Maharashtra State BoardHSC Commerce 11th
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The distribution of a sample of students appearing for a C.A. examination is: Marks 0 – 100 100 – 200 200 – 300 300 –400 400 – 500 500 –600 No. of students 130 150 190 220 280 130 Help C.A. - Mathematics and Statistics

Sum

The distribution of a sample of students appearing for a C.A. examination is:

Marks 0 – 100 100 – 200 200 – 300 300 – 400 400 – 500 500 – 600
No. of students 130 150 190 220 280 130

Help C.A. institute to decide cut-off marks for qualifying an examination when 3% of students pass the examination.

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Solution

To decide cut off marks for qualifying an examination when 3% of students pass, we have to find P97.
We construct the less than cumulative frequency table as given below:

Marks No. of students
(f)
Less than cumulative frequency
(c.f.)
0 – 100 130 130
100 – 200 150 280
200 –  300 190 470
300 –  400 220 690
400 –  500 280 970
500 –  600 130 1100 ← P97
Total 1100  

Here, N = 110
P97 class = class containing `((97"N")/100)^"th"` observation

∴ `(97"N")/100=(97xx1100)/100` = 1067
Cumulative frequency which is just greater than (or equal) to 1067 is 1100.
∴ P97 lies in the class 500 –  600
∴ L = 500, h = 100, f = 130, c.f. = 970

∴ P97 = `"L" + "h"/"f" ((97"N")/100 - "c.f.")`

= `500 + 100/130 (1067 - 970)`

= `500 + 10/13 (97)`

= 500 + 74.62
= 574.62 ≈ 575
∴ The cut-off marks for qualifying an examination is 575.

Concept: Relations Among Quartiles, Deciles and Percentiles
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APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board
Chapter 1 Partition Values
Miscellaneous Exercise 1 | Q 7 | Page 21
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