Statistics — Karnataka (SSLC) Class 9 Mathematics Solutions (Free)

Mean: add up every value in the data set and divide by how many values there are — the familiar "average." Median: sort the data from smallest to largest and pick out whichever value lands exactly in the middle. Mode: whichever value shows up more often in the data than any other value does. Given data: 16, 19, 16, 24, 27, 16, 31 (7 values) Mean: Sum = 16 + 19 + 16 + 24 + 27 + 16 + 31 = 149 Mean = 149 / 7 ≈ 21.29 Median: Arrange in order: 16, 16, 16, 19, 24, 27, 31 With 7 values (odd), the median is the middle (4th) value: 19 Mode: 16 appears three times — more than any other value. Mode = 16

For grouped data, the mean uses the midpoint (x) of each class interval, weighted by its frequency (f): Mean = Σ(f × x) / Σf | Class | Midpoint (x) | Frequency (f) | f × x | |-------|------|------|-------| | 0–10 | 5 | 6 | 30 | | 10–20 | 15 | 10 | 150 | | 20–30 | 25 | 15 | 375 | | 30–40 | 35 | 9 | 315 | | 40–50 | 45 | 4 | 180 | Σf = 6 + 10 + 15 + 9 + 4 = 44 Σ(f × x) = 30 + 150 + 375 + 315 + 180 = 1050 Mean = 1050 / 44 ≈ 23.86 The mean of the grouped data is approximately 23.86.

Sorting the 16 recorded marks into the given class intervals: 40–50: 41, 49, 49 → frequency 3 50–60: 52, 59, 52, 59, 52 → frequency 5 60–70: 61, 61, 69, 61, 69 → frequency 5 70–80: 71, 71, 71 → frequency 3 | Class Interval | Frequency | |---|---| | 40–50 | 3 | | 50–60 | 5 | | 60–70 | 5 | | 70–80 | 3 | | **Total** | **16** | A bar graph of this data would show four bars, with the two middle intervals (50–60 and 60–70) tallest at a height of 5 each, and the two outer intervals shorter at a height of 3 each — showing marks clustering toward the middle of the range.

Given data: 4, 8, 12, 16, 20 (already in ascending order, 5 values) Mean: Sum = 4 + 8 + 12 + 16 + 20 = 60 Mean = 60 / 5 = 12 Median: With 5 values (odd), the median is the middle (3rd) value: 12 Both the mean and median equal 12. Why: this data set increases by a constant amount (4) between consecutive values, making it a symmetric arithmetic sequence. For any such evenly-spaced sequence, the values above the middle term exceed it by exactly the same total amount that the values below the middle fall short of it — so the average always lands exactly on the middle value.

Given data: 68, 74, 79, 84, 79, 89, 84, 94, 79, 74 (10 values) Mean: Sum = 68+74+79+84+79+89+84+94+79+74 = 804 Mean = 804/10 = 80.4 Median: Arrange in order: 68, 74, 74, 79, 79, 79, 84, 84, 89, 94 With 10 values (even), median = average of the 5th and 6th values = (79 + 79)/2 = 79 Mode: 79 appears three times — the most frequent value. Mode = 79 Range: Range = Highest − Lowest = 94 − 68 = 26 Students scoring above the mean (80.4): 84, 89, 84, 94 — that's 4 students.

Sorting the 15 recorded ages into the given class intervals: 20–25: 23, 24, 24 → frequency 3 25–30: 26, 27, 25, 29, 26, 28, 27 → frequency 7 30–35: 30, 34, 35, 31, 30 → frequency 5 | Class Interval | Midpoint (x) | Frequency (f) | f × x | |---|---|---|---| | 20–25 | 22.5 | 3 | 67.5 | | 25–30 | 27.5 | 7 | 192.5 | | 30–35 | 32.5 | 5 | 162.5 | Σf = 3 + 7 + 5 = 15 ✓ Σ(f × x) = 67.5 + 192.5 + 162.5 = 422.5 Mean = 422.5 / 15 ≈ 28.17 The mean age is approximately 28.17 years.

For a data set with an ODD number of values (n), the median is the single middle value once the data is sorted — the value at position (n + 1)/2. With 11 values, the median is at position (11 + 1)/2 = 6th position after sorting. Arranging the data in ascending order: 19, 27, 38, 46, 52, 58, 63, 71, 80, 84, 95 Counting to the 6th value: 19 (1st), 27 (2nd), 38 (3rd), 46 (4th), 52 (5th), 58 (6th) The median is 58. Unlike the mean, the median depends only on which value falls in the middle position, not on how large or small the extreme values are — which is why an outlier like 95 in this data set has no effect on the median at all, even though it does pull the mean upward.