Question 1 :
The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequencies f1 and f2.
Solution :
x 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120 |
midpoint (x) 10 30 50 70 90 110 |
f 5 f1 10 f2 7 8 |
fx 50 30f1 500 70f2 630 880 |
Mean = 62.8
Σf = 5 + f1 + 10 + f2 + 7 + 8
Σf = 30 + f1 + f2
50 = 30 + f1 + f2
f1 + f2 = 20 -----(1)
Σfx = 50 + 30f1 + 500 + 70f2 + 630 + 880
Σfx = 2060 + 30f1 + 70f2
Σfx/Σf = (2060 + 30f1 + 70f2)/50 = 62.8
2060 + 30f1 + 70f2 = 62.8(50)
2060 + 30f1 + 70f2 = 3140
30f1 + 70f2 = 3140 - 2060
30f1 + 70f2 = 1080
3f1 + 7f2 = 108 -----(2)
(2) - 3(1) :
-4f2 = -48
f2 = 12
Substitute f2 = 12 in (1).
f1 + 12 = 20
f1 = 8
So, the missing frequencies are 8 and 12.
Question 2 :
The diameter of circles (in mm) drawn in a design are given below
Calculate the standard deviation.
Solution :
32.5 - 36.5 36.5 - 41.5 40.5 - 44.5 44.5 - 48.5 48.5 - 52.5 |
mid 34.5 39 42.5 46.5 50.5 |
f 15 17 21 22 25 |
d = x - A -8, 64 -3.5, 12.25 0, 0 4, 16 8, 64 |
fd -120 -59.5 0 88 200 |
Σfd = -120 - 59.5 + 0 + 88 + 200
Σfd = 108.5
Σfd/Σf = 108.5/100 = 1.085
Σfd2 = 15(64) + 17(12.25) + 21(0) + 22(16) + 25(64)
= 960 + 208.25 + 0 + 352 + 1600
= 3120.25
Σfd2/Σf = 3120.25/100
= 31.2025
N = Σf = 100
Standard deviation = √31.20 - (1.085)2
= √31.20 - 1.18
= 5.47 (Approximately)
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Aug 31, 24 06:59 PM
Aug 31, 24 06:48 PM
Aug 31, 24 07:14 AM