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Question 1 1.5 points Save Answer If the mean and standard deviation of a data set are 76 and 12 respectively, then which value is Unusual (using Range rule of thumb)? Formulae Sheet STAT-101 Minimum "usual" value = x - 2s P(x) = - n! p* qn-x (n-x )ix! Ex E(x-2)2 S = n n-1 Margin of Error, F _ upper CL-lower C.L. Maximum "usual" value = x + 2s Variance = npq 2 Point estimate = Upper CL.+ Loer CL Relative Frequency = Frequency of the class Total Frequency E =ExP(x); E=np P (A or B) = P(A) + P(B) - P (A and B) Class-Width = (Max Value- Min Value)No. of classes 7 = O A. 105 O B. 85 O C. 95 O D. 65Question 2 1.5 points Save Answer The following frequency distribution represents the amount of sugar in mg/It for a sample of bottled drinks: Class Limit Frequency 5-9 3 10-14 S 15-19 2 20-24 8 25-29 14 30-34 8 Then the relative frequency of the class 15-19 is: Formulae Sheet STAT-101 Minimum "usual" value = x - 2s P(x) = n! (n- xe )have! p* qn -x I = E(x-7)2 S = n-1 Margin of Error, E = " upper CL.-lower C.L. Maximum "usual" value = x + 2s Variance = npq 2 Point estimate = Upper C.L.+ Loer C.L Relative Frequency = Frequency of the class Total Frequency E =ExP(x); E =np P (A or B) = P(A) + P(B) - P (A and B) Class-Width = (Max Value- Min Value)/No. of classes10-14 15-19 20-24 8 25-29 14 30-34 8 Then the relative frequency of the class 15-19 is: Formulae Sheet STAT-101 Minimum " usual" value = x - 2s P(x) = n! (n-x) p* qn-x E(x-7)2 S = n n-1 Margin of Error, E _ upper CL.-lower C.L. Maximum "usual" value = x + 2s Variance = npq 2 Point estimate = - Upper C.L.+ Loer C.L Relative Frequency = Frequency of the class Total Frequency E =EXP(x): E = np P (A or B) = P(A) + P(B) -P (A and B) Class-Width = (Max Value- Min Value)/No. of classes O A. 0.05 O B. 0.2 O C. 0.5 OD.2Question 3 1.5 points Save Ans For a sample data set, if )(x - X)- = 81 and n =10, then the variance and the standard deviation for this data are given respectively by Formulae Sheet STAT-101 Minimum "usual" value = x - 2s P(x ) = = 7! (n-x)lx: P* qn-x Ex E(x-7)2 n S = n-1 Margin of Error, E = upper C.L.-lower C.L. Maximum "usual" value = x + 2s Variance = npq 2 Point estimate = Upper C.L.+ Loer C.L Relative Frequency = Frequency of the class Total Frequency E =EXP(x): E = np P (A or B) = P(A) + P(B) - P (A and B) Class-Width = (Max Value- Min Value)/No. of classes Z = OA. 3 & 9 OB. 1 & 1 O C.4 &2 O D.9 & 3Question 4 1.5 points In an experiment of 16 trials and each trial has only two possible outcomes, if the probability of success i trial is 0.5, then the standard deviation of Binomial probability distribution is Formulae Sheet_STAT-lll \"\"3\"" \"\"3' mm \"1\"\" mewwmm Hm P= Wm \"mm 0&2 08-16 0&4 Question 5 1.5 points Save Ansy If P(Z -1.65) = 0.0495, then P(-1.65 1.96) is equal to TABLE A-2 (continued) Cumulative Area from the LEFT .01 .02 .03 .04 .05 .06 07 08 .09 0.0 5000 5040 5080 5120 .5160 5199 .5239 5279 5319 .5359 0.1 5398 5438 .5478 .5517 .5557 .5596 5636 5675 5714 5753 0.2 5793 5832 5871 5910 5948 5987 6026 6064 6103 6141 0.3 6179 6217 6255 6293 633 6368 .6406 .6443 .6480 6517 0.4 6554 6591 6628 6664 6700 6736 6772 6808 .6844 6879 0.5 6915 6950 6985 .7019 7054 7088 7123 .7157 7190 7224 0.6 7257 .729 