Jessica wants to calculate her grade in her accounting class.she has four test grades (82, 65, 71, and 77) and a homework grade

Answer:

a) P(X=2)= 0.29

b) P(X<2)= 0.59

c) P(X≤2)= 0.88

d) P(X>2)= 0.12

e) P(X=1 or X=4)= 0.24

f) P(1≤X≤4)= 0.59

Step-by-step explanation:

a) To find P(X=2), we calculate: P(X=2)= 1 - P(X=0) - P(X=1) - P(X=3) - P(X=4) which equals 1 - 0.41 - 0.18 - 0.06 - 0.06, resulting in 0.29

b) For P(X<2), we sum P(X=0) and P(X=1): 0.41 + 0.18 yields 0.59

c) To obtain P(X≤2), we add P(X=0), P(X=1), and P(X=2): 0.41 + 0.18 + 0.29 equals 0.88

d) To calculate P(X>2), we find P(X=3) + P(X=4): 0.06 + 0.06 gives us 0.12

e) For P(X=1 or X=4), we use the union of probabilities: P(X=1) + P(X=4) which is 0.18 + 0.06, resulting in 0.24

f) P(1≤X≤4) is found by adding P(X=1), P(X=2), P(X=3), and P(X=4): 0.18 + 0.29 + 0.06 + 0.06 results in 0.59

Answer:

The anticipated number of tests required to identify 680 acceptable circuits is 907.

Step-by-step explanation:

For any circuit, there are two potential results: it either passes the test or it fails. The likelihood of passing is independent between circuits. Therefore, we apply the binomial probability distribution to address this scenario.

Binomial probability distribution

This distribution calculates the chance of obtaining exactly x successes across n trials, where x has only two possible outcomes.

To find the expected number of trials to achieve r successes with a probability p, the formula is given by:

Circuits from a specific factory pass a certain quality evaluation with a probability of 0.75.

Thus, to determine the expected number of tests needed for 680 acceptable circuits, let’s denote this as E where r = 680.

The expected number of tests necessary to find 680 acceptable circuits is 907.

Answer:

They anticipate the following ratio:

x/20 = 160/8, where x represents the population

Next

x = 20×160/8 = 400