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).
It’s 222829299292929292 ☺️
Answer:
-2 2/3
Step-by-step explanation:
I am pretty sure
2/8=3/12
The reason for this is because the other answers are not proportionate between both numerator and denominator. All of them except for 2/8=3/12 are multiplied by different numbers for both the numerator and denominator.