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Q.

Calculate Mean and Median from the following frequency distribution :

| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |

|---|---|---|---|---|---|---|

| No. of students | 5 | 12 | 15 | 25 | 8 | 3 | 2 |

OR

Calculate Mode from the following frequency distribution :

| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |

|---|---|---|---|---|---|

| No. of students | 12 | 18 | 27 | 20 | 17 | 6 |

Manipur CohsemCOHSEM Manipur Higher Secondary 1st Year (Commerce) 2024Subjective· 8mImportance★★★★★est
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Mean ≈ 30.14 and Median = 31.2 for the given frequency distribution. [OR: Mode = 25.625 for the alternative distribution.]

Part A — Mean and Median:

MarksNo. of students (f)Mid-point (x)fxCumulative frequency (cf)
0–1055255
10–20121518017
20–30152537532
30–40253587557
40–5084536065
50–6035516568
60–7026513070
TotalN = 70Σfx = 2110

Mean (Direct method): Mean = Σfx / N = 2110 / 70 = 30.14 (approx.)

Median: N/2 = 70/2 = 35. The cumulative frequency first reaching/exceeding 35 is 57, in the class 30–40 — so 30–40 is the Median class.

L (lower limit of median class) = 30; cf (cumulative frequency of the class before median class) = 32; f (frequency of median class) = 25; h (class width) = 10

Median = L + [(N/2 − cf)/f] × h = 30 + [(35 − 32)/25] × 10 = 30 + (3/25) × 10 = 30 + 1.2 = 31.2

OR — Part B: Mode (for the alternative frequency distribution):

| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 | …

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