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Apr 3, 2024

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1. Your professor gives one of two types of exams, easy tests and hard tests. e 50% of the time he gives hard tests, 50% of the time he gives easy tests Easy tests have an average score of 80 and a standard deviation of 5. e Hard tests have an average score of 50 and a standard deviation of 15. Both types of tests produce normally distributed scores P(S > 95|H)P(H) P(z>3)x 5 (d) You have another friend of unknown ability. She reports that she got a score above 95 on the test. What is the probability it was a hard test? 0.5 P(S>9) P(z>3)x5+Pz>3)x5 — © K (b) (c) You have a friend of unknown ability. She reports that she got a score above 80 on the test. What is the probability it was an easy test? What is the probability of getting a score above an 80 on a hard test? S —50 _ 80— 50 P(S > 80|H) = P ( P(S>80)=P(S>80NH)+P(S>8NE) P(S > 80) = P(S > 80|H)P(H) + P(S > 80|E) P(E) P(S >80) = (1— F.(2)) x .5+ P(S > 80|E) x .5 S—80 _ 80 —80 P(S > 80|H) = (1 — F.(2)) x.5+ P(S > 80|E) x.5 N———r P(S > 80|E)P(E) 5% .5 2. Your professor gives one of two types of exams, easy tests and hard tests. o 75% of the time he gives hard tests, 25% of the time he gives easy tests o Easy tests have an average score of 85 and a standard deviation of 5. e Hard tests have an average score of 70 and a standard deviation of 10. > |H)=P(z>2)=1—Fz(2) What is the probability of getting a score above an 80 on a random test given by the professor? e Both types of tests produce normally distributed scores (a) What is the average score on all tests given by the professor? 75 x 70 4 .25 x 85 (b) What is the probability of getting a score above a 90 on a test given by the professor? 5 > TIH) =P(z>0)=1-F,(0) =.0228 =5 P(Score > 90) = P(Score > 90|Easy) x.25 + P(Score > 90|Hard) x.75 — — P(z>1) P(z>2) P(Score > 90) ~ .16 x .25 +.025 x .75 = 05875 The exact answer if using CDF the of the z distribution is 0.1586553 x .25 + 0.02275013 x .75 = 0.05672641 (c) What is the probability of getting a score below 807 P(Score < 80) = P(z < —1)x.25+P(z < 1)x.75 ~ .16x.25+.84x.75 = .67 The exact answer would is actually 0.6706724 (d) You have a friend of unknown ability. She reports that she got a score P(S>80) 0228x5+5x%x5 P(Easy|Score > 90) = 1 (e) What is the probability it was a hard test? P(Hard|Score > 90) = 1 — P(Easy|Score > 90) = .301 3. You work for a small insurance company. Your company is currently insuring 900 different cars for $10,000. If a car you are insuring gets in a crash you have to pay $10,000. Each car you insure has a 5% chance of crashing each year. If 50 cars you insure crash in a given year, you have to pay 50 x $10,000 = $500, 000 in insurance payouts that year. (a) What is the expected value of the amount your company would pay out? Answer Let C be the number of insured cars that crash, and therefore 10,000C is the amount that gets paid out. C' is a binomial random variable. E[10,000xC] = 10,000x E[C] = 10,000xnxp = 10,000x900x.05 = 450, 000 (b) What is the standard deviation? o = /10,000 x 900 x .05 x (I — .05) = 65, 383.48 (c) What is the probability your company has to pay out more than $550,000? Answer Convert to a z score: above 90 on the test. What is the probability it was an easy test? 25 x P(z> 1) 25x P(z>1) +.75 x P(z > 2) 10,000 x C' — 450,000 _ 550,000 — 450, 000 P(10,000xC > 550,000) = P ( 65.383.48 65,383.48 P(10,000 x C > 550,000) = P(z > 1.5294) ~ .063 (d) If your company pays out an average of $550,000 per year over the next 4 years (or more) it will go bankrupt. What is the probability this will happen? Answer The average, C' over the course of 4 years will have half the standard deviation. 10,000 x C' — 450,000 _ 550,000 — 450, 000 P(10,000xC > 550,000) = P ( ) ( 65,383.48/v/4 65,383.48/v/4 P(10,000 x C' > 550,000) = P(z > 3.0589) =.0397/.0567 = .699 4. You have a falafel cart and you sell falafel every weekday near Washington Square Park during lunch time. Your daily revenue is normally distributed with a mean of $200 and a standard deviation of $50. (a) Suppose there is another location that might be worth switching to. You plan to experiment with selling there for awhile, and then use a hypothesis test to determine whether you should switch. If the new location has a normally distributed revenue with a true mean of 210 and a standard deviation of 50, how many days would you have to try selling there to have a power of 70%. Use an a = .05 (significance level). Answer: You are trying to determine whether to switch, so a one-sided test makes sense. Your null hypothesis is that the new location has the same average revenue as the old place ($200). Your o = .05, so your cutoff to reject the null hypothesis, §* will be determined by the function: g —200 e 50 —50/\/5 =1.64 - 7" =200+ 1.64 x n We also know that our power needs to 50%, if the mean revenue in the new location is $210, our sample mean has to be greater than 3* 50% of the time: PY>y)=25 ‘We can convert both sides of the inequality to z scores: Y —210 _ 5* —210 > ) = P( 50/vn > 50/yn ) ° We know that P(z > 0) = .5, so we need: Lt 0y =210 Plugging this value into the equation above, we can find the n that will allow us to have 50% power: 210 = 200+ 1.64 x 2L 5 10v/n = 164 x 50 > n > (5 x 1.64)2 = 67.24 n 7 5. There are two candidates running for president. Candidate A and Canditate B. In a poll of 400 voters, 208 support A and 192 support B. 4

