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# Intro Prob & Stat for Business STT 315

MSU

GPA 3.72

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This 23 page Class Notes was uploaded by Aryanna Jerde on Saturday September 19, 2015. The Class Notes belongs to STT 315 at Michigan State University taught by Staff in Fall. Since its upload, it has received 88 views. For similar materials see /class/207819/stt-315-michigan-state-university in Statistics and Probability at Michigan State University.

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Date Created: 09/19/15

1 NOTE THIS ASSIGNMENT DEALS WITH THE POWER CURVE AND METHODS OF DESIGNING A HYBRID TEST INFORMATION FROM A PRELIMINARY SAMPLE DETERMINES A NEEDED SAMPLE SIZE IN ORDER THAT OUR HYBRID TEST ACHIEVE A GIVEN ALPHA AND A PARTICULAR BETA For a particular test not specified here is a plot of Preject H0 l mu as mu varies between 25 and 31 I will tell you that the boundary between H0 and H1 for this test is mu0 27 Preject H0 l mu which is IDEALLY 0 on H0 and 1 on H1 1 133 05 39 x 04 I EE 239 23 29 313 31 H0 gtlt Hl mu gt a Give a fairly accurate numerical value for alpha Illustrate what you are doing in the curve ANS Alpha is the height above the boundary value 111110 27 which is by inspection of this curve alpha 015 b Give a fairly accurate numerical value for the power and beta at mul 28 Illustrate what you are doing in the curve ANS power is the height at 28 which is around 047 So beta the chance of failing to reject whn mu is 28 is 053 0 Identify the null and alternative hypotheses ANS H0 is where reject probability is low so it is mu less or equal 27 H1 mu gt 27 where reject probability is high d In the above plot overlay another curve representing Preject H0 l mu for a BETTER test of these hypotheses with the SAME ALPHA ANS Overlay a curve that is higher everywhere on H1 and lower everywhere on H0 except at muo where it is the same alpha 01 ie having better performance for the same alpha 2 For another test the plot of Preject H0 l mu as mu varies between 25 and 31 has a shape different from 1 Preject H0 l mu which is ideally 0 on H0 and 1 on H1 1 03 i k 26 2 28 29 30 31 H1 gtlt H1 mu gt a Give a fairly accurate numerical value for alpha Illustrate what you are doing in the curve ANS This is evidently a two sided test and alpha is the height of curve at 111110 275 which is alpha 012 just be close b Give a fairly accurate numerical value for power and beta at mul 285 Illustrate what you are doing in the curve ANS Power at mul 285 is the height there It is around 078 just be close Beta 1 power 022 c Identify the null and alternative hypotheses ANS H0 mu 275 H1 mu is not equal to 275 d Overlay on the above plot a curve representing a BETTER test for the SAME value of alpha ANS Overlay a curve everywhere higher than the given one except having the same height at muo 275 preserving alpha 3 Calculate sample standard deviation s for the data 4 5 ll 12 ANS The sample mean is XBAR 4511124 8 s root 482 582 1182 1282 41 root50 3 4 a A particular onesided ztest of an H0 to the left of H1 and having alpha 005 rejects H0 if the test statistic EXCEEDS which tabled value ANS zalpha z005 1645 b Consider a TWOsided ttest NOT ztest of H0 mu 16 ounces with n 10 and alpha 01 Evaluate the tthreshold for rejection applicable if the population is normal ANS DF 9 and talpha 2 t005 1833 0 Refer to 4b If the test statistic turns out to equal l9 what action is taken by the test ANS Two sided test rejects H0 if ABSOLUTE VALUE of test statistic exceeds threshold 1833 So it rejects H0 1 Refer to 4b An initial sample of n0 10 has sample standard deviation s0 61 What is the recommended total sample size n to achieve alpha 01 and at mu 165 beta 0025 Refer to page 314 Remember the test is TWOsided ANS For the two sided test t0 talpha 2 1833 but t1 tbeta t0025 2262 in all cases even for the two sided From the formula of page 314 we use s from a preliminary sample in place of sigma which is typically not known n t0 t1 sOmuO mul 2 1833 2262 61 16 165 2 2496 6 Refer to 4d If the sample mean of all n 2496 is XBAR 1623 what is the value of the Hybrid Test Statistic What is the action taken by the test ANS The Hybrid Test Statistic 1623 16 61 root2496 1884 Notice that the HYBRID TEST employs XBAR from all n 2496 and also employs root2496 but it sticks with 50 61 from the initial sample of 10 This twosided test rejects H0 if the ABSOLUTE VALUE of the HYBRID test statistic exceeds 1833 determined from the