a. Let 
be a random variable representing the weight of a ball bearing selected at random. We're told that 
, so
where 
. This probability is approximately
b. Let 
be a random variable representing the weight of the 
-th ball that is selected, and let 
be the mean of these 4 weights,
The sum of normally distributed random variables is a random variable that also follows a normal distribution,
so that
Then
c. Same as (b).
For a known standard deviation, the confidence interval for sample size = n is
where
x = average
n = sample size
= stad. deviation
z = contant that reflects confidence interval
Let a = x
Let b =
From the given information,
a - b = 0.432 (1)
a + b = 0.52 (2)
Add (1) and (2): 2a = 0.952 => a = 0.476
Subtract (2) from (1): -2b = -0.088 => b = 0.044
Therefore, the confidence interval may be written as
(0.476 - 0.044, 0.476 + 0.044), or as
(0.476
0.044)
Zero, because ANYTHING to the power of zero equals 1.
The unknown number is X;
x/6+2=9;
X/6=7;
X=42;
The answer is 42
