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Detailed Chapter 15 Mean Median and Frequency Polygon ICSE Solutions for Class 9 Mathematics
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Class 9 Mathematics Chapter 15 Mean Median and Frequency Polygon ICSE Solutions PDF
Question 1. Fill in the blanks:
(i) The difference between the maximum and the minimum observations in a set of data is called the _______ of the data.
(ii) The number of observations in a particular class interval is called the _______ of the class interval.
Answer:
(i) The difference between the maximum and the minimum observations in a set of data is called the range of the data.
(ii) The number of observations in a particular class interval is called the frequency of the class interval.
In simple words: The range tells us how spread out the data is from the smallest to the largest value. Frequency tells us how many times a specific type of observation appears in a group.
🎯 Exam Tip: Remember that 'range' is simply the difference between the highest and lowest values, while 'frequency' counts how often something appears within a class.
Question 2. Fill in the blanks :
(i) Lower limit of the class interval 26 - 33 is ..........
(ii) Upper limit of the class interval 21 - 25 is ..........
(iii) The class mark of the class interval 20 - 29 is ..........
(iv) The class mark of the class interval 9.5 - 19.5 is ..........
(v) If the class-marks of a distribution are 32, 39, 46, 53; then the width of each class interval is ..........
(vi) The mid-point of a class interval is called its ..........
(vii) If the class marks of a distribution are 28, 34, 40, 46, 52; then the class intervals are of length ..........
(viii) If the class marks in a frequency distribution are 19.5, 26.5, 33.5, 40.5, 47.5, 54.5, 61.5, then the class size of the distribution is ..........
Answer:
(i) Lower limit of the class interval 26 - 33 is 26.
(ii) Upper limit of the class interval 21 - 25 is 25.
(iii) The class mark of the class interval 20 - 29 is \( \frac{20+29}{2} = \frac{49}{2} = 24.5 \).
(iv) The class mark of the class interval 9.5 - 19.5 is \( \frac{9.5+19.5}{2} = \frac{29}{2} = 14.5 \).
(v) If the class-marks of a distribution are 32, 39, 46, 53; then the width of each class interval is C.I. \( = 39 - 32 = 7 \). The class interval width is the difference between two consecutive class marks.
(vi) The mid-point of a class interval is called its class mark.
(vii) If the class marks of a distribution are 28, 34, 40, 46, 52; then the class intervals are of length C.I. \( = 34 - 28 = 6 \).
(viii) If the class marks in a frequency distribution are 19.5, 26.5, 33.5, 40.5, 47.5, 54.5, 61.5, then the class size of the distribution is class size \( = 26.5 - 19.5 = 7 \).
In simple words: The lower and upper limits define a class. The class mark is the middle point of a class. The width of a class interval or class size is the difference between two consecutive class marks or limits.
🎯 Exam Tip: Remember that class mark is the average of the upper and lower limits, and class width can be found by subtracting any two consecutive class marks or lower limits.
Question 3. Fill in the blanks :
(i) The range of the set of data 5, 8, 15, 21, 7, 10 is ..........
(ii) The range of the set of data 15, 13, 14, 17, 19, 16, 14, 15 is ..........
(iii) The class mark of the class interval 10 - 19 is ..........
(iv) The class mark of the class interval 9.5 - 16.5 is ..........
Answer:
(i) The range of the set of data 5, 8, 15, 21, 7, 10 is 21 - 5 = 16. (The numbers listed in the source for solution were 10-5=5, which is incorrect for the given data set.)
(ii) The range of the set of data 15, 13, 14, 17, 19, 16, 14, 15 is 19 - 13 = 6.
(iii) The class mark of the class interval 10 - 19 is \( \frac{10+19}{2} = \frac{29}{2} = 14.5 \).
(iv) The class mark of the class interval 9.5 - 16.5 is \( \frac{9.5+16.5}{2} = \frac{26}{2} = 13 \).
In simple words: The range of a data set is simply the difference between its largest and smallest value. The class mark is found by adding the two limits of a class and dividing by two.
🎯 Exam Tip: Always identify the maximum and minimum values correctly from the given data set to calculate the range. For class marks, correctly sum and divide the limits.
Question 4. Which of the following variables are continuous and which are discrete :
(a) size of shoes,
(b) number of pages in a book,
(c) distance travelled by a train,
(d) time,
(e) daily temperature,
(f) intelligence quotient,
(g) number of goals scored in a hockey-match,
(h) height,
(i) weight,
(j) number of rooms in apartments,
(k) number of wage earners in a factory,
(l) number of car accidents.