initial sample size 10 So it does reject H0 since 1884 gt 1833 5 Typically p 050 of ecustomers shop after 6 pm EST We decide to test the null hypothesis H0 p lt or 05 versus H1 p gt 05 with alpha 005 a An initial sample of 100 ecustomers finds 54 who shop after 6 pm Determine the numerical value of the test statistic based upon pHAT which you must identify Use the form of the test statistic which estimates sd by rootp0 q0 Reduce the test statistic to a number and be sure to say if it is positive or negative ANS pHAT 54100 054 Test statistic pHAT 05 root05 05 root100 054 0505 10 04 20 08 positive b Determine the rejection threshold of a ztest for 5a and state which action reject H0 or fail to reject H0 is taken ANS For this one sided test we reject H0 for values of the test statistic greater than the z for which PZ gt z alpha 005 That value is z 1645 sec DF in nity in ttable Since TS 08 does not exceed 2 1645 we fail to reject H0 c We ve failed to reject in b but maybe it is really true that pHAT gt 05 and we simply did not gather enough data to gain the needed precision to reject H0 What 11 do we need in order to utilize a Hybrid Test for which alpha 005 but also for p 053 beta 008 ANS See pg 319 20 1645 and 21 141 since PZ gt141 008 Continue sampling to n z0 rootpO q0z1rootp1q1p0p12 1645 root05 05 141root053 0475 53 2 2589 round up d If we do continue sampling to n 25 89 as recommended in c and if we find 13382589 customers who etransact after 6 pm what is the Hybrid Test Statistic and what action is taken ANS The Hybrid TS HTS is pHATfmal p0rootp0 q0 nFINAL and the test rejects H0p lt 05 if this HTS gt 1645 the same value of z used by the initial test HTS 13382589 05root05 05 2589 171 Since this exceeds 1645 the test based on the more stringent criteria and consequently larger sample rejects H0 e Sketch the general appearance of Prej H0 l mu for this one sided test having alpha 005 and at p 053 beta 008 Clearly identify alpha beta p0 pl null and alternative hypothesis as recognizable entities in your sketch ANS The general shape ofl 6 A test statistic for a ztest evaluates from the data to 314 a If the hypothesis is H0 p 04 and the alternative is H1 p is not 04 What is the numerical value of pSIG ANS For a two sided test pSIG is P Z gt 314 2 PZ gt 314 2 05 04992 Always pSIG is the probability of data more disagreeable with H0 than is our data SAM b Refer to 6a If pSIG 022 and alpha 032 What action is taken by the ztest and why ANS Always the rule is to reject H0 if pSIG lt alpha In this case we reject H0 ASSIGNMENT BELOW Assignment due 102606 in recitation l A test of H0 mean income lt 29 versus Hl mean income gt 29 has alpha 005 and for mu 295 has beta 01 a Sketch the general appearance of a plot of Preject H0 l mu vs mu Clearly indicate alpha and beta 29 295 as recognizable entities in your sketch which should blow up for detailed view the vicini of the interval 28 31 b Overlay on a the plot of Preject H0 l mu vs mu for the IDEAL case in which we could census the entire population ie actuall know if H0 1s true or not c Overlay on a the likely shape of a better test larger POWER also having alpha 005 but based on a larger sample size ans To maintain the same alpha our better test must also pass through the point mu 29 at height alpha 005 But everywhere on H0 in this case left of 29 the plot must be lower than our first plot yet higher than our first plot for values mu in H1 ie mu gt 29 The better plot more closely approximates the ideal plot and is generally available using a test having a larger sample size It is also achievable using a test based on a better statistic such as a regression estimate We won t go there although it is much the same as what we are doing just use the appropriate estimate of sd of the estimate in defining the test statistic eg TS muHATregr mu0 sy root1rhoHAT2n d Assume a preliminary sample of 100 is available from which the sample sd is s 61 Give the sample size n needed for a z based Hybrid Test with alpha 005 and at mu 295 beta 01 Clearly identify z0 z1 mu0 mu1 ans The formula of pg 314 gives 11 z0z1 s0 mu0 mu1 2 16451282 61 29 295 2 1276 where z0 zalpha z005 1645 positive since H0 lt H1 z1 zbeta z01 1282 even had test been 2 sided Had we used the s0 061 originally given but changed by me we d have gotten 11 13 which being less than our 11 of 100 would have indicated we already had more thna enough samples to achieve both the desired alpha and beta e If the sample of larger size d is obtained and we find that the