5

MNM Corporation gives each of its 500 employees an aptitude test: The scores on the test are normally distributed with mean of 75 and standard deviation of 15. A si...

Question

MNM Corporation gives each of its 500 employees an aptitude test: The scores on the test are normally distributed with mean of 75 and standard deviation of 15. A simple random sample of 36 is taken from the population of 500 employeesUsing only the appropriate statistical table in your textbook, what is the probability that the average aptitude test score in the sample will be less than 78.69? ANSWER: (Report your final answer to 4 decimal places; using conventional rounding rules)Using only the

MNM Corporation gives each of its 500 employees an aptitude test: The scores on the test are normally distributed with mean of 75 and standard deviation of 15. A simple random sample of 36 is taken from the population of 500 employees Using only the appropriate statistical table in your textbook, what is the probability that the average aptitude test score in the sample will be less than 78.69? ANSWER: (Report your final answer to 4 decimal places; using conventional rounding rules) Using only the appropriate statistical table in your textbook, four percent of the samples ofn = 36 taken from this population should have mean test score above what value? ANSWER: (Report your final answer to 4 decimal places, using conventional rounding rules)



Answers

Find the sample size required to estimate the population mean. The Wechsler IQ test is designed so that the mean is 100 and the standard deviation is 15 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of college professors. We want to be 99\% confident that our sample mean is within 4 IQ points of the true mean. The mean for this population is clearly greater than $100 .$ The standard deviation for this population is less than 15 because it is a group with less variation than a group randomly selected from the general population; therefore, if we use $\sigma=15$ we are being conservative by using a value that will make the sample size at least as large as necessary. Assume then that $\sigma=15$ and determine the required sample size. Does the sample size appear to be practical?

So in this question, we are given sample salaries in thousands of dollars for employees, and in part, they were asked to find the sample mean and the sample standard deviation. So for the sample mean we basically take the some effects divided by the number of elements. Sample standard deviation. We take the summer off squares, we divide by and minus one and take the square root. So if we take all of the's values here, our sample mean yes, 41.5385 So Mm hmm. 0.5 green 85 the unit ISS thousands off dollars and our sample standard deviation. So we basically take each value. We subtract the mean from it, we square it and then we summit and then we divide by the number of elements. So the sample standard deviation is 5.3169 same unit off thousands off dollars. So this is our answer to part. So now, in part B were asked to, we're told that each employee receives a 5% race and were asked to find the sample mean and standard deviation. So with the 5% race for us to find the sample mean and standard TV should. So for 5% rates were basically going to take each of these values, and we're going to multiply that by 1.5 and then we're going to use our sample mean and standard deviation formulas and calculate the values. So when we multiply everything by 1.5, for example, 42 come 44.1 and just another example 45 will become 47.25 So we do that for all the values and here, our sample mean is going to be 43.615 thousands off dollars and our sample standard deviation using this formula is going to be 5.583 thousands of dollars. So that's our answer to part. Think now in Part C were asked to take the salaries and divided by 12 to get the monthly salary, whereas to find the mean and standard deviation for the salaries. So, for example, if we take 42 and we divide by 12, we get 3.5. So we're going to do the same thing for all our values here, and after doing that, we will calculate again the same using same formula. The sample mean and simple center, activations or sample. Mean here is 3.46 15 So 3.4615 thousands off dollars and our sample standard deviation is point 0.44 mhm in the unit ears again, thousands off dollars. So we have our answer for part C Now in party, what we're asked is what we can conclude from the results off A, B and C. So basically what we can conclude from our results is when there was a 5% raise in salaries, there was also actually a 5% raise in the mean and standard deviation. So 5% brace in the data for salaries also caused it 5% increase in the sample mean and the sample standard deviation and mean the same thing when the salaries were divided by 12. Then the sample standard deviation and the mean we're also divided by 12, which we can actually see here. So and we actually take our original samples mean and center division, and we divide by 12. We actually get thes values, and so that is our conclusion for parties. Whatever change in the day. If all of the data changes consistently in a particular manner, so does the sample mean and standard division. They basically changed by that same consistent amount.

