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Suppose X is a standard normal random variablePr (X < 1.645) is approximately (1 decimal place accuracy): C 0.95 0.01 0.05 0.99 B.0.90 H. 0.89 0.94 0.009 L. 0.04...

Question

Suppose X is a standard normal random variablePr (X < 1.645) is approximately (1 decimal place accuracy): C 0.95 0.01 0.05 0.99 B.0.90 H. 0.89 0.94 0.009 L. 0.04 G.0.98 is tossed 3 times. The probability that head appears 2.A fair coin most once iS; 8 8 20.10 0.09N.None of theseENone400 and the variance is 100 population . mean is 400 ( /x Suppose the of History: Then the standard deviation is: for & test C.20 D. 10 160,000 B. 10,000E: none

Suppose X is a standard normal random variable Pr (X < 1.645) is approximately (1 decimal place accuracy): C 0.95 0.01 0.05 0.99 B.0.90 H. 0.89 0.94 0.009 L. 0.04 G.0.98 is tossed 3 times. The probability that head appears 2.A fair coin most once iS; 8 8 2 0.10 0.09 N.None of these ENone 400 and the variance is 100 population . mean is 400 ( /x Suppose the of History: Then the standard deviation is: for & test C.20 D. 10 160,000 B. 10,000 E: none



Answers

A normally distributed population has mean 1,214 and standard deviation 122 . a. Find the probability that a single randomly selected element $X$ of the population is between 1,100 and 1,300 b. Find the mean and standard deviation of $x-$ for samples of size $25 .$ c. Find the probability that the mean of a sample of size 25 drawn from this population is between 1,100 and 1,300

This question, We're told that he normally distributed population has a mean of 57,800 and a standard deviation of 7:15. And were asked for us for the probability, since this is a normally distributed population probability that a single randomly selected element is between 58 1,057 1000. So we convert this to our standard normal random variable And we when we do that conversion, we get c between .27 and -1.27 And we look for these two values in our table, so we get .6064 -14-3, Which gives us .4641. We're party part B find the mean and standard deviation for samples of size hundreds, So are mean Is basically our population mean, which is 57,000 800. Our standard deviation is basically 750 over squared of 100 which is 75 Part C. Find the probability that the mean of a sample of 600 Between 58,057,000, and we're going to use, our answer is calculated in part B to convert our X to the standard normal random variable, so we can see between 2.67 and -10.67. That's just probably dizzy, less than 2.67, Which is .9962.

In problem three were given a distribution of values 77-81. And the probability for each of those values Question a asks us to determine the probability that we would get exactly 80 and that's going to be just reading it straight from the table. The probability of 80 is 0.40 as we can see right there. Problem be asks for the probability of a value greater than 80. Now. It's important to take note to the fact that we are just talking about greater than 80 and not 80. Exactly. So the only value that's greater than 80 is 81 and the probability of getting an 81 is 0.10 part C asks for the probability of At most 80 being less than or equal to 80. So that would be from 80 down. Now there's kind of two ways that you can go about this. Um you could add up before values 77, 78, and 80 those four probabilities or really the only thing that we're excluding is the 81. That's the only value that does not satisfy this. So we don't want the 10%. Instead we want The 90% or 0.90. And again you could have added the .15.15 would be .3 Plus .2 would be .5 and then plus .4 would be .90. Now D. E and F. We're going to require a bit more work. We want to determine the mean of the probability distribution. So this is very similar to the idea of doing a weighted average. And so for part D the mean of this distribution is going to be the sum of each of the possible values Multiplied by its probability. So 77 times .15 Plus, times .15 plus, 79 Times .20 Plus. And I'm gonna jump down the line here 80 Times .40 plus 81 Times .10. And so I'm just going to take a moment and plug these into the calculator. And so 77 times 770.15 plus 78 times 780.15 plus, 79 times 790.20 plus, 80 times .4 plus 81 times .1. And I'm sorry, I think you have to do that again. I didn't did not get the correct answer, but I can see what I did wrong, 79.15 should be the correct answer. For part E we want the variants of the probability distribution. And then for part f we're going to convert that to the standard deviation. So, for variants you need to um take and and this is very similar to your variance formula. Just when you're dealing with a data set, we're going to take each possible value and subtract the mean square that and multiply it by its probability. So again, it's kind of like doing the idea of a weighted average. So this is going to be for the variants sigma squared sub x X, just representing the probability distribution that we're talking about, We have a possible value of 77 minus the mean squared, multiplied by that weight 0.15 I'm going to move this over a little bit, so I have some more space here, then we add to that 78 -79.15 Squared, multiplied by .15 plus 80 -79.15 Squared. And that is, I'm sorry, I missed the 79. Let me get that in there. So 79 -79.15, multiplied by .20 Plus. Now the 80 -79.15 Squared, And it's probability is .40. And then lastly, 81 -79.15 squared, Multiplied by .10. So then you carefully enter this into the calculator. It does help if you're able to do a little bit of mental math. So, for instance, I just knew that 77 -79.15 would be negative 2.15. So I entered that into my calculator squared Times .15. Just make sure that you put the negative 2.15 in parentheses with the square on the outside, or don't even worry about the negative. That can work too. Okay. Yeah, yeah, So I have an answer for the variants of 1.5275, again, just entering everything that we have there into the calculator. Now, the nice thing for part F, we want the standard deviation for the probability distribution, we're just going to take the square root of the variance, And that gives you approximately 1.236, Yeah.

