Answer: 95% confidence interval = 20,000 ± 2.12
( 19228.736 , 20771.263 ) OR ( 19229 , 20771 )
Step-by-step explanation:
Given :
Sample size(n) = 17
Sample mean = 20000
Sample standard deviation = 1,500
5% confidence
∴ = 0.025
Degree of freedom () = n-1 = 16
∵ Critical value at ( 0.025 , 16 ) = 2.12
∴ 95% confidence interval = mean ±
Critical value at 95% confidence interval = 20,000 ± 2.12
( 19228.736 , 20771.263 ) OR ( 19229 , 20771 )
Answer:
a) It can be used because np and n(1-p) are both greater than 5.
Step-by-step explanation:
Binomial distribution and approximation to the normal:
The binomial distribution has two parameters:
n, which is the number of trials.
p, which is the probability of a success on a single trial.
If np and n(1-p) are both greater than 5, the normal approximation to the binomial can appropriately be used.
In this question:
So, lets verify the conditions:
np = 201*0.45 = 90.45 > 5
n(1-p) = 201*(1-0.45) = 201*0.55 = 110.55 > 5
Since both np and n(1-p) are greater than 5, the approximation can be used.