Answer:
First question: LCL = 522, UCL = 1000.5
Second question: A sample size no smaller than 418 is needed.
Step-by-step explanation:
First question:
Lower bound:
0.36 of 1450. So
0.36*1450 = 522
Upper bound:
0.69 of 1450. So
0.69*1450 = 1000.5
LCL = 522, UCL = 1000.5
Second question:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.

In which
z is the zscore that has a pvalue of
.
The margin of error is:

The project manager believes that p will turn out to be approximately 0.11.
This means that 
95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The project manager wants to estimate the proportion to within 0.03
This means that the sample size needed is given by n, and n is found when M = 0.03. So






Rounding up
A sample size no smaller than 418 is needed.
It is to be solved by reminder thorem
f(x)/(x-k) will have reminder f(k),
so, f(2) = 5*(2^4) + 8 *(2^3) +4* (2^2) -5(2) +67
=5*16 + 8*8 +4*4 -5*2 +67
=80 + 64 + 16 -10 +67
= 217
Answer:
1 6/7
Step-by-step explanation:
1 & 6/7
Stuck on the same thing, tried reviewing and reviewing but now I’ll just try finding the answer key