In: Statistics and Probability
Find the mean for the following frequency tables. (Round your answers to one decimal place.)
(a) |
|
(b) |
|
(c) |
|
Solution:-
Solution:
a)
Class (1) |
Frequency (f) (2) |
Mid value (x) (3) |
d=x-A/h=x-74.5/10 A=74.5,h=10 (4) |
f⋅d (5)=(2)×(4) |
49.5 - 59.5 | 2 | 54.5 | -2 | -4 |
59.5 - 69.5 | 3 | 64.5 | -1 | -3 |
69.5 - 79.5 | 6 | 74.5=A | 0 | 0 |
79.5 - 89.5 | 14 | 84.5 | 1 | 14 |
89.5 - 99.5 | 5 | 94.5 | 2 | 10 |
--- | --- | --- | --- | --- |
n=30 | ----- | ----- | ∑f⋅d=17 |
Mean
=A+∑fd/n⋅h
=74.5+17/30⋅10
=74.5+0.5667⋅10
=74.5+5.6667
=80.2
b)
Class (1) |
Frequency (f) (2) |
Mid value (x) (3) |
d=x-A/h=x-74.5/10 A=74.5,h=10 (4) |
f⋅d (5)=(2)×(4) |
49.5 - 59.5 | 50 | 54.5 | -2 | -100 |
59.5 - 69.5 | 30 | 64.5 | -1 | -30 |
69.5 - 79.5 | 15 | 74.5=A | 0 | 0 |
79.5 - 89.5 | 1 | 84.5 | 1 | 1 |
89.5 - 99.5 | 0 | 94.5 | 2 | 0 |
--- | --- | --- | --- | --- |
n=96 | ----- | ----- | ∑f⋅d=-129 |
Mean
=A+∑fd/n⋅h
=74.5+-129/96⋅10
=74.5+-1.3438⋅10
=74.5+-13.4375
=61.1
c)
Class (1) |
Frequency (f) (2) |
Mid value (x) (3) |
d=x-A/h=x-74.5/10 A=74.5,h=10 (4) |
f⋅d (5)=(2)×(4) |
49.5 - 59.5 | 14 | 54.5 | -2 | -28 |
59.5 - 69.5 | 32 | 64.5 | -1 | -32 |
69.5 - 79.5 | 15 | 74.5=A | 0 | 0 |
79.5 - 89.5 | 21 | 84.5 | 1 | 21 |
89.5 - 99.5 | 2 | 94.5 | 2 | 4 |
--- | --- | --- | --- | --- |
n=84 | ----- | ----- | ∑f⋅d=-35 |
Mean
=A+∑fd/n⋅h
=74.5+-35/84⋅10
=74.5+-0.4167⋅10
=74.5+-4.1667
=70.3