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Topics include binomial-to-normal approximation for a sample of investors, properties of the sample mean including expected value and variance, and joint discrete distributions with conditional expectation, conditional variance, marginal distributions, and independence testing. Additional questions address linear regression diagnostics, residual assumptions, and computation of fitted regression parameters, variance estimates, confidence intervals, and interpretation of interval width.",{"@graph":69,"@context":123},[70,84,106],{"@type":71,"itemListElement":72},"BreadcrumbList",[73,77,79,82],{"item":74,"name":75,"@type":76,"position":8},"https://docshare.wps.com","Home","ListItem",{"item":78,"name":9,"@type":76,"position":14},"https://docshare.wps.com/document/",{"item":80,"name":20,"@type":76,"position":81},"https://docshare.wps.com/document/exam/",3,{"item":83,"name":65,"@type":76,"position":19},"https://docshare.wps.com/document/cs1-specimen-questions-and-solutions-july-2020/398785/",{"url":83,"name":65,"@type":85,"image":86,"author":91,"headline":65,"publisher":94,"fileFormat":97,"inLanguage":63,"description":67,"dateModified":98,"datePublished":99,"encodingFormat":97,"isAccessibleForFree":100,"interactionStatistic":101},"DigitalDocument",{"url":87,"@type":88,"width":89,"height":90},"https://docshare.wps.com/thumbnails/cs1-specimen-questions-and-solutions-july-2020/398785.png","ImageObject",300,407,{"name":92,"@type":93},"sagouaw","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-10-28","2026-09-27",true,{"@type":102,"interactionType":103,"userInteractionCount":105},"InteractionCounter",{"@type":104},"ViewAction",11,{"@type":107,"mainEntity":108},"FAQPage",[109,115,119],{"name":110,"@type":111,"acceptedAnswer":112},"How is the distribution of X determined in Q1?","Question",{"text":113,"@type":114},"X is modeled as binomial with parameters (300, 0.4). For approximation, X is then taken as approximately normally distributed with mean 120 and variance 72.","Answer",{"name":116,"@type":111,"acceptedAnswer":117},"What is the variance of the sample mean Xbar in Q2?",{"text":118,"@type":114},"Var(Xbar) equals sigma2/n, showing that the sample mean’s variance is reduced by a factor of 1/n compared with the variance of an individual observation.",{"name":120,"@type":111,"acceptedAnswer":121},"Why does Q4 suggest the trainee may have made an error?",{"text":122,"@type":114},"The regression slope indicates a positive relationship, while the sample correlation coefficient indicates a strong negative relationship. This contradiction suggests an error in the trainee’s calculations or reporting.","https://schema.org",{"og:url":83,"og:type":125,"og:title":65,"og:site_name":95,"og:description":67},"article",{"robots":127,"canonical":83},"index,follow",{"doc_id":129,"site_id":62},398785,1790578870,{"code":4,"msg":5,"data":132},{"doc_id":129,"user_id":133,"nickname":92,"user_avatar":134,"doc_module":4,"category_id":19,"category_name":20,"doc_title":65,"doc_description":67,"doc_content":135,"file_id":136,"file_url":137,"file_type":138,"file_size":139,"view_count":105,"is_deleted":4,"is_public":8,"is_downloadable":8,"audit_status":8,"page_count":34,"language":140,"language_code":63,"site_id":62,"html_lang":63,"table_of_contents":141,"faqs":142,"seo_title":143,"seo_description":67,"update_tm":144,"read_time":145,"preview_page_count":4},3985747862151,"https://ap-avatar.wpscdn.com/davatar_a8503ba1806abce46bf441b54a3ca4cd","CS1 Specimen Questions and Solutions  \nJuly 2020  \nQ1  \nA survey showed that 40% of investors invest in at least two companies in order to diversify their risk. Let X be the random variable denoting the number of investors who have invested in more than one company in a random sample of 300 investors.  \n(i) Determine the distribution of X. [2]  \n(ii) Calculate an approximate probability that X is greater than 100. [2][Total 4]  \nSolution  \n(i) Xfollows a binomial (300, 0.4) distribution [0.5]  \nSo, approximately Xfollows a normal distribution [0.5]  \nwith E(X) = 300*0.4 = 120 and V(X) = 120*0.6 = 72 [1]  \n(ii) Using continuity correction [0.5] P(X>100) = P(X>=100.5) = 1-Phi((100.5-120)/sqrt(72)) = 1-Phi(-2.298)= Phi(2.298) = 0.989 [1.5]  \nQ2  \nLet 􀜺1, 􀜺2,…, 􀜺􀯡 be a random sample consisting of independent random variables with mean µ and variance 􀟪 2 . Consider the sample mean:  \n􀴤 ∑1 􀜺􀯜  \n􀜺 =  􀝊   \nIn your answer you may denote µ by mu, 􀟪2 by sigma2 and  by Xbar.  \n􀴤  \n(i) Write down the expected value of 􀜺 . [1]  \n􀴤  \n(ii) Derive the variance of 􀜺 . [2]  \n􀴤  \n(iii) Comment on the variance of variable 􀜺 as compared to the variance of 􀜺􀯜 . [1]  \nAn actuary is interested in exploring the difference in the size of claim losses from two insurance portfolios, and can take samples of claims from these portfolios.  \n(iv) Explain how the answer to part (iii) can affect the precision of the actuary’s comparison. [2]  \n[Total 6]  \nSolution  \n(i) E(Xbar) = mu [1]  \n(ii) V(Xbar) = V(sum_over_1_to_n(X_i)/n) = sum_over_1_to_n(V(X_i))/ n^2 because of independence [1]  \n= n*sigma2/n^2 = sigma2/n [1]  \n(iii) The variance of the sample mean is smaller compared to the variance of individual variables by a factor of 1/n. [1]  \n(iv) Individual values are less precise than the average of a sample. The actuary should take large samples and compare the means to improve the precision of the comparison.  \n[2]  \nQ3  \nX and Y are discrete random variables with joint distribution as follows:  \n\n|  | X = 0 | X = 1 | X = 3 |\n| --- | --- | --- | --- |\n| Y = –1 | 0.08 | 0.03 | 0.00 |\n| Y = 0 | 0.03 | 0.12 | 0.20 |\n| Y = 3 | 0.11 | 0.11 | 0.06 |\n| Y = 4.5 | 0.04 | 0.20 | 0.02 |\n\n(i) (a) Identify which one of the following options gives the correct value of the expectation 􀜧 (􀜻 | 􀜺 = 1) : [2]  \n(A1) 1.3789  \n(A2) 2.6087  \n(A3) 3.1398  \n(A4) 4.0945  \n(b) Identify which one of the following options gives the correct value of the variance var(􀜺 | 􀜻 = 3) : [2]  \n(A1) 1.2487  \n(A2) 0.9832  \n(A3) 1.9388  \n(A4) 2.2235  \n(ii) Calculate the probability functions of the marginal distributions for X and Y.  \n[2]  \n(iii) Determine whether 􀜺 and 􀜻 are independent.  \n[2]  \n[Total 8]  \nSolution  \n(i)(a) E(Y | X = 1)  \n= ∑􀳤 􀝕 􀜲 (􀜻 = 􀝕 | 􀜺 = 1)  \n=∑􀯬 􀝕 􀯉~~ ~~(􀯒􀯉=(􀯬􀯑,=􀯑1)=~~ ~~1)