Answer:
0.1507 or 15.07%.
Step-by-step explanation:
We have been given that the manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .016 millimeters. To be acceptable the diameter needs to be between 21.97 and 22.03 millimeters.
First of all, we will find z-scores for data points using z-score formula.
, where,
z = z-score,
x = Sample score,
= Mean,
= Standard deviation.



Let us find z-score of data point 22.03.



Using probability formula
, we will get:

Therefore, the probability that a randomly selected ball bearing will be acceptable is 0.1507 or 15.07%.
Answer:

Step-by-step explanation:
Given

Required
Solve
Using sine rule, we have:

This gives:

So, we have:

In radical forms, we have:


Take LCM

Rewrite as:

Hence:

Ask every fifth person that exits the school
(systematic random)
30/54 =x/1,140
30*1140= 34200
34200/54 = 633.33
ANSWER:
Estimated- 633