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The average number of daily emergency room admissions at a hospital is 85 with a standard deviation of 37_ In a simple random sample of 30 days, what is the probabi...

Question

The average number of daily emergency room admissions at a hospital is 85 with a standard deviation of 37_ In a simple random sample of 30 days, what is the probability that the mean number of daily emergency admissions is between 75 and 95?8612212813889970.8990

The average number of daily emergency room admissions at a hospital is 85 with a standard deviation of 37_ In a simple random sample of 30 days, what is the probability that the mean number of daily emergency admissions is between 75 and 95? 8612 2128 1388 9970 .8990



Answers

Hospital Noise Levels Noise levels at various area urban hospitals were measured in decibels. The mean of the noise levels in 84 randomly selected corridors was 61.2 decibels, and the standard deviation of the population was $7.9 .$ Find the $95 \%$ confidence interval of the true mean.

Here for the solution. The mean of the noise level is 84 corridors was 81.2 describes the standard deviation was 7.9 does an equal to 84 x bar equal to 61.2 and s equal to 7.9. The confidence level is 95%. So alpha called to one minus 0.95 or ALF. I called to 0.5 or alpha by two by dividing both sides by two. So it will be also by two equal to 0.25 or by adding 0.5 both sides. We get 0.5 minus Alpha by to call to 0.5, minus 0.25 or 0.5 minus alpha by two equal to 0.4750 on From here we get set Alpha by two equal to 1.96 Hence, the required confidential level is that it's X bar minus Jack Alpha by two s by under route and is less than mu is less than s bar X bar that alpha by two s by under root end. So here by substituting the values. Or we can say that 61.2 minus 1.96 back in 7.9, divided by who underwrote 80 for less than mu is less than 61.2. Bless one 0.96 in brackets and 0.9 you are under Route 84 or it is 59.5 is less than mu is less than 62.9. New exist between the 59.5 and 62.9. So this is the definition for the solution, please call.

And 68. I'm given a situation in which I have a normal distribution center to 5.3, with the standard deviation of 2.1. I want to represent that. Visually, I don't have to fill out all my standard deviation pieces, but I can at least do one on the side of a mean And now I can look to my question of 68 it says. What is the probability of spending more than two days in recovery? Um, as a probability statement. The probability of spending more than two days, which is X being greater than two on the picture would look something like this. What is the probability of spending more than two days? The reason it all the pictures So you can see at the end when you get your probability. Um, does the new Miracle Probability match the picture? And normally it's Does the picture show less than 50% of more than 50% Journalist. All you need to know to check your answer so we can go to our normal CDF function here are lower bound is where we start shading in to here. Upper bound is where we stopped in this case, we don't stop. So we're gonna insert Nun, Nun, Nun Nun to represent infinity, our mean is 5.3. Standard deviation is 2.1 type. But in we get the answer, 0.94 to 0. And as you can see in the picture, about 94% of the areas shaded That looks good so we could check their work. That is option D.

So now we're looking at finding a confidence interval for a mean and we're told that the sample size is 24. We're told that the main of those 24 numbers came toe Ah, 41.6 decibels. And we know that the standard deviation and it sounds like the standard deviation is for those 24 24 rooms. So that is a sample standard deviation was 7.5 decibels. And we wanna find a 95% confidence interval for the mean for the true mean sound level in decibels. And so we know that this is a small sample. So we're going to use our T star value and we'll have us over the square root of an So we know that the 41.6 is our X bar. And then if we have 20 sample size of 24 we know that our degrees of freedom is 23. And so if we look on our table and we look for the degrees of freedom here for 23 then look for the level of confidence to be 95% and that value we get here, I looked it up right is 2.69 and then we have our sample standard deviation of 7.5 and we have our sample size of 24. So we have 41.6 plus or minus and let me quick type it in 2.69 Time 7.5 divided by the square root of 24. And when I do that, I get that margin of air is about 3.17 So if we want to write that as an interval, which we normally would and I'm gonna quick store that value, I'm going to store that value and hit stow on my calculator. I'm just going to store it as X and so I'm gonna take my 41.6 and subtract away X, and that gives me my 38.4 decibels and I will change that to adding, and that gives me 44 point eight. So we are 95% confident that the actual means decibel level is somewhere in here. Now we have a chance. It's not because we're only 95% confident, but it's a pretty good chance that it's in that level and we're finished

For this problem, we need to calculate a confidence interval for the mean to calculate a confidence in trouble. We need three different things. The mean which we're told us 41.6 the standard deviation, which we're told if seven and the T value, which we can figure out using the degrees of freedom and the confidence interval present. The degrees of freedom for this problem is equal to 24 minus one 23 in the confidence interval percentages in 95%. Using our tea table and these pieces of information, we can figure out that the T value is 2.69 Using these three components, we can now write our confidence interval the confidence central formula. It's a sample mean plus or minus T value times the standard deviation over the square root of end or, in this case, 41.6 plus or minus 2.69 times seven over the square root of 24 multiplied out. That should equal 38 0.644 less than or equal to mu is less than or equal to 44 0.556 That should be your 95% confidence interval for this problem


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