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).
Answer:
mn
Step-by-step explanation:
I have no clue how to do that if u do no how to do other math can u help me I need alot of help
1024 is the answer to this problem.
500. But 1000 cannot go in 2, the result would be 0.002.