7324 .7357 7389 7422 7454 7486 .7517 7549 0.7 7580 .7611 .7642 .7673 7704 .7734 7764 .7794 .7823 .7852 0.8 7881 7910 .7939 7967 .7995 8023 8051 8078 .8106 8133 0.9 8159 8186 8212 .8238 .8264 8289 .8315 8340 .8365 .8389 1.0 8413 8438 .8461 .8485 .8508 8531 8554 8577 .8599 .8621 1.1 8643 8665 8686 8708 .8729 .8749 8770 8790 8810 8830 1.2 8849 8869 .8888 .8907 .8925 8944 8962 .8980 8997 .9015 1.3 9032 9049 .9066 9082 9099 9115 131 9147 9162 9177 1.4 9192 9207 9222 .9236 9251 .9265 9279 9292 9306 9319 1.5 9332 9345 9357 .9370 9382 9394 9406 9418 9429 .94 41 1.6 9452 9463 .9474 9484 9495 * .9505 9515 .9525 .9535 9545 1.7 9554 9564 9573 9582 9591 .9599 9608 9616 9625 9633 1.8 9641 9649 9656 .9664 9671 .9678 9686 9693 .9699 .9706 1.9 9713 9719 9726 .9732 9738 .9744 9750 .9756 9761 9767 2.0 9772 9778 9783 9788 9793 9798 9803 .9808 9812 9817 2.1 982 9826 9830 9834 9838 9842 9846 9850 9854 9857 O A. 0.9520 O B. 0.9350 O C.-0.9345 O D. 0.9255Question 6 Find the original data from the stem-and-leaf plot: Stem Leaves 2 247 3 5 5 4 0 O A. 12, 12, 14, 17, 33, 35, 40 O B. 12, 14, 17, 33, 35, 35, 40 O C. 12, 12, 14, 17, 33, 35, 35, 40 O D. 12, 12, 14, 17, 20, 33, 35, 35, 40Question 1' 1.5 points - The data regarding the gender and majors based on a sample of 100 undergraduate students at a university are given as follows: Major STAT] 01 STATZOI If a student is randomly selected, then the probability of selecting a STATED] major, given that the student selected is female, is: O A.o_4 O B, (15 O C.o_3 O D.0_5 Question 8 1.5 points Save Answer Suppose that a researcher asked 100 students who plan to take a new course in their next semester. The results can be categorized as a frequency distribution as follows: Course Frequency MATH150 30 STAT101 50 STAT201 20 The probability of selecting a student who is taking MATH150 is: O A. 0.70 O B. 0.50 O C. 0.20 O D. 0.30Question 9 The random variable X has the following probability distribution: X O 2 3 P(X) 3k 0.1 0.2 2k The value of k is O A. 0.24 O B. 0.70 O C. 0.14 O D. 0.20Students have exam scores that are normally distributed with mean 78 and standard deviation 8. What is the probability that a randomly selected student has an exam score greater than 90'? _ABLE 731-2 (continued) Cumulative Area from the LEFT 2 .99 .91 .92 .93 .94 .95 .95 .97 .99 .99 9.9 .5999 .5949 .5999 .5129 .5159 .5199 .5239 .5279 .5519 .5559 9.1 .5399 .5439 .5479 .5517 .5557 .5595 .5555 .5575 .5714 .5753 9.2 .5793 .5932 .5971 .5919 .5949 .5997 .5925 .5954 .5193 .5141 95 .5179 .5217 .5255 .5293 .5351 .5359 .5495 .5443 .5499 .5517 9.4 .5554 .5591 .5529 .5554 .5799 .5735 .5772 .5999 .5944 .5979 9.5 .5915 .5959 .5995 .7919 .7954 .7999 .7123 .7157 .7199 .7224 9.5 .7257 .7291 .7324 .7557 .7399 .7422 .7454 .7495 .7517 .7549 9.7 .7599 .7511 .7542 .7575 .7794 .7734 .7754 .7794 .7923 .7952 9.9 .7991 .7919 .7939 .7957 .7995 .9923 .9951 .9979 .9195 .9133 9.9 9159 9195 9212 9239 9254 9299 9515 9349 9555 9599 1.9 .9413 .9439 .9451 .9495 .9599 .9531 .9554 .9577 .9599 .9521 1.1 9543 9555 9595 9799 9729 9749 .9779 9799 9919 9959 1.2 .9949 .9959 .9999 .9997 .9925 .9944 .9952 .9999 .9997 .9915 1.3 .9932 9949 .9955 .9992 .9999 .9115 .9131 .9147 .9152 9177 1.4 .9192 .9297 .9222 .9235 .9251 .9255 .9279 .9292 .9395 .9519 1.5 .9532 .9345 .9557 .93 79 .9392 .9394 .9495 .9419 .9429 .9441 1.5 .9452 .9453 .9474 .9494 .9495 7 .9595 .9515 .9525 .9535 .9545 1.7 .9554 9554 .9573 .9592 .9591 .9599 .9599 9515 .9525 9535 1.9 .9541 .9549 .9555 .9554 .9571 .9579 .9595 .9593 .9599 .9795 1.9 .9713 .9719 .9725 .9732 .9739 .9744 .9759 .9755 .9751 .9757 2.9 .9772 .9779 .9793 .9799 .9793 .9799 .9993 .9999 .9912 .9917 2.1 .9921 9925 .9939 .9954 .9939 .9942 .9945 .9959 .9954 .9957 O 5- 9.9332 0 3- 9.5530 0 C- 9.5909 0 0.9.0559 Question 11 Given the following cumulative frequency table, the frequency of the class 40-49 is 9. Cumulative frequency Less than 20 4 Less than 30 Less than 40 Less than 50 15 Less than 60 20 O True O FalseQuestion 12 1 points Save Answer The Mean, Median and Mode of a distribution are 5, 4 & 2 respectively. It is most likely the distribution is Positively Skewed. O True O FalseQuestion 13 1 points If P (A) = 0.65, P (B) = 0.40, and P (A and B) = 0.25, then P (A or B) = 0.90. Formulae Sheet STAT-101 Minimum "usual" value = x - 2s P(x) = (n-x)Ix! p* qn-x I = Ex E(x-)2 S = n-1 Margin of Error, E = upper CL-lower C.L. Maximum "usual" value = x + 2s Variance = npq 2 Point estimate _ Upper C.L.+ Loer C.L Relative Frequency = Frequency of the class 2 Total Frequency E =_ xP(x): E = np P (A or B) = P(A) + P(B) - P (A and B) Class-Width = (Max Value- Min Value)/No. of classes Z = O True O FalseQuestion 14 1 points Save Answer If the mean (E) of binomial distribution is equal to 8 and the probability of success in one trial is p = 0.4, then the number of trails is n= 20. Formulae Sheet STAT-101 Minimum "usual" value = x - 2s P(x) = (n-xe )Ix! p* qn-x I = Ex E(x-7)2 S = n n-1 Margin of Error, E = upper CL-lower C.L. Maximum " usual" value = x + 2s Variance = npq 2 Point estimate = Upper CL.+ Loer C.L Relative Frequency = Frequency of the class Total Frequency E =ExP(x): E=mp P (A or B) = P(A) + P(B) - P (A and B) Class-Width = (Max Value- Min Value)/No. of classes O True O FalseQuestion 15 1 points Save Answer The z-score of normal distribution is -3, the mean of the distribution is 32 and the standard deviation of normal distribution is 4, then the value of X for a normal distribution is 20. Formulae Sheet STAT-101 Minimum "usual" value = x - 2s n! P(x) = (-x Mix! Ex . S = Ex-7)2 n n-1 Margin of Error, E = upper CL-lower C.L. Maximum "usual" value = x + 2s Variance = npq 2 Point estimate = Upper C.L.+ Loer C.L Relative Frequency = Frequency of the class Total Frequency E =ExP(x); E=np P (A or B) = P(A) + P(B) - P (A and B) Class-Width = (Max Value- Min Value)/No. of classes X- M O True O FalseThe following frequency distribution represents the amount of sugar in my/It for a sample of bottled drinks: Class Limit Frequency 5-9 3 10-14 5 15-19 2 20-24 8 25-29 14 30-34 8 Then the relative frequency of the class 15-19 is: Formulae Sheet STAT-101 Minimum "usual" value = x - 2s P(x ) = - n! (n-se )ice! p* qn-x 2X S= [(x-1)2 X = = n-1 Margin of Error, E = upper CL-lower C.L. Maximum "usual" value = x + 2s Variance = npq 2 Point estimate = - Upper C.L.+ Loer C.L Relative Frequency = Frequency of the class Total Frequency E =EXP(x): E = np P (A or B) = P(A) + P(B) - P (A and B) Class-Width = (Max Value- Min Value)/No. of classes Z = : O A. 0.05 O B. 0.2 O C. 0.5 O D. 2