overall sample mean for this larger sample is XBARfinal 2903 what is the value of the Hybrid Test Statistic and what action is taken by the test ans HTS XBARfinal mu0s0 rootnFlNAL 2 A test of H0 mean income 29 versus H1 mean income is not 29 has alpha 005 and for mu 295 has beta 01 a Sketch the general appearance of a plot of Preject H0 l mu vs mu Clearly indicate alpha and beta 29 295 as recognizable entities in your sketch which should blow up for detailed view the vicini of the interval 28 31 u l 1 quot b Overlay on a the plot of Preject H0 l mu vs mu for the IDEAL case in which we could census the entire population ie actuall know if H0 is true or not c Overlay on a the likely shape of a better test larger beta also having alpha 005 but based on a larger sample size H 39 7 quot 1 1 d Assume a prelimina sample of 100 is available from which the sample sd is s Give the sample size n needed for a z rec476706nb l These notes for the chapter 8 assignment do not provide the solutions They offer insights into some practical issues raised by the problems For large positive 2 the probability zTailz PZ gt z is rather well approximated by 22 1 zTailz E a 2 and in fact the ratio of the left and right sides tends to 1 as 2 tends to infinity So you have the means to evaluate pSlG for really large 2 Some of these exercises have standard scores like 10 or 13 Here 3 denotes the constant 2718281828 As shown below this is not so accurate for z 196 which should give 0025 But is better for z 2576 which from ttable should give 0005 When we move up to z 50 nearly the largest entry of the ztable we get very close to the 00000003 obtained from the table Ne 15 2 71828182845905 1 1 1 2 39 2 2 e 196V2n 2576V2n 50 V27r 00298168 000561079 297344 X 1077 8 2 In the broader sense we are comparing 5 5 with zero The appropriate standard score is a whopping 13749 Whether by test or by C1 the evidence is surely very strong We shall likely use the Z methods for CI and for testing the hypothesis of no difference between the two makes This is because 40 is a fairly large sample size and we have no reason to believe the population scores are normal In addition to a CI it seems a little silly to use a 95 CI when the data is so convincing you should calculate pSIG which will be tiny indeed for the appropriate two sided test What is the population Since the drivers have been selected at random from some population of drivers the statistical analysis will apply to that popula tion of drivers rec476706nb 2 Could drivers do best with the first car they drive Perhaps one39s senses are sharpest for the first drive To blunt such criticisms one might arrange it so that each brand is driven first in a randomly selected one half of the 40 tests Then the population is all possible allocations of first driven brand for all pos sible drivers The score is still 1 Mazda time Nissan time If we were using only a fixed set of 40 professional drivers the population would be all possible allocations of first driven to these fixed drivers It is known that the z test and CI perform well in all three of the applications above 5 23Sqrt4o 13749 8 4 The main point is that this z test is one sided because we are seeking to show that the program incentives reduce consumption The data is very con vincing Do state the null and alternative hypotheses and the form and evalua tion of the test statistic and conclusion of the test You should also report pSIG 02 01Sqrt60 154919 8 6 Why are we interested in d proportion invested HK after Oct 15 pro portion invested HK before Oct 15 It seems we are interested in investor39s habits regardless of the size of their portfolios The sample size 25 is perhaps a little small to justify z methods As long as you keep 5 4 sd 2 and the boundary of the null and alternative hypotheses in the same units you need not worry over how to record these percentage scores It is not about 0 1 data however for we are simply recording scores that are expressed as percents Do state the null and alternative hypotheses and the form and evaluation of the test statistic and conclusion of the test You should also report pSIG The evidence seems overwhelming rec476706nb 3 4 2Sqrt25 10 8 8 The sampling unit is TV program The score is d rating by men rat ing by women These scores 1 are believed to be normally distributed so t methos are appropriate You must determine n D and sd yourself Do state the null and alternative hypotheses and the form and evaluation of the test statistic and report pSIG Remember