Using normal distribution

critical probability

critical value for the critical probability is CV = 1.96
standard error

margin of error = CV*SE = 1.96 * 0.5 = 0.98
95% confidence interval is between μ + 0.98 and <span>μ - 0.98, which is (19.62, 21.58)</span>
Answer:
Probability = 0.58
Step-by-step explanation:
This problem is solve by using Baye's Probability.
Let P(A) = Probability that operator attended training course = 50% = 0.5
P(B) = Probability that operator not attended training course = 50% = 0.5
Also P(Q) = Probability that operator meet their production quotas
Then, P(Q|A) = 90% = 0.9
P(Q|B) = 65% = 0.65
P(A|Q) = ?
Then by Baye's Theorem,

⇒ 
⇒ P(A|Q) = 0.58
which is required probability.
First, convert the percentage into a decimal by removing the % sign and dividing by 100.
40% => 0.4
The word 'of' in mathematics can mean multiplication.
0.4 * 126 = 50.4
The value of the digit 5 in 75 is in the ones place, because there are seven tens, and five ones.