7. You find a penny on the street. You want to test if the penny is fair, meaning it has an equal probability of landing on heads or tails. You flip it 100 times. (a) You decide that if 60 or more of the flips land on heads or 60 or more of the flips land on tails, you will conclude that the penny is not fair. Answer (a) Construct a 95% confidence interval for the support received by A. Answer: P(H > 60|H < 40) =1 — P(40 < H < 60) Let p be the actual support for candidate A, and p be our estimate based We know that E[H] = 50 and oy — vI00 X 5 X 5 = 5 Therefore on our sample: . 208 P= 700 =2 P(40§H§60)=P(@§z5@)=P(—25z§2) p is essentially a binomial random variable, divided by n, so its variance is the following: (b) What is the probability of making a Type 1 error if the penny is fair. Answer. Let p be the probability of a penny landing on heads. If the Var(p) = Var (Binomial RV/n) = 1 /n2Var(Binomia.l RV) = %np(l—p) _ p(1-p) penny is fair, then p = .5. The number of heads is a binoqxial random n n variable, therefore the expected value of the number of heads is 100p = 50 We don’t know p, but we have an estimate of it with p, so we can estimate and the standard deviation is the standard deviation p by plugging it into the formula: Vnp(1—p) =+/100 x 5x (1—5) =v25=5 ~ [52(1-52) VVar(p) = a0~ .025 6 We know that p will be approximately normally distributed around p with a standard deviation of approximately .025: P(p—2x.025<p<p+2x.025)=.95 (using 1.96 instead of 2 is fine) This inequality can be rearranged into a 95% confidence interval for p: P(p—2x.025<p<p+2x.025)=.95 P(52—2x.025 <p< .52+2x .025) = .95 — P(47 < p < .57) = .95 (b) Estimate the probability that candidate B wins (gets more than 50% of the vote) Al : W d to find the iated with p = .5 nswer: e need to i © 2 score assoclatec Wil p We know that if n is relatively large, than a binomial random variable is 5—p 5— .48 approximately normally distributed. This means that approximatel 95% z= \/Tr(fi =025 8 of the distribution falls within 2 standard deviations of the expected value or mean. Our rule is that we reject the null hypothesis if we get 60 or 1— F(8) = F(—.8) = .2118 more heads or 40 or fewer heads. Candidate B has approximately a 21 percent probability of winning the z> G()—TSO =2o0rz< 4 ; 50 =— election based on our sample. R e Al ] (c) What is the probability of making a Type 2 error if the penny has a 55 percent chance of landing on heads? Answer: We need to find the probability of getting less than 60 or more than 40 heads if the probability of getting a head is 55%. The number of heads is approximately normal (large n), so we need to find the z scores associated with 40 and 60 heads to estimate this probabiliy. The mean number of heads if p = .55 is 55, and the standard deviation will be: V/np(1 —p) = /100 x .55 x (1 — .55) =4.97 ~ 5 60_55~1andz> 40-55 497 © 497 "~ P(Type 2 Error|p = .55) = F(1)-F(-3) = (1-F(-1))-F(-3) = (1-.1583)—.0013 ~ .84 Therefore the probability of making a Type 2 error if p = .55 is 84%. (Again a precise answer using the z table is acceptable and preferred). z <

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