Howdy! I’m Professor Curtis of Aspire Mountain Academy here with more statistics homework help. Today we’re going to learn how to find the probability of accepting a batch in manufacturing. Here's our problem statement: Among 7897 cases of heart pacemaker malfunctions, 340 were found to be caused by firmware, which is software programmed into the device. If the firmware is tested in three different pacemakers randomly selected from this batch of 7897, and the entire batch is accepted if there are no failures, what is the probability that the firmware in the entire batch will be accepted? Is this procedure likely to result in the entire batch being accepted?
OK, we have two parts. The first part is asking for the probability. To get the probability of the firmware actually passing the inspection, the probability that the lot is going to be accepted is going to be determined by all three of those randomly selected pacemakers passing inspection. So the first one has to pass, and the second one has to pass, and the third one has to pass. Well, the probability of the first one passing is simply one minus the probability of that first one failing, because there's only two options — I mean, the part either passes or it fails. So there is our failure rate for the first one; 1 minus the failure is going to give us the pass. We can produce the same thing for each of the other two pieces that we’re going to test.
So now here's a probability of acceptance. And the probability that the first one is going to fail is going to be equal to 340/7897 because you can see from our problem statement that 340 of those failures were caused by the firmware, and that's what we’re testing for with our acceptance test. So 340 is the part over the whole 7897 — that's the probability that the first one fails. So that's what we put in here. And is simply probability talk for multiplication. So when you see and, that simply translates into multiplication. The probability that the second one is going to fail is going to be just the same as the probability of the first; it's the same failure rate for all the different pieces, so I can just write the same thing here for the second piece. And again translates into multiplication. And then we find that the failure of the third piece is going to have the same probability as the other two pieces. So now you got this written.
Now what we have to do is simplify this and solve. So notice how you have the same expression multiplied by itself three times, so I can just write this as that expression raised to the third power. And then I can go and solve for that expression. So when I do that, 340 divided by 7097 gives me 0.0431. I subtract that from 1 and get 0.9569. I raise that to third power, and I get 0.8763. This is my probability that the lot will be accepted. I’m asked to round to three decimal places.
And then here on the second part, it’s a drop-down menu. So “this procedure is <blank> to result in the entire batch being excepted.” There’s three options. Well, it’s not going to be certain because certain means the probability is 100%, and that’s not what we have. It’s not going to be unlikely, because unlikely means that the probability is less than 50%; we have more than 50% here. So we’re going to choose likely, because the probability is greater than 50%. I check my answer. Well done!
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9/6/2020 07:15:30 am
I wish my university explained it this way!! THANK YOU
9/21/2020 02:46:03 am
YOU ARE THE BEST I hope u all the best
3/1/2021 06:29:07 pm
Thank you so much! So much easier to understand this way!
9/15/2021 09:34:17 pm
Thank you so much I tried the same formula for a different problem and it worked first try!
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Frustrated with a particular MyStatLab/MyMathLab homework problem? No worries! I'm Professor Curtis, and I'm here to help.