Answer:
a.

b-Check illustration below
c.(-0.0517,0.0177
Step-by-step explanation:
a.let
denote processes 1 & 2.
For
: T1=10,n1=200
For
:T2=20,n2=300
Therefore

b. To test for hypothesis:-
i.

ii.For a two sample Proportion test

iii. for
(0.5 alpha IS 0.025),
reject
if
iv. Do not reject
. The noncomforting proportions are not significantly different as calculated below:

z=-0.78
c.
for the p1-p2 is given as:

=(-0.0517,+0.0177)
*CI contains o, which implies that proportions are NOT significantly different.
The maximum occurs when the derivative of the function is equal to zero.

Then evaluate the function for that time to find the maximum population.

Depending on the teacher, the "correct" answer will either be the exact decimal answer or the greatest integer of that value since you cannot have part of a rabbit.
(a) 0.059582148 probability of exactly 3 defective out of 20
(b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.
(a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So
0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)
= 0.05^3 * 0.95^17 * 20! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)
= 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)
= 0.05^3 * 0.95^17 * 20*19*3
= 0.000125* 0.418120335 * 1140
= 0.059582148
(b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.
The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)
= 0.05*0.95^3*24/(1!3!)
= 0.05*0.857375*24/6
= 0.171475
So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
Here are all of your answers:
1) 8,640,000,000
2) 864,000,000,000,000
3) 864,000,000,000
4) 86,400,000,000,000
I hope this helps! If it does, please rate as Brainliest :D