Answer: A variable is continuous if it can take any value between two given values, like decimals or fractions. A variable is discrete if it can only take specific, separate values, usually whole numbers, like counts.
(a) size of shoes - discrete (shoe sizes are usually in fixed steps like 7, 7.5, 8)
(b) number of pages in a book - discrete (you can't have half a page)
(c) distance travelled by a train - continuous (distance can be any value, like 10.5 km)
(d) time - continuous (time can be measured very precisely, like 2.34 seconds)
(e) daily temperature - continuous (temperature can take any value, like 25.7 degrees Celsius)
(f) intelligence quotient - continuous (IQ scores can vary across a continuous scale)
(g) number of goals scored in a hockey-match - discrete (you can only score whole goals, not half a goal)
(h) height - continuous (height can be any value, like 165.2 cm)
(i) weight - continuous (weight can be any value, like 50.8 kg)
(j) number of rooms in apartments - discrete (you count rooms in whole numbers)
(k) number of wage earners in a factory - discrete (you count people in whole numbers)
(l) number of car accidents - discrete (you count accidents in whole numbers)
In simple words: If you can count it in whole numbers, it's discrete. If you can measure it with decimals and it can be any value in between, it's continuous.
🎯 Exam Tip: Think if the variable can be 'counted' (discrete) or 'measured' (continuous) to classify it correctly. Values like 'number of items' are almost always discrete.
Question 5. The class marks of a distribution are 2.03, 2.23, 2.43, 2.63, 2.83, 3.03 and 3.23 Determine the class size and the class boundaries.
Answer: The class marks given are 2.03, 2.23, 2.43, 2.63, 2.83, 3.03, and 3.23.
First, we find the class size. The class size is the difference between any two consecutive class marks.
Class size \( = 2.23 - 2.03 = 0.20 \). This is a consistent difference for all pairs.
Now, we find the class boundaries. The lower class boundary is found by subtracting half the class size from the class mark, and the upper class boundary is found by adding half the class size to the class mark.
Half the class size \( = \frac{0.20}{2} = 0.10 \).
For the first class mark (2.03):
Lower boundary \( = 2.03 - 0.10 = 1.93 \)
Upper boundary \( = 2.03 + 0.10 = 2.13 \)
So, the first class interval is 1.93 - 2.13.
Similarly, for the other class marks, the boundaries will be:
For 2.23: \( 2.23 - 0.10 = 2.13 \) and \( 2.23 + 0.10 = 2.33 \)
\( \implies \) Class interval: 2.13 - 2.33
For 2.43: \( 2.43 - 0.10 = 2.33 \) and \( 2.43 + 0.10 = 2.53 \)
\( \implies \) Class interval: 2.33 - 2.53
For 2.63: \( 2.63 - 0.10 = 2.53 \) and \( 2.63 + 0.10 = 2.73 \)
\( \implies \) Class interval: 2.53 - 2.73
For 2.83: \( 2.83 - 0.10 = 2.73 \) and \( 2.83 + 0.10 = 2.93 \)
\( \implies \) Class interval: 2.73 - 2.93
For 3.03: \( 3.03 - 0.10 = 2.93 \) and \( 3.03 + 0.10 = 3.13 \)
\( \implies \) Class interval: 2.93 - 3.13
For 3.23: \( 3.23 - 0.10 = 3.13 \) and \( 3.23 + 0.10 = 3.33 \)
\( \implies \) Class interval: 3.13 - 3.33
The class boundaries are 1.93 - 2.13, 2.13 - 2.33, 2.33 - 2.53, 2.53 - 2.73, 2.73 - 2.93, 2.93 - 3.13, and 3.13 - 3.33.
In simple words: We first find how wide each class is by looking at the difference between the middle points. Then, we add and subtract half of this width from each middle point to get the start and end values for each class.
🎯 Exam Tip: Remember that the class size is constant for a given set of class marks. Class boundaries are always found by adding and subtracting half of the class size from the class mark.
Question 6. The following are the monthly rents (in rupees) of 30 shops :
42, 49, 37, 82, 37, 75, 62, 54, 79, 84, 75, 63, 44, 74, 44, 36, 69, 54, 48, 74, 39, 48, 45,61,71,47,38, 80,51,31.
Using the class intervals of equal width in which one class interval being 40-50 (excluding 50). construct a frequency table for the above data.
Answer: The given data shows monthly rents for 30 shops.
We first find the highest and lowest rent values.
Highest rent \( = 84 \)
Lowest rent \( = 31 \)
Range \( = 84 - 31 = 53 \).
