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The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of 5 minutes and a standard deviation of 3 minutes.Find the ...

Question

The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of 5 minutes and a standard deviation of 3 minutes.Find the probability that it takes at least 8 minutes to find a parking space. (Round your answer to four decimal places.)

The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of 5 minutes and a standard deviation of 3 minutes. Find the probability that it takes at least 8 minutes to find a parking space. (Round your answer to four decimal places.)



Answers

The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of 5 minutes and a standard deviation of 3 minutes.

Find the probability that it takes at least 8 minutes to find a parking space. (Round your answer to four decimal places.)

70 to 70% of the time. It takes more than how many minutes to find a parking space. I'm still dealing with a normal distribution Have a mean of five and a Senate aviation of two. And what I'm actually looking for here is the probability that X is less than some value is 70% or 0.70 So what I'm gonna do here is using my inverse norm a potion of my calculator with 0.70 as my area, I mean, is five must enter. Deviation is too. Insert this into your calculator and you'll get 6.5 of which is option D.

70. So we have the length of time that it takes to find a parking spot at 9 a.m. To follow a normal distribution with mean a five and a standard deviation of two based upon the given information and numerically justified. Would you be surprised if it took less than one minute to find a parking space? So would you be surprised if it took less than one minute? Well, we can use our normal CDF command here to figure out the probability because right now we really can't make a new miracle justification. So we need to know some sort of probability. Then we could make a new miracle justification, So we need to have a lower an upper bound mean instead of deviation. In this case, the values that we're looking at is anything from negative infinity all the way up to one so we can insert negative infinities we're gonna use negative, uh, the upper bound where we're going to stop Is that once that basically we're looking for the probability between the far left side of our distribution and all the way into one, we have a meaning five standard deviation of two. We complaint that in and we would get a probability of 0.2, I said. Now would we surprised? Would we be surprised if it took less than one minute in this case? I think yes, we would be surprised because the probability of finding a spot in less than one minute is around 2% which seems pretty unlikely. So that's option a.

Okay, so this question is given a normal distribution of times it takes for this guy can to get to work. It's a normal distribution with mean of 25 standard deviation of 4.5. So yep. Stand deviation of 4.5 and he has 35 minutes to get to work before he is considered late. So and we want to see what percent of times, if it does indeed follow Roman distribution. What percent of times is this guy can going to be late for work. So basically 35 minutes or more is equal to late. And so there are two ways to approach this. One way is basically the second way is basically the first way, but with one extra step. So I'm gonna explain both ways, but I'm only going to do the first way on this whiteboard. Um, okay, So the way that I'm thinking is just calculate the percent off. Um, so let's draw this bell curve real quick with the 25 in the middle. Just calculate the percent of times it's 35 minutes or onward. So, like this, using the normal CDF function on your graphing calculator And if you don't have a graphing calculator, you can just search search a calculator up online. There are tons of them. So, yes, he did normal cdf um, remember to get to normal cdf at least 70 84. You go two seconds, and then you click the bars button, so get normal CDF of our minimum, which is 35 our maximum, which could just be any really large numbers so we can do 10,000 on that are mean andr. Standard deviation. And we should get using this method. We should get around 0.13 um, which goes to 1.3%. So that means that Ken is one late 1.3% of the time, which is actually pretty impressive everything about it. But also, you want to be late 0% of the time in most jobs. So, yeah, that's there's a little food for thought. The second way to calculate it is just take the easy score of 35 in relation to those normal distribution and then just do normal CDF off the Z score 1000 0 and one, and then you should get the same answer mainly because taking these quarters skills everything down to like our normal distribution was your own one. So, yeah, 1.3% is the answer because that's the percent of data points above 35 and that's the percent of times that can will be late.

74 i Q is normally distributed with the mean of 100 a standard deviation of 15. Supposed one individual is randomly chosen. Let X because you are an individual. Hey, X could be mined by human evolution. Flip mean 100 a center deviation of 15 Be Find the probability that the person has an I Q. Greater than 1 20 All right, that is 1 20 included photograph and write a probability statement. So have a probability statement. Uh, quick sketch where I want to mark the mean and I want to do obscenity vision on each side of that. My boundary value of the value that I'm looking at here is 1 20 and I want to know the probability greater than 1 20 In order to stop that, I can use my normal CDF function calculator. Ah, lower bound. Here was the 1 20 My upper bound was infinity. But understand certain 9999 mean ended up being 100 center deviation was 15. I can play that in to find my very small area 0.0 in on 12 say small because I could see my graph that it's a small area that I shaded in. So my answer, my picture should, um, be similar. See, kill me in S A is an organization whose members have the top 2% of all accuse. Find a minimum. I Q needed to qualify for this organization and sketch the graft. So want to think about what's going on here, in which we have 100 as the center and the standard deviation being 15. So this is organization, which they contain the top 2% of all IQ's. I want to find that minimum value because I'm looking for that minimum value and looking for this little boundary about you that separates the left side of the draft from the right side of Well, in order to figure this out, I want to focus on the left side of this information because generally your calculators are left side functions. So the area to the left of this bulan has actually point not eight. So I'm looking for here's which value in which X is less than K K is gonna be. My mystery value gives us an area of 0.98 So there's my probability statement in order to solve this invert norm feature in which I answered the area to the left area to the left 0.98 Here the mean is 100 the standard deviation is 15. Tanaka found that boundary value to be 130.8 d the middle, 50% of all I used ball between which to values sketched the graph and rightly probability statement. I know that have a mean of 100 Senate aviation 1 15 and I want to figure out that the middle 50% fall between which to values. Well, in this case, I'm gonna treat you like a separate process. Um, because I can use my inverse norm calculator function too soft for this. Um, I can solve for my first question, Mark, I'll call this question mark number one, and then I could solve for question mark number. So my probability statements to help me answer these for question mark number one. Essentially, we're looking for where the probability where X is less than okay. One to equal. So you think the area with left because my calculators left side function here? Uh, the area to the left of this value is 25 because I have a symmetrical distribution with 50% in the middle. I have 25 1 that left have 25 on the right because into higher curve equals one. So the first thing I'm looking for here is which value gives me an area 25 so I can use my inverse norm. A 0.25 my area Mean was 100. My standard deviation. 15. This gives me 89.88 So I have my first value. And then from our second value, I think about this question number two, the total area to the left of that I used actually 75%. So it'll look like probability. That X is less than Kate, too. Would be £0.7. So my inverse norm statement will look very similar. Except I'll insert of 75 for my area and I'll get a value of 1 10.12


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