pSIG represents the chance of getting more evidence against the null hypothesis than this sample has provided if the true pd is equal to the bounbdary of the null and alternative 8 10 This is a two sample problem The target is pNikon pMinolta the differ ence between the mean ratings of the two camera models in the population of all photographers comprising the population from which each sample of 30 was selected We39ll apply the Z method with the caution that the two sample sizes of 30 each are a little lower than we would like This could be a problem if some few photographers give one or the other camera model unusually low or high scores such as photographers who cannot abide the lack of any capabil ity to mechanically preview the diaphram closure on the Minolta Do state the null and alternative hypotheses and the form and evaluation of the test statistic and conclusion of the test You should also report pSIG The evi dence seems weak 132 30 2 2 8578Sqrt 138621 8 12 I find the statement of the problem vague We are not clearly told what the nature of the chart is These mis givings aside it is intended to be a two sample problem Apply the Z method with the caution that the two sample sizes of 35 each are a little lower than we would like This could be a problem if some few investment spikes up or down unduly affect the means Do state the null and alternative hypotheses and the form and evaluation of the test statistic and conclusion of the test You should also report pSIG The evi dence is not overwhelming although some test at modest a may reject a hypothesis of no difference rec476706nb 4 9002 8002 35 35 3200 zsoo Sqrt 196521 8 14 You are being invited to apply a two sample t test for small sample sizes n1 n2 13 The two sample means and sd are to be determined from the data given Do state the null and alternative hypotheses and the form and evaluation of the test statistic and conclusion of the test You should also report pSIG hote117 11 14 25 9 18 36 19 22 24 16 31 23 17 11 14 25 9 18 35 19 22 24 15 31 23 Length hotel 13 NSqrt1312 SqrtMeanhote1quot2 Meanhote1 quot2 762166 8 16 You are being invited to apply a one sided two sample Z test Do state the null and alternative hypotheses and the form and evaluation of the test statistic You should report pSIG Remember What you hope to quotprovequot is the opposite of the null hypothesis Here quotprovequot only means that the oppo site of What you promote the null hypothesis rarely yields sample data more disagreeable With the null hypothesis than your data is 4612 5602 183869 105022 S rt q 100 80 10 141 rec476706nb 5 I IN THE FOLLOWING PROBLEMS THE CALCULATIONS HAVE BEEN CORRECTED FROM THE ORIGINAL POSTING IN WHICH A SYSTEMATIC ERROR AN INAPPROPRIATE SQUARE OF pq VALUES WAS INDUCED AND PROPAGATED BY CUT AND PASTE TO BE VERY CLEAR ON THIS POINT THERE IS A SQUARE JUST ABOVE BECAUSE WE NEED VARIANCES NOT SDS UNDER THE SQUARE ROOT IN THE 01 DATA CASES BELOW ALL VARIANCES ARE GIVEN IN TERMS OF pq VALUES 8 30 You are being invited to apply a one sided two sample z test Do state the null and alternative hypotheses the form and evaluation of the test statis tic and conclusion reached by the test You should report pSIG Remember what you hope to quotprovequot is the opposite of the null hypothesis Here quotprovequot only means that the opposite of what you promote the null hypothesis rarely yields sample data more disagreeable with the null hypothesis than your data is The evidence is overwhelming Ns50 1000 1950 2500 085 078 035 015 078022 In1 8578SqrtT 2500 own 499822 1 095222 499822 V 2 7r mus 300108 X 10 7 In6 8 32 You are being invited to apply a one sided two sample z test Do state the null and alternative hypotheses and the form and evaluation of the test statistic and the decision reached You should report pSIG Remember what you hope to quotprovequot is the opposite of the null hypothesis Here quotprovequot only means that the opposite of what you promote the null hypothesis rarely yields sample data more disagreeable with the null hypothesis than your data is The evidence is insubstantial in comparison with the very large sample sizes With such values as 13 we are wary of using z for sample sizes like 30 but can be comfortable with such large sample sizes as we have here 019 081 013 087 5000 2060 In2 19 13 05 Sqrt Out2 108032 rec476706nb 6 8 34 You are being invited to apply a one sided two sample Z test Do state the null and alternative hypotheses and the form and evaluation of the test statistic You