Answer:
(7.7981, 8.5419). The distribution of percentage elongation can be any distribution with finite mean and variance.
Step-by-step explanation:
We have a large sample size n = 56 research cotton samples. Besides, and
. A 95% large-sample CI for the true average percentage elongation is given by
, i.e.,
or equivalently (7.7981, 8.5419). The distribution of percentage elongation can be any distribution with finite mean and variance because we have a large sample.
Answer:
The expression is not completely factored.
Step-by-step explanation:
The expression of can be broken up into two binomials:
Together, you have this:
** There is alot of missing information to this question, but at least you know how to work this out.
I am joyous to assist you anytime.
Answer:
Step-by-step explanation:
For the disease having force of mortality µ.
we apply exponential distribution law
probability of dying within 20 years
p( x <20) = = .1
= .9
For the disease having force of mortality 2µ.
we apply exponential distribution law
probability of dying within 20 years
p( x <20) = 1 -
= 1 - ( .9)²
= 1 - .81
= .19
or 19%