All right, read the variable X is normally distributed with me and me equals 10 and sigma equals two. We wish to use the normal distribution table to find the following probabilities. A through e. This question is challenging your understanding. Uh normal variables or random variables that are normally distributed to solve. Let's first review relative relevant information related to this distribution before proceeding. So a Z table map Z scores on two probabilities. That is probably greater than zero equals peanut, implies that peanut is the area here in purple under the Z score to the right of ours. The not so as an example, the probably is greater than zero, for instance, is 00.5. For standard normal very variable Z because the mean is zero and we have symmetry. The segues nicely into the two theorems. We're gonna be using to solve this problem. That is a cemetery of the normal curve. And the fact that the total area of the normal curve is one. So because random variable X is not a standard normal, but instead of the normal, we have to convert to standard normal Z scores in order to compute probabilities associated. So for instance to probably X greater than X equals 10.5, we can first computers the Z score to a teensy not equal zero. That's by converting our Z score to our X. Using our sigma and dinner deviation We have X equals Z, not sigma plus new or zero times two plus 10 equals time. And be. Now I'm probably executed mexicans 100.95 gives you score of 1.65. That's access 1.65 times two plus 10 equals 13.3. Next in c we have the Z squared zero. That's probably that's the falls between zero and zero is zero. However, we can't have a problem. Is this one? So it must be that Xena is equal to zero itself, that X is equal to 10 for part D. We note that we have our random variable X attracting 10. So we subtract away than me and you 10 that we're looking for probably less than negative. Z equals one minus 10.9 5/2 0.25. That's you know, as much as 1.96. Thus X is again converting by multiplying by the standard deviation 1.96 times two equals plus. Um I have put 3.9 to notice how we didn't add 10 because we already subtracted 10 away. E assault identically to G. This time we have a Z score .005 giving the 2002.33 that access plus or -4.66

All right, read the variable X is normally distributed with me and me equals 10 and sigma equals two. We wish to use the normal distribution table to find the following probabilities. A through e. This question is challenging your understanding. Uh normal variables or random variables that are normally distributed to solve. Let's first review relative relevant information related to this distribution before proceeding. So a Z table map Z scores on two probabilities. That is probably greater than zero equals peanut, implies that peanut is the area here in purple under the Z score to the right of ours. The not so as an example, the probably is greater than zero, for instance, is 00.5. For standard normal very variable Z because the mean is zero and we have symmetry. The segues nicely into the two theorems. We're gonna be using to solve this problem. That is a cemetery of the normal curve. And the fact that the total area of the normal curve is one. So because random variable X is not a standard normal, but instead of the normal, we have to convert to standard normal Z scores in order to compute probabilities associated. So for instance to probably X greater than X equals 10.5, we can first computers the Z score to a teensy not equal zero. That's by converting our Z score to our X. Using our sigma and dinner deviation We have X equals Z, not sigma plus new or zero times two plus 10 equals time. And be. Now I'm probably executed mexicans 100.95 gives you score of 1.65. That's access 1.65 times two plus 10 equals 13.3. Next in c we have the Z squared zero. That's probably that's the falls between zero and zero is zero. However, we can't have a problem. Is this one? So it must be that Xena is equal to zero itself, that X is equal to 10 for part D. We note that we have our random variable X attracting 10. So we subtract away than me and you 10 that we're looking for probably less than negative. Z equals one minus 10.9 5/2 0.25. That's you know, as much as 1.96. Thus X is again converting by multiplying by the standard deviation 1.96 times two equals plus. Um I have put 3.9 to notice how we didn't add 10 because we already subtracted 10 away. E assault identically to G. This time we have a Z score .005 giving the 2002.33 that access plus or -4.66


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