. In a class test, marks obtained by 120 students are given in the following frequency
distribution. If it is given that mean is 59, find the missing frequencies x and y and
mode.
Answers
Answer:
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Step-by-step explanation:
x = 29 and y = 1 are the frequencies
Step-by-step explanation:
Class Mid-point Frequency
0 - 10 5 1
10 - 20 15 3
20 - 30 25 7
30 - 40 35 10
40 - 50 45 15
50 - 60 55 x
60 - 70 65 9
70 - 80 75 27
80 - 90 85 18
90 - 100 95 y
It is given that the class has 120 students and is also equal to the sum of frequencies.
Therefore 1 + 3 + 7 + 10 + 15 + x + 9 + 27 + 18 + y = 120
Therefore x + y =30 .... (i)
Also we know that the mean of the data = 59.
Therefore (5×1) + (15×3) + (25×7) + (35×10) + (45×15) + (55x) + (65×9) + (75×27) + (85×18) + (95y) = 120×59
⇒ 5 + 45 + 175 + 350 + 675 + 55x + 585 + 2025 + 1530 + 95y = 7080
⇒ 55x + 95y = 1690
∴ 11x + 19y = 338 .... (ii)
Multiplying equation (i) by 11 we get
11x + 11y = 330 .... (iii)
Subtracting (iii) from (ii) we get 8y = 8. Therefore y = 1 .... (iv)
Using the value of y from (iv) in (i) we get x + 1 = 30 ∴ x = 29 .... (v)