should report pSIG Remember what you hope to quotprovequot is the opposite of the null hypothesis Here quotprovequot only means that the oppo site of what you promote the null hypothesis rarely yields sample data more disagreeable with the null hypothesis than your data is 8 36 You are being invited to apply a two sided two sample Z test I am wary of the values 0075 and 0072 which strain the CLT but for the sample sizes of 1000 are probably ok Do state the null and alternative hypotheses and the form and evaluation of the test statistic You should report pSIG Remember what you hope to quotprovequot is the opposite of the null hypothesis Here quotprovequot only means that the opposite of what you promote the null hypothesis rarely yields sample data more disagreeable with the null hypothe sis than your data is The evidence is weak 0075 0925 0072 0928 In3 075 072Sqrt 100 100 own 0257057 8 38 You are being invited to apply a two sided two sample Z test Do state the null and alternative hypotheses and the form and evaluation of the test statistic and decision reached You should report pSIG Remember what you hope to quotprovequot is the opposite of the null hypothesis Here quotprovequot only means that the opposite of what you promote the null hypothesis rarely yields sample data more disagreeable with the null hypothesis than your data is Taking the orientation East West I give below the approximation of PZ gt m where m is the test statistic You need it to determine pSIG for this two sided test Re calculate showing what m is equal to As you can see the evidence is far more overwhelming than the a 005 test reveals You might look at a 99 CI for p1 p2 East West in this case 0551 0449 0483 0517 1000 1000 39 In4 m 551 483Sqrt STT Week 101606 Read chapter 7 except testing population variances pg 301 Most important skills are given in pink Answers in yellow 1 Each package of a shipment is scored by X its content weight We wish to test the hypothesis that muX is at least 128 pounds versus the alternative that muX lt 128 The test is to have signi cance level alpha 001 The test will be based on a with replacement sample of 50 packages from the shipment N Is this is a one sided or a two sided test I w A 7 0 11 Meieqn n 1111 an39 1391 thy 1 m 1139 391quot I Suppose that a sample of 50 packages is selected with replacement from the shipment and we nd XBAR 122 this suggests an underweight shipment since 122 is well below 128 with sample sd s 11 What is the value of the test statistic for this data and what action re39ect H0 or do not re39ect H0 is taken b the test 39 quot39 1 I This test might be part of a contract allowing the receiver to literally reject the entire shipment if the null hypothesis is rejected According to the way the test has been designed what percentage of shipments meeting the muX 128 standard will be falsely re39ected I Evaluate the probability pSIG that we d see an XBAR at least as low as the 122 our samle ave iftrul mu 128 Camila it h lni39z I h What if the test is fed a sample of 50 from a shipment with muX 120 How freuentl would it re39ect such a trul underwe39 ht shiment 2 Suppose that the sample size in 1 had been 11 5 and the population distribution is normal As in I suppose XBAR 122 and SK 11 for the sample of 5 from a normal population quotw 7 J 139 M IU Gal Jilin mm m l 1 1i I Suppose that a sample of 5 packages is selected from the shipment and we nd XBAR 122 this suggests an underweight shipment since 122 is well below 128 with sample sd s 11 What is the value of the test statistic for this data and what action reject H0 or do not reject H0 is taken by the test I This test might be part of a contract allowing the receiver to literally reject the entire shipment if the null hypothesis is rejected According to the way the test has been designed what percentage of shipments meeting the muX 128 standard will be falsely rejected vw quott mi nl 393 3 A restaurant is accustomed to having around 20 of order for sh A check of a with replacement sample of 40 orders nds 12 for sh I Give the form and evaluate it of test statistic for a test of the hypothesis that the fraction of orders for sh is 02 versus the alternative that p is not 02 17 atistic when d we 7 M I From a and d what action is taken by the test Assignment due in recitation 101906 from your book Use ztest in all cases KEEP IN MIND THAT THE PROBLEMS FROM YOUR TEXTBOOK ARE INTENDED TO BE SOMEWHAT VAGUE ENCOURAGING YOU TO THINK ABOUT THE KIND OF TEST TO USE I VE OFFERED SOLUTIONS