All right, We're getting some statistics about the population. Means score on the critical reading, math and writing sections of the S A T. And we're also given that the population standard deviation for all three of these categories is 100. Now, for each of these parts were goingto excuse me, we're giving a sample size a section, and then we're supposed to find the probability that our point estimate for the mean will fall under a certain interval. So for part A, our sample size is 90. We're looking at the creek reading section and our error, the error. We're trying to find the probability of its plus or minus 10. Really? Technically, none of this writing and math and critical reading stuff the scores don't matter because we're only looking at the error. So yeah, keep that in mind anyway. So let's start with our standard deviation of the sampling distribution. Ah, I I think it's safe to say that 90 is small enough compared to all the people who take the S a T that we can just assume better. Our population sizes relatively infinite, so we're just going to use that formula, so it's gonna be 100 over the square root of 90. This equals about 10.54 Now we're going to Z score the upper and lower bounds. So negative. 10. Get See. Score to approximately 0.95 on lowered snake. It's positive. 10. Not negative. Over 10.54 Okay. Looking at our table since by central limit their, um this is big enough that we can just assume it's normal. Billy, on the lower end of zero point 1711 Probability on the upper end is 0.82 uh, 89 We find our probability by subtracting the upper limit minus the lower limit. And this equals zero. Oops. 0.6578 All right, party sample sizes still 90. Now, we're looking at the math section and we're still looking for an air of 10. But you probably noticed, in part A. We never used this bit of information here. Like the population mean never came into it, so we don't need to worry about it. Fact, we don't even need to worry about it here or here. It's completely irrelevant to the question anyway, so let's do this. Our standard deviation of the sampling. Distribution is also 10.54 And you might notice this means literally. All the math is the same. So for a gravity sake, I'm not gonna write it out again. This probability is 0.6578 again. All right. We were looking at the writing section. Not that it matters. Our sample size is 100 we're still looking for an error of plus or minus 10. So let's find our standard deviation with the sampling distribution. This time, it will be different. Uh, well, this isn't even approximately. It's ah 100 divided by 10. So this is 10. All right, let's see. Score the upper and lower. But if the standard deviation is 10 and we're looking for an error of plus or minus 10 batches means not ours. E lower is negative one and ours ear upper is one. So this meat So what? Z equals negative one. Our probability is 0.1587 And when Z equals one, our probability is equal to 0.84 13 It's probability lower probability, upper so probably upper minus probability. Lower equals 0.8413 13 There we go, minus 0.1587 This equals 0.68 to 7. And this should make sense because of the 68 95 99.7 role that 68% uh, is congruent with what we know about the set of points between negative one and one standard deviation. That means now the problem asks us to analyze the standard deviation and compare it two parts A and B, and we notice it's larger. And this makes sense because we looked at a larger sample. So yeah, that means we're having more. We have a wider standard deviation here for four hour point estimate for the meat, so they don't lead to a larger probability, and there you have it.

So we're assuming we have a normal distribution and we are given that they mean it's supposed to be 498 with a standard deviation of 100. So this would be a score of about 5 98 right here, and this would be a score of about 3 98 there. And we want to find what's the likelihood of having a score for this particular type of? I think it was an exam between 405 100. We're supposed to assume a normal population again, so we would convert both these into Z values. We take the score minus the mean yeah, divided by the standard deviation and the score minus the main, divided by the standard deviation. And that will give us our Z values and that differences negative. 98 negative night. Excuse me. Negative 98 divided by 100. So that's going to be a Z score of negative 1000.98 and this is the score is only two divided by 100. So 1000.2 and picture wise that 500 would be about here. So it's again very close to a zero Z value and the 400 is going to be about here. So we're finding that area, really? And I'm gonna use my calculator. I could use a table as well, but I'm going to use my normal CTF. And when I hit on my calculator second and distribution normal CDF, I'm going to use this as my low number negative 0.98 My upper number is going to be 0.2 and I'm leaving the mean for the standard normal distribution at zero and standard deviation of one. And when I do that, I find out that that probability is 0.34444 So I don't recall if it said percent. If it said percent, we'd get at about 34% chance of that happening or the probability is about 340.34 Then we had to look at Part B, and it said if you had a total of 300 people, how many of those would you find would have a score that's over 700? So let's find the probability of getting a score over 701st, and we can see that it's not real likely. Here's at 600 roughly and 700 would be about here. This would be at 6 98 so I can see that that probability I'll kind of color in blue is not going to be very big, which means you're not going to have a huge number here. So let's convert this into a Z score again and our Z score. What we got minus what we're assuming and think that was 4 98. It wasn't it. Yep, 4 98 and then divided by 100 and so 700 minus the 4 98. That difference is two oh two. And then we're dividing that by 100. So it's going to be two point Oh two. So that's our Z value, or about two standard deviations higher than the mean. And again I'm going to use my normal CDF here. Second and distribution go to normal CDF, and I'm going to use this as the low number 2.2 You could also look this up in a Z table and then for the upper value. I'm just going to put in 1000 and leave the mean and standard deviation at zero and one. And when I type that in. I find out that that percent is 0.217 approximately. And so now I'm going to multiply that by 300 to find out what 2.17% is of 300 times 300 and I find out that that comes out to be 6.5, so there would be probably six or seven, six or seven people I would anticipate to have a score that is greater than, uh, 700.

