Afirm is accustomed to training operators who do certain tasks on a production line. those operators who attend the training course are known to be able to meet their production quotas 90% of the time. new operators who do not take the training course only meet their quotas 65% of the time. fifty percent of new operators attend the course. given that a new operator meets her production quota, what is the probability that she attended the program?

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.

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;

answer: b

step-by-step explanation: