|
x |
P(x) |
|
1 |
0.25 |
|
2 |
0.30 |
|
3 |
0.45 |
(a) We have mean as
or
or
. The variance would be
or
or
. The standard deviation would be
or
or
.
(b) We have the new distribution as below.
| y | p(y) |
| 6 | 0.25 |
| 7 | 0.30 |
| 8 | 0.45 |
We have mean as
or
or
. The variance would be
or
or
. The standard deviation would be
or
or
.
(c) We have
,
and
. This means that the mean increased by 5 units, the variance and
standard deviation did not change. It complies with the usual rule
that
and
.
(d) The new distribution would be as below.
| z | p(z) |
| 5 | 0.25 |
| 10 | 0.30 |
| 15 | 0.45 |
We have
or
or
. The variance would be
or
or
. The standard deviation would be
or
or
.
(e) We have
,
and
. This means that the mean increased by 5 times, the variance
increased by 25 (= 5 squared) times, and standard deviation
increased by 5 times. It again complies with the usual rule that
and
.
Consider the following discrete probability distribution. x P(x) 1 0.25 2 0.30 3 0.45 Calculate the...
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A discrete random variable X follows the geometric distribution
with parameter p, written X ∼ Geom(p), if its distribution function
is
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