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Assemble computer desk Irom kit is random variable having the If the: time to assemble an casy probabilities normal = distribution with mean 55.8 minutes and standa...

Question

Assemble computer desk Irom kit is random variable having the If the: time to assemble an casy probabilities normal = distribution with mean 55.8 minutes and standard deviation [2.2 minutes, what are the _ that this kind of desk can be assembled in (5 pts) a) less than 49.7 minutes;(5 pts)6) betwcen 61,9 and 74.1 mninuic%;

assemble computer desk Irom kit is random variable having the If the: time to assemble an casy probabilities normal = distribution with mean 55.8 minutes and standard deviation [2.2 minutes, what are the _ that this kind of desk can be assembled in (5 pts) a) less than 49.7 minutes; (5 pts) 6) betwcen 61,9 and 74.1 mninuic%;



Answers

Manufacturing time The assembly time in minutes for a component at an electronic manufacturing plant is normally distributed with a mean of $\mu=55$ and standard deviation $\sigma=4 .$ What is the probability that a component will be made in less than one hour?

Problem 14. The whole time of a color in a customer service center Has a mean value of 23.8 seconds. And the standard division of 4.6 seconds, we want to find the probability that the meantime in assemble of 100 of 1200 calls, does somebody equals 1200 calls Will be within 4.5 seconds of the population. Mean myths get the probability four, the distribution to be within 4.5 of them in which means we will add a .5 to this value And subtract open five from this value. This means X will be between 23, 23.3 seconds and 24.3 seconds. The probability for them to be between -3.67 65 and 3.765. Which means we want to get the area Between -3.765 and 3.765. He wants to get this area due to symmetry, we can get the area to the left of -3.765 and subtracted from multiply it by two and subtract it from one, multiplied by two to get this area And subtract the results from one. To get the shaded area, which means it would be equal 1 -2 deployed by the probability four. Is it to be smaller than -3.765. We do. So because we have the standard normal distribution tables that gives us the area to the left of the school. This will be the most easiest way to get the shaded area Equals 1 -2, multiplied By the area. Let's enter the tables by the value of -3.7 eight or 7 seven. Where the street 0.77 we can find -3.7 here. 77 is given here point Or it then it's all point. Or or it it equals one minus to buy. Oh, it It equals 0.999 84, or at equals 99.98 4%. This is the percentage, or the probability to have the coal hold between 4.5 seconds of the mean By taking a sample of 1200 goods.

So to solve this problem, we're going to need to calculate four different T scores. One for each, um, specific number of minutes that we're looking at stone part. They were looking at the probability that somebody does less than 17 minutes. I used X to represent the number of minutes that they dio or do use the stair climber. So probability that it's less than 17 minutes. So we need a Z score for 17 minutes. Part B. We're looking for between 20 and 28. We need to one for 20 and one for 28 for part C, we're gonna need one for 30. I get dizzy score. We can use this formula. So Z score for 17 minutes, be given as X minus mu. So exes are 17 minus mu, which is the population, I mean, which is given us 20 just We divide all of that by the standard deviation, which is five Consol. This to find that disease scored 17 is negative 170.6. Now we do the same thing for a next see score. So the Z score for 20 minutes would be using the same formula. We're just gonna plug in 20 instead of 17. So right in that red bubble we're gonna put 20 instead. And then we calculated again we should get zero z score for 28 plugging in 28th this time. Okay, calculate that and you should get a C score of 1.6 and then our Z score for 30 should be too. So here's our Z scores. Oops, These air Z scores and now we can use them to find these probabilities. So each of these three scores represents of Mark on our standard normal curve. So our first the score of negative 0.6, we fall somewhere like here and then the corresponding value on RZ table represents the area to the left or the probability of a value being less than this value. So we look up negative 0.6 on RZ table. You should get points to seven. So 0.27 is the probability that somebody does less than 17 minutes. That's our answer for a part B is a little bit more complicated. Now we're looking for what's between two values. We have our first one, which again the corresponding values could be everything to its left. We also have a second one. And again, the corresponding value on RZ table will be everything to the left of that. Since we just want this middle section, we're going to take the corresponding value for our higher value. Sorry. 28. We're gonna take the course the Z score for the 28. So 1.6 and its corresponding probability, which again is going to cover this entire area to the left, all that green, and then we're subtracting off the same thing. But for are last year values. So the Z score of 20 everything to the left of that and that should just give us a middle region. This is just a generalized picture because the Z score of zero would actually line up or like here. But this gives you the general concept. So we look up our 1.6 on our, um, de table. You should get a probability of 0.945 and then we look up, are devalue for 20 on the Z table, and you should get a probability of 200.5 source subtracting 0.5. Then you can compute that, and you should get a probability of 0.445 and that's our answer for Part B. And then for part C, this is gonna be similar to part A. Except this time we want the area to the right instead of to the left, because we're looking for things that are greater than rather than less than because this represents a probability we're still gonna have our line. And normally the corresponding value gives us the area to the left. The probability that that it's less than that because we're working with probability, we can just do one minus this value, and that should give us the opposite, which is everything to the right instead, and that's what we're looking for. So we have our one minus, and then we look up our Z score or Z score for thirties to look up to on RZ table or again. You can use a graphing utility. You should get 0.977 We have one minus 10.977 gives us 0.0 to 3, and that's the probability that somebody's on the climber for more than 30 minutes.

Here, we are told that the time that customer spends waiting at a check in counter at an airport Is a random variable with mean of 8.2 minutes and standard deviation of 1.5 minutes than a random sample, 49 Customers is observed. And we are asked to find some probabilities for part A. We are asked to find the probability That the waiting time is less than 10 minutes. This is the average waiting time of the sample. So we have not been told much about the shape of the distribution of the overall population of waiting times. However, since our sample size is 49 and we're dealing with the sample average, we can say that the sample averages are approximately normally distributed, and we can also say that the mean of sample averages is the population mean Which is 8.2 and the standard deviation of sample averages is equal to the population. Standard deviation over the square root of the sample size, which is equal to 1.5 divided by seven Or .2143. So sample averages are approximately normally distributed, They have a mean of 8.2 and a standard deviation of .2143. Now, if we standardize the probability expression, it's the probability that set is less than or equal to 10 -8.2, which is the mean of the sample averages Divided by .2143, which is the standard deviation of the sample averages and this is Very close to one. So we're almost guaranteed that the sample average will be less than 10. Now, for B, we want the probability that the waiting time is between five and 10 minutes. So this is The probability that the waiting time is less than 10, Which we've already calculated to be one minus the probability that the waiting time Yes, listen, record of five, and this also comes out to effectively one. So nearly all of the distribution is between five and 10, and then for part C, were asked for the probability that the waiting time Is less than six minutes. This is equal to the probability that said is less than or equal to you 6 -8.2, Provided by 2.21, 4, 3. And this turns out to be approximately zero. So the probability that the average waiting time of the sample of 49 customers Is less than six minutes is essentially zero.

Mhm. So in this question we told, the machine produces fasteners whose length must be written in .5" of 22. So 22 plus minus 220.5 is where their length should lie, Length are normally distributed with a mean of 22" and a standard deviation of .17". And were asked for the probability that randomly selected fastener will have an acceptable ones. Because the probability that it lies Between 21.5 and 22 0.5. So basically that will be our standard Normal Variable -2.94 Hands 2.94. So we take property The less than 2.9 for minus properties the less than -2.94. So that's .9984 -1016. Which gives us .996 ft answer to party Would be first told the machine produces 20 fasteners per hour. The length of each one is independent. So we know here there's a binomial random variable, there's 10 Equals to 20 fasteners, assuming that they're all independent by the probability that they will have acceptable length, so the probability that each one As an acceptable language, .996 ft. So using our binomial distribution, we basically are looking for probability that all 20 will be acceptable, so that's Equal to 20 factorial, zero factorial and the pictorial P to the power of X. Q. to the power of N -6, which gives us .9379.


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