Answer:
(a) 0.0833 or 8.33%
(b) 0.40 or 40%
Step-by-step explanation:
Parts line one (n1) = 1,000 parts
Defects line one (d1) = 100 parts
Parts line two (n2) = 2,000 parts
Defects line two (d2) = 150 parts
Total number of parts (n) = 3,000 parts
a. Probability of a randomly selected part being defective:
The probability is 0.0833 or 8.33%
b. Probability of a part being produced by line one, given that it is defective:
The probability is 0.40 or 40%.
Answer:
For this case assuming that the random variable is X
And replacing n = 24 we got:
And we notate the distribution we got:
Step-by-step explanation:
Previous concepts
The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".
The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.
The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."
Solution to the problem
For this case assuming that the random variable is X
And replacing n = 24 we got:
And we notate the distribution we got: