, 4, 4, 4, 5
\( \therefore \) Number of observations(n) = 9 (i.e., odd)
\( \therefore \) Median is the middle most observation Here, the 5th number is at the middle position, which is 3.
\( \therefore \) The median of the given data is 3. iii. Given data in ascending order: 0, 2, 2, 3, 3, 4, 4, 4, 5 Here, the observation repeated maximum number of times = 4
\( \therefore \) The mode of the given data is 4.
In simple words: First, calculate the mean by summing all goals and dividing by the number of matches. Then, arrange the goals in order to find the middle value for the median. Finally, identify the most frequent goal count for the mode.

🎯 Exam Tip: When asked for multiple measures of central tendency, calculate each one systematically. Always arrange data for median and mode, and double-check calculations for the mean.

Question 7. The calculated mean of 50 observations was 80. It was later discovered that observation 19 was recorded by mistake as 91. What Was the correct mean?
Solution: Here, mean = 80, number of observations = 50 \[ \text{Mean} = \frac{\text{The sum of all observations}}{\text{Total number of observations}} \]
\( \therefore \) The sum of all observations = Mean \( \times \) Total number of observations The mean of 50 observations is 80 Sum of 50 observations = 80 \( \times \) 50 = 4000 One of the observation was 19. However, by mistake it was recorded as 91. Sum of observations after correction = sum of 50 observation + correct observation – incorrect observation = 4000 + 19 - 91 = 3928
\( \therefore \) Corrected mean \[ = \frac{\text{Sum of observations after correction}}{\text{Total number of observations}} \] \[ = \frac{3928}{50} \] = 78.56
\( \therefore \) The corrected mean is 78.56.
In simple words: To correct the mean, first find the total sum of the original observations. Then, adjust this sum by subtracting the incorrect value and adding the correct value. Finally, divide the new sum by the total number of observations to get the corrected mean.

🎯 Exam Tip: For corrected mean problems, always calculate the initial sum correctly before adjusting. Clearly show the steps of subtracting the wrong value and adding the right one.

Question 8. Following 10 observations are arranged in ascending order as follows. 2, 3, 5, 9, x + 1, x + 3, 14, 16, 19, 20. If the median of the data is 11, find the value of x.
Solution: Given data in ascending order: 2, 3, 5, 9, x + 1, x + 3, 14, 16, 19, 20.
\( \therefore \) Number if 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.
\( \therefore \) Median = \( \frac{(x+1)+(x+3)}{2} \)
\( \therefore \) 11 = \( \frac{2x+4}{2} \)
\( \implies \) 22 = 2x + 4
\( \implies \) 22 - 4 = 2x
\( \implies \) 18 = 2x
\( \implies \) x = 9
In simple words: Since the data is already ordered and there are 10 observations, the median is the average of the 5th and 6th values. Set this average equal to the given median (11) and solve the resulting equation for x.

🎯 Exam Tip: When given an ordered dataset with variables and the median, correctly identify the middle term(s). For an even number of observations, remember to average the two middle terms. Show all algebraic steps clearly.

Question 9. The mean of 35 observations is 20, out of which mean of first 18 observations is 15 and mean of last 18 observations is 25. Find the 18th observation.
Solution: \[ \text{Mean} = \frac{\text{The sum of all observations}}{\text{Total number of observations}} \]
\( \therefore \) The sum of all observations = Mean \( \times \) Total number of observations The mean of 35 observations is 20
\( \therefore \) Sum of 35 observations = 20 \( \times \) 35 = 700 ,..(i) The mean of first 18 observations is 15 Sum of first 18 observations = 15 \( \times \) 18 = 270 ...(ii) The mean of last 18 observations is 25 Sum of last 18 observations = 25 \( \times \) 18 = 450 ...(iii)
\( \therefore \) 18th observation = (Sum of first 18 observations + Sum of last 18 observations) - (Sum of 35 observations) = (270 + 450) - (700) ... [From (i), (ii) and (iii)] = 720 - 700 = 20 The 18th observation is 20.
In simple words: First, calculate the total sum of all 35 observations, then the sum of the first 18 and the last 18. The 18th observation is found by adding the sums of the first and last 18 observations and subtracting the total sum of all 35 observations, since the 18th observation is counted twice (once in the first 18 and once in the last 18).

🎯 Exam Tip: This question tests understanding of overlapping sums. The key is to recognize that the middle observation is counted in both partial sums. Ensure your arithmetic for the sums is correct before the final subtraction.

Question 10. The mean of 5 observations is 50. One of the observations was removed from the data, hence the mean became 45. Find the observation which was removed.
Solution: \[ \text{Mean} = \frac{\text{The sum of all observations}}{\text{Total number of observations}} \]
\( \therefore \) The sum of all observations = Mean \( \times \) Total number of observations The mean of 5 observations is 50 Sum of 5 observations = 50 \( \times \) 5 = 250 ...(i) One observation was removed and mean of remaining data is 45. Total number of observations after removing one observation = 5 - 1 = 4 Now, mean of 4 observations is 45.
\( \therefore \) Sum of 4 observations = 45 \( \times \) 4 = 180 ...(ii)
\( \therefore \) Observation which was removed = Sum of 5 observations – Sum of 4 observations = 250 – 180 ... [From (i) and (ii)] = 70
\( \therefore \) The observation which was removed is 70.
In simple words: Calculate the total sum of the initial 5 observations. Then, calculate the total sum of the remaining 4 observations. The difference between these two sums will be the value of the observation that was removed.

🎯 Exam Tip: This is a common mean-based problem. Always start by calculating the total sum of observations for both initial and modified datasets. The difference directly gives the missing or removed value.

Question 11. There are 40 students in a class, out of them 15 are boys. The mean of marks obtained by boys is 33 and that for girls is 35. Find out the mean of all students in the class.
Solution: Total number of students = 40 Number of boys = 15
\( \therefore \) Number of girls = 40 – 15 = 25 The mean of marks obtained by 15 boys is 33 Here, sum of the marks obtained by boys = 33 \( \times \) 15 = 495 ...(i) The mean of marks obtained by 25 girls is 35 Sum of the marks obtained by girls = 35 \( \times \) 25 = 875 ...(ii) Sum of the marks obtained by boys and girls = 495 + 875 ... [From (i) and (ii)] = 1370
\( \therefore \) Mean of all the students \[ = \frac{\text{Sum of the marks of all the students}}{\text{Total number of students}} \] \[ = \frac{1370}{40} \] = 34.25
\( \therefore \) The mean of all the students in the class is 34.25.
In simple words: First, find the total marks for boys and the total marks for girls by multiplying their respective mean marks by the number of students in each group. Then, add these two total sums to get the combined total marks and divide by the total number of students in the class to find the overall mean.

🎯 Exam Tip: This is a weighted mean problem. Remember to calculate the total sum for each subgroup first (mean \( \times \) count) before combining them. The overall mean is the total sum divided by the total number of items.

Question 12. The weights of 10 students (in kg) are given below: 40, 35, 42, 43, 37, 35, 37, 37, 42, 37. Find the mode of the data.
Solution: Given data in ascending order: 35, 35, 37, 37, 37, 37, 40, 42, 42, 43
\( \therefore \) The observation repeated maximum number of times = 37
\( \therefore \) Mode of the given data is 37 kg
In simple words: To find the mode, arrange the weights in order and then identify the weight that occurs most frequently in the list.

🎯 Exam Tip: Always order the data to easily count frequencies and identify the mode. Ensure you list all modes if the data is bimodal or multimodal.

Question 13. In the following table, the information is given about the number of families and the siblings in the families less than 14 years of age. Find the mode of the data.

Solution: Here, the maximum frequency is 25. Since, Mode = observations having maximum frequency
\( \therefore \) The mode of the given data is 2.
In simple words: The mode is the value with the highest frequency. In this table, the number of siblings '2' has the highest frequency (25 families), making it the mode.

🎯 Exam Tip: For frequency distribution tables, the mode is simply the data value (or class interval) corresponding to the highest frequency. Look for the largest number in the frequency row.

Question 14. Find the mode of the following data.

Solution: Here, the maximum frequency is 9. Since, Mode = observations having maximum frequency But, this is the frequency of two observations.
\( \therefore \) Mode = 35 and 37
In simple words: The mode is the value that appears most often. In this case, both 35 marks and 37 marks have the highest frequency of 9 students, so both are considered modes.

🎯 Exam Tip: It is possible for a dataset to have more than one mode (bimodal or multimodal). Always check if multiple values share the highest frequency. Clearly list all modes if they exist.

Maharashtra Board Class 9 Maths Chapter 7 Statistics Practice Set 7.5 Intext Questions And Activities

Question 1. The first unit test of 40 marks was conducted for a class of 35 students. The marks obtained by the students were as follows. Find the mean of the marks. 40, 35, 30, 25, 23, 20, 14, 15, 16, 20, 17, 37, 37, 20, 36, 16, 30, 25, 25, 36, 37, 39, 39, 40, 15, 16, 17, 30, 16, 39, 40, 35, 37, 23, 16. (Textbook pg, no. 123)
Solution: Here, we can add all observations, but it will be a tedious job. It is easy to make frequency distribution table to calculate mean.

\[ \text{Mean} (\bar{x}) = \frac{\Sigma f_i x_i}{N} \] \[ = \frac{956}{35} \] = 27.31 marks (approximately)
\( \therefore \) The mean of the mark is 27.31.
In simple words: To find the mean marks for a large number of students, create a frequency distribution table. Multiply each mark (x_i) by the number of students who scored it (f_i), sum these products ( \( \Sigma \)f_ix_i ), and then divide by the total number of students (N).

🎯 Exam Tip: For problems with many observations, organizing data into a frequency distribution table is essential for accuracy and efficiency in calculating the mean. Double-check your products and sums.