Answer: (751.05, 766.95)
Step-by-step explanation:
We know that the confidence interval for population mean is given by :-
,
where 
=population standard deviation.
= sample mean
n= sample size
z* = Two-tailed critical z-value.
Given : 
n= 42
We know that from z-table , the two-tailed critical value for 99% confidence interval : z* =2.576
Now, the 99% confidence interval around the true population mean viscosity :-
![759\pm (2.5760)\dfrac{20}{\sqrt{42}}\\\\=759\pm (2.5760)(3.086067)\\\\=759\pm7.9497=(759-7.9497,\ 759+7.9497)\]\\=(751.0503,\ 766.9497)\approx(751.05,\ 766.95)](https://tex.z-dn.net/?f=759%5Cpm%20%282.5760%29%5Cdfrac%7B20%7D%7B%5Csqrt%7B42%7D%7D%5C%5C%5C%5C%3D759%5Cpm%20%282.5760%29%283.086067%29%5C%5C%5C%5C%3D759%5Cpm7.9497%3D%28759-7.9497%2C%5C%20759%2B7.9497%29%5C%5D%5C%5C%3D%28751.0503%2C%5C%20766.9497%29%5Capprox%28751.05%2C%5C%20766.95%29)
∴ A 99% confidence interval around the true population mean viscosity : (751.05, 766.95)
The difference between -8 F and 14 F is 22 F.
Answer:
P(A) = 0.1620 , P(A|B) = 0.0707
Step-by-step explanation:
Total people = 10 + 15 + 20 = 45 , Sophomores = 10, Juniors = 15 , Seniors = 20
P(A) , exactly 3 juniors selected = (15 c 3) (30 c 2) / (45 c 5) = (455 x 435) / 1221759 = 0.1620
P (B) , exactly 2 seniors selected = (20 c 2) (25 c 3) / (45 c 5) = (190 x 2300)/ 1221759 = 0.3576
P(A|B), exactly 3 juniors & exactly 2 seniors selected = P (A∩B) / P (B)
(15 c 3) (20 c 2) / (45 c 5) = (455 x 190) / 1221759 = 0.0707
Answer is Indians only answers super simple questions
