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प्रश्न
For one term, absentee record of students is given below. If mean is 15.5, then the missing frequencies x and y are.
Number of days | 0 - 5 | 5 - 10 | 10 - 15 | 15 - 20 | 20 - 25 | 25 - 30 | 30 - 35 | 35 - 40 | TOTAL |
Total Number of students | 15 | 16 | x | 8 | y | 8 | 6 | 4 | 70 |
विकल्प
x = 4 and y = 3
x = 7 and y = 7
x = 3 and y = 4
x = 7 and y = 6
MCQ
उत्तर
x = 7 and y = 6
Explanation:-
Construct a table as follows:
Class interval | Frequency (fi) | Midpoint (xi) | fixi |
0 - 5 | 15 | 2.5 | 37.5 |
5 - 10 | 16 | 7.5 | 120 |
10 - 15 | x | 12.5 | 12.5x |
15 - 20 | 8 | 17.5 | 140 |
20 - 25 | y | 22.5 | 22.5y |
25 - 30 | 8 | 27.5 | 220 |
30 - 35 | 6 | 32.5 | 195 |
35 - 40 | 4 | 37.5 | 150 |
Total | 70 | 12.5x + 22.5y + 862.5 |
Mean = `(12.5x + 22.5y + 862.5)/70`
`⇒ 15.5 = (12.5x +22.5y + 862.5)/70`
`⇒ 15.5 xx 70 = 12.5x +22.5y + 862.5`
⇒ 12.5x + 22.5y = 222.5
⇒ 125x + 225y = 2225
⇒ 5x + 9y = 89 .....(i)
Also,
x + y + 57 = 70
x + y = 13 ......(ii)
Multiplying equation (ii) by 5 and then subtracting from (i) as,
5x + 9y = 89
5x + 5y = 65
4y = 24
y = 6
Substituting the value of y in equation (ii), we get
x + y = 13
⇒ x + 6 = 13
⇒ x = 7
Hence x = 7 and y = 6
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