BASED ON CONVENTIONAL INTERPRETATIONS FOR MY OWN PROBLEMS AND TEST QUESTIONS I TRY TO BE VERY EXPLICIT ABOUT WHAT TEST I AM ASKING FOR 716 Except change the language of the problem to whether the new engine increases the milespergallon a One or two sided m Q tw 39 J 61 L vvul he39 I 7 I 139 393 iat iLerIFl39I n39w b Use tentry for 005 or 0025 11 u W11 W39me h Law j H H 7 w wa 718 a One or two sided Eff J m mdh a s xgyzh m c Form 7 51quot r J 7 d Reject H0 or not 0 Form and eval Tl 728 Change the language to that it still controls at least 42 a One or two sided 7 39 quot391 lessquot 7 0 Form and evaluation of test statistic J 7 7 l 7111 1 1 111 2h M L um 3mg Recitation assignment due 92106 1 RV X is normally distributed with E X 23 and Var X 16 RV Y is normally distributed with E Y 21 and Var Y 9 X Y are independent rv a Determine the expectation E W we may just refer to it as the mean and standard deviation of W X 7Y Remember Var Y Var Y Sketch the densit for W and identify the mean and sd in your sketc 7 E W E X 7 E Y 23 7 l6 additivity of expectation even 1 independence Var W Var X VarY Var X l2 Var Y 16 9 additivity of variance for sums of indep rv b SameasaforU5X2Y77Y EU5EX2EY77EY523221721 VarUVar5XY725VarXVarY25l69 Because 2Y is m I i of 7Y it would be i to calculate Var U as Var5X Var2Y VarY Variance may not add for sums of dependent rv 2 Independence means that for every pair of values X py l X p ll in the missing lette a K 39l l mu For continuous models such as the normals py l X pX y pX where pX is the probability density for X evaluated at value X an eXample being eXpression 42 evaluated at X 95 say and pX y joint density for X Y something we ve not discussed For the question above what letter would you put in p I if things were formally like the discrete case where py l X PY y l X X Why 3 Z denotes a rv having the standard normal distribution a PZ 12 I continuous models distribute probabilities by area under a curve and there is zero area under the standard normal density above the single point z 12 b PZlt0 05 the zdensity is symmetric about its mean which is zero In the table 2 09 00 00000 PO lt Z lt 009 c E Z 0 the standard normal Z has E Z 0 and sd Z l d sd Z 1 see c e What familiar rV has the distribution of W 15 Z 100 Hint calculate the expectation and sd of W EW10015EZ100 Var W Var15 Z 152 1 Sd W 15 Same as IQ which is modeled as normal mean 100 sd 15 f P0 lt Z lt 148 04306 08 z 14 04306 P0 lt Z lt148 g PZ gt 213 PZ gt 0 7 P0 lt Z lt 213 05 7 04834 draw picture 03 z 21 04834 P0 lt Z lt 219 h PZ lt 132 PZ lt 0 P0 lt Z lt132 05 04066 02 z 1 3 04066 P0 lt Z lt 213 i Find the 84Lh percentile on Same as z with P0 lt Z lt z 034 picture it So enter 034 to the inside body of the ztable and read off the value of z You may not nd 034 in the body ofthe ztable so choose the closest table entry 02 Z 09 So 2 099 is the 84th percentile of the standard normal ie PZ lt 099 N 084 Recall the Rule of Thumb says around 68 of any normal falls within one sd of its mean j Use i to nd the 843911 percentile of sales ifmean of sales is 3400 and sd is 944 Sales mu z sigma 3400 099 944 k Find the 21St percentile of IQ We need 2 with PZ lt z 021 This is the same as nding 2 with PZ gt z 021 and taking its negative This is the same as nding 2 with P0lt Z lt z 05 7 021 029 and negating it 01 z 08 So 2 081 remember we had to negate it ie PZ lt 081 N 021 4 Models for income neXt quarter have it normally distributed with mean 34 and sd 7 Give the standard score for income X 41 z Ximusigma4l 7347 E Fquot Give the standard score for X 29 to two decimals z 29 7 34 7 57 07l Use b to determine Pincome gt 29 Pincome gt 29 PZ gt 07D 05 P0 lt Z lt 071 05 01 O i ll picture it a 11 B a E D u 0010 1quot 13 D H O D 3 D 5quot H B O O B FD 5 g U D U 4 N 1 Q 27 D H D quotU A O N N V H 0 O So the 58Lh percentile of ncome is 34 020 7 5 The fraction of employees who ve had clashes with customers is 014 A with replacement sample of 64 employees has a number X who ve had such clashes s the name for th 7 ll L 1 2 1 Li u39 What i b Write the expression for PX X pX and evaluate it to determine pl3 x pX 1p 64 13 511 01413 08651 Determine the mean and sd of rv X mean of binomial n p is equal to np 64 014 i quot3 O sd ofbinomial is equal to rootn p q root64 0 14 086 m d Use c and Ztable to give the normal appr0Ximation of pl3 employ the 05 pl3 7 P z in 125 7 896 27759 125 7 896 27759 PZ in 128 164 PZ in 0 164 7 PZ in 0 128 04495 7 03997

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