So in this question, were given sample annual salaries in thousands of dollars and in part they were asked to find the sample mean and the sample standard deviation, which is given by this for Mila over here. So to find the sample mean and standard deviation, we basically use thes formulas. They substitute all the values of X, and our sample means is equal to 41 point 6923 thousands of dollars. Our sample standard deviation is 5.99 rounded to two decimal places. And so this is our answer to part in now for part B, we're told that each employing takes $1000 race. So you get $1000 races. Since all of these air in thousands of dollars, we're gonna add one. So the soldiers on 41 36 all the way up to 48 and then we're going to calculate our sample mean and standard deviation, for example, mean in this case is 42.6923 and our sample standard deviation actually stays the same at five point 99 Now, in part C, we're told that each employee takes a pay cut off $2000. So all of our values, you are going to decrease by two. So this is going to become 37 38 so on. And in that case, our sample mean is basically going to decrease by two from the initial. So we're gonna have 39.69 on our sample. Standard deviation remains the same at 5.99 So we have our answers to part A B and sink. Now let's move on to part Dean. So in part D were asked what we can conclude from the results of A, B and C So basically what? We can conclude this, that when we had or subtract a constant value from the data when we add or subtract a constant valley from our data, our mean well increase or decrease. Why the same amount So say we add or subtract an amount key from the data. Are minnow increase or decrease by the same amount? What the standard deviation remains the seen? Because the variation the data is constant. It just increases or decreases all the data just cool up or down by the same amount. So the standard deviation remains the same, but to me increases or decreases by the value that we had or subtract the data.


Similar Solved Questions

5 answers
Young 5 double-slit experiment, Ine *uuun Uatk Innqe located 0.034 the sd2 Ue centrdl pnont innae separation between the Jits i2 1.5 What is the rarelenath the Iiqht Eono HsedncDndiSqenih dalt trineCenlral bricht IrineoGauitlh dnlaeetaeEic nDouble :It
Young 5 double-slit experiment, Ine *uuun Uatk Innqe located 0.034 the sd2 Ue centrdl pnont innae separation between the Jits i2 1.5 What is the rarelenath the Iiqht Eono Hsednc Dndi Sqenih dalt trine Cenlral bricht Irineo Gauitlh dnlaeetae Eic n Double :It...
5 answers
Apply Green's Theorem to evaluate the integral.fxoy x)dx + (y + 9x)dy C: The circle (X ~ 9)2 + (Yy - 92 =6Sxoy x)dx + (y 9x)dy = (Type an exact answer; using T as needed )
Apply Green's Theorem to evaluate the integral. fxoy x)dx + (y + 9x)dy C: The circle (X ~ 9)2 + (Yy - 92 =6 Sxoy x)dx + (y 9x)dy = (Type an exact answer; using T as needed )...
4 answers
8 0 transports ue QUESTION Itofcyoe ; fatty ligands acids for orator breaks the receptor down membrane blood oni the stream ipunoq liver the lipoprotein chylomicrons: asedil
8 0 transports ue QUESTION Itofcyoe ; fatty ligands acids for orator breaks the receptor down membrane blood oni the stream ipunoq liver the lipoprotein chylomicrons: asedil...
5 answers
Let $alpha$ be the largest root of$$f(x) equiv e^{x}-x-2=0$$Find an interval $[a, b]$ containing $alpha$ and for which the bisection method will converge to $alpha$. Then estimate the number of iterates needed to find $alpha$ within an accuracy of $5 imes 10^{-8}$.
Let $alpha$ be the largest root of $$ f(x) equiv e^{x}-x-2=0 $$ Find an interval $[a, b]$ containing $alpha$ and for which the bisection method will converge to $alpha$. Then estimate the number of iterates needed to find $alpha$ within an accuracy of $5 imes 10^{-8}$....
1 answers
Show thatthe functionf (x) /xis uniformly continuous on the set A [a, 0) , where a is a positive constant
Show thatthe functionf (x) /xis uniformly continuous on the set A [a, 0) , where a is a positive constant...
5 answers
Bopanna and Middelkoop are playing series of Tennis matches at Wolfkran Open Tournament; France_ Let the corresponding probability mass function table is given below; 0 1 2 3 4 5 20 20 30 (i) Find the value of Find the probability that any of the player wins at least six matches_ Determine the cumulative distribution function of the given data for record purposerandom variable that represents the occur
Bopanna and Middelkoop are playing series of Tennis matches at Wolfkran Open Tournament; France_ Let the corresponding probability mass function table is given below; 0 1 2 3 4 5 20 20 30 (i) Find the value of Find the probability that any of the player wins at least six matches_ Determine the cumul...
5 answers
Answer "true" or "false" to the following statements:To ensure the same precision, a population of 1 million would require a much larger random sample than would a population of 100,000.
Answer "true" or "false" to the following statements: To ensure the same precision, a population of 1 million would require a much larger random sample than would a population of 100,000....
5 answers
CUL70r9 Valt tnemMhe Jeun 2-Iu 351 Tru Ank qureiae 7o72 4 07910 0984) 046
CUL70r9 Valt tnemMhe Jeun 2-Iu 351 Tru Ank qureiae 7o72 4 07910 0984) 046...
5 answers
Get 𝑡^2 ∗ 𝑡𝑒^𝑡 answer: 𝑡^2 + 4𝑡 + 6 + (2𝑡 − 6)𝑒^𝑡
get 𝑡^2 ∗ 𝑡𝑒^𝑡 answer: 𝑡^2 + 4𝑡 + 6 + (2𝑡 − 6)𝑒^𝑡...
5 answers
If limsup(x_n)<sup(x_n)prove x_n is bounded and it attains its supremum. i.e there exists n_0 s.t x_(n_0)=supx_n
if limsup(x_n)<sup(x_n) prove x_n is bounded and it attains its supremum. i.e there exists n_0 s.t x_(n_0)=supx_n...
5 answers
CUESIION 24Calculate AH? for: 2C2H4(g) H2O(e) ~> C4HgOH(C)Using:2C02 2H2O() ~ > C2Ha(g) 302(g) C4HgOH(E) 602(g) ~> 4C02 SHzO()AH = +I4LkJAH? = -1534.7 kJ1587.5ki-12875k12wSh-IBSk
CUESIION 24 Calculate AH? for: 2C2H4(g) H2O(e) ~> C4HgOH(C) Using: 2C02 2H2O() ~ > C2Ha(g) 302(g) C4HgOH(E) 602(g) ~> 4C02 SHzO() AH = +I4LkJ AH? = -1534.7 kJ 1587.5ki -12875k 12wSh -IBSk...
4 answers
QUESTION 14Using Newton's forward interpolation formula find the value of y when x = 4.6 from the following data: (4 1) , (6 3) , (8 , 8) , (10 20) Y = 1.523 y = 1.535 y =1.512 y =1.556
QUESTION 14 Using Newton's forward interpolation formula find the value of y when x = 4.6 from the following data: (4 1) , (6 3) , (8 , 8) , (10 20) Y = 1.523 y = 1.535 y =1.512 y =1.556...
5 answers
In both prokaryotes and eukaryotes, initiation Of translation usually begins immediately upstream of a Shine-Dalgarno sequence. requires the action of release factors which cause the small and large subunits of a ribosome to associate with one another: requires the action of initiation factors. begins at a UAG codon downstream of the 5' UTR of an mRNA molecule. begins only after ubiquitin has been added to the mRNA molecule to be translated
In both prokaryotes and eukaryotes, initiation Of translation usually begins immediately upstream of a Shine-Dalgarno sequence. requires the action of release factors which cause the small and large subunits of a ribosome to associate with one another: requires the action of initiation factors. begi...

-- 0.022360--