The problem states that one class interval is 40-50 (excluding 50), which means \( 40 \le \text{rent} < 50 \). This interval has a width of 10. We will use this width for all classes.
Starting from the lowest rent (31), we create class intervals of width 10.
The frequency table is as follows:
| Class interval | Tally marks | Frequency |
|---|---|---|
| 30 - 40 | \( \cancel{||||} \text{ } | \) | 6 |
| 40 - 50 | \( \cancel{||||} \text{ } \cancel{||||} \text{ } ||| \) | 8 |
| 50 - 60 | \( ||| \) | 3 |
| 60 - 70 | \( |||| \) | 4 |
| 70 - 80 | \( \cancel{||||} \text{ } | \) | 6 |
| 80 - 90 | \( ||| \) | 3 |
| Total | 30 |
In simple words: We sort the rents into groups of 10. For example, the 30-40 group includes all rents from 30 up to, but not including, 40. Then, we count how many rents fall into each group and mark it with tally marks.
🎯 Exam Tip: When constructing frequency tables, always ensure class intervals are mutually exclusive (no overlap) and exhaustive (cover all data points). Carefully count each data point into the correct interval.
Question 7. Construct a frequency table for the following set of data of weights (in gms) of 30 oranges using equal class intervals one of them being 50-60 (60 not included).
45, 55, 30, 85, 75, 85, 40, 60, 65, 40, 60, 75, 70, 60, 70, 85, 85, 80, 35, 45, 40, 50, 60, 65, 55, 45, 30, 80, 85, 75.
Answer: The given data shows the weights of 30 oranges in grams.
First, we find the greatest and lowest weights.
Greatest weight \( = 85 \) gms
Lowest weight \( = 30 \) gms
The problem states that one class interval is 50-60 (60 not included), meaning \( 50 \le \text{weight} < 60 \). This interval has a width of 10. We will use this width for all classes.
Starting from the lowest weight (30), we create class intervals of width 10.
The frequency table is as follows:
| Class interval (weight in gms) | Tally marks | Frequency |
|---|---|---|
| 30 - 40 | \( ||| \) | 3 |
| 40 - 50 | \( \cancel{||||} \text{ } | \) | 6 |
| 50 - 60 | \( ||| \) | 3 |
| 60 - 70 | \( \cancel{||||} \text{ } | \) | 6 |
| 70 - 80 | \( \cancel{||||} \) | 5 |
| 80 - 90 | \( \cancel{||||} \text{ } || \) | 7 |
| Total | 30 |
In simple words: We organize the weights of the oranges into groups of 10 grams, starting from 30 grams. We then count how many oranges fall into each group and use tally marks to show this count. Each class includes the lower limit but not the upper limit.
🎯 Exam Tip: When given an example class interval (like 50-60, 60 not included), use it to determine the class width and the type of interval (inclusive lower, exclusive upper) for all other classes.
Question 8. Given below are the Systolic Blood Pressure readings (mm Hg.) made on a person for 30 days. Construct a frequency distribution for the data using class intervals of equal size, one of them being 150-160 (160 not included).
165, 159, 190, 150, 171, 182, 165, 141, 135, 155,192,152,155,145, 145,160, 175, 179, 178, 120,125, 128, 152, 135, 142, 128, 140, 130, 130, 160.
Answer: The data shows systolic blood pressure readings for 30 days.
First, we identify the highest and lowest blood pressure readings.
Highest BP \( = 192 \)
Lowest BP \( = 120 \)
Range \( = 192 - 120 = 72 \).
The problem states that one class interval is 150-160 (160 not included), meaning \( 150 \le \text{BP} < 160 \). This interval has a width of 10. We will use this width for all classes.
Starting from the lowest BP (120), we create class intervals of width 10.
The frequency table is as follows:
| Class interval (BP in mm Hg) | Tally marks | Frequency (no. of days) |
|---|---|---|
| 120 - 130 | \( |||| \) | 4 |
| 130 - 140 | \( |||| \) | 4 |
| 140 - 150 | \( \cancel{||||} \) | 5 |
| 150 - 160 | \( \cancel{||||} \text{ } | \) | 6 |
| 160 - 170 | \( |||| \) | 4 |
| 170 - 180 | \( |||| \) | 4 |
| 180 - 190 | \( | \) | 1 |
| 190 - 200 | \( || \) | 2 |
| Total | 30 |
In simple words: We group the blood pressure readings into intervals of 10 units, like from 120 to just below 130. Then we count how many readings fall into each group to create the frequency table. This helps us see how often different BP levels occurred.
🎯 Exam Tip: When data points fall exactly on a class boundary, remember to place them in the class where the lower limit is inclusive and the upper limit is exclusive (e.g., 160 goes into 160-170, not 150-160).
Question 9. Present the following data of the %marks of 60 students in the form of a frequency table with 10 classes of equal width, one class being 40-49.
41 17 83 63 54 92 60 58 70
06 67 82 30 44 57 49 34 73
54 63 36 52 32 75 60 38 09
79 28 30 42 93 43 80 03 32
57 67 24 64 63 11 35 82 10
38 00 41 60 32 72 53 92 88
62 55 60 33 40 57
Answer: The data lists the percentage marks of 60 students.
First, we find the highest and lowest percentage marks.
Highest percentage of marks \( = 93 \)
Lowest percentage of marks \( = 00 \)
Range \( = 93 - 00 = 93 \).
The problem asks for 10 classes of equal width, with one class being 40-49. This means the class interval 40-49 includes both 40 and 49 (inclusive limits). The width is \( 49 - 40 + 1 = 10 \). So, all classes will have a width of 10.
Starting from the lowest mark (00), we create class intervals of width 10.
The frequency table is as follows:
| Class intervals (% age marks) | Tally marks | Frequency (no. of students) |
|---|---|---|
| 0 - 9 | \( |||| \) | 4 |
| 10 - 19 | \( ||| \) | 3 |
| 20 - 29 | \( || \) | 2 |
| 30 - 39 | \( \cancel{||||} \text{ } \cancel{||||} \text{ } | \) | 11 |
| 40 - 49 | \( \cancel{||||} \text{ } || \) | 7 |
| 50 - 59 | \( \cancel{||||} \text{ } |||| \) | 9 |
| 60 - 69 | \( \cancel{||||} \text{ } \cancel{||||} \text{ } | \) | 11 |
| 70 - 79 | \( \cancel{||||} \) | 5 |
| 80 - 89 | \( \cancel{||||} \) | 5 |
| 90 - 99 | \( ||| \) | 3 |
| Total | 60 |
In simple words: We sort all the students' marks into 10 groups, each covering a range of 10 marks (like 0-9, 10-19, and so on). Then, we count how many students fall into each group and list the numbers in a table. This way, we can easily see how many students got marks in different ranges.
🎯 Exam Tip: When given class intervals like '40-49', it means both 40 and 49 are included (inclusive intervals). Make sure your tallying accurately reflects this and that your total frequency matches the total number of data points.
Question 10. Thirty 16 year old boys were tested to find their pulse rate. This following figures were obtained for the number of beats per minute. Using the class intervals 51-55, 56-60, etc., of equal width, prepare a frequency table.
55, 72, 70, 66, 74, 70, 74, 53, 57, 62, 71, 58, 68, 75, 79, 68, 63, 59, 54, 51, 61, 66, 78, 73, 59, 52, 66, 60, 72, 56.
Answer: The data lists the pulse rates of thirty 16-year-old boys.
First, we find the highest and lowest pulse rates.
Highest pulse rate \( = 79 \)
Lowest pulse rate \( = 51 \)
Range \( = 79 - 51 = 28 \).
The problem specifies class intervals like 51-55, 56-60, which are inclusive class intervals. The width of each class is \( 55 - 51 + 1 = 5 \). All classes will have a width of 5.
Starting from the lowest pulse rate (51), we create class intervals of width 5.
The frequency table is as follows:
| Pulse rate Per minute | Tally marks | No. of boys (f) |
|---|---|---|
| 51 - 55 | \( \cancel{||||} \) | 5 |
| 56 - 60 | \( \cancel{||||} \text{ } | \) | 6 |
| 61 - 65 | \( ||| \) | 3 |
| 66 - 70 | \( \cancel{||||} \text{ } || \) | 7 |
| 71 - 75 | \( \cancel{||||} \text{ } || \) | 7 |
| 76 - 80 | \( || \) | 2 |
| Total | 30 |
In simple words: We group the pulse rates into small ranges, for example, from 51 to 55 beats per minute. Each group has a width of 5. Then, we count how many boys have a pulse rate in each group to create the frequency table. This helps us see how common different pulse rates are.
🎯 Exam Tip: When class intervals are inclusive (e.g., 51-55), ensure you correctly calculate the class width by including both the lower and upper limits in your count. Double-check that all data points are assigned to a class.
ICSE Solutions Class 9 Mathematics Chapter 15 Mean Median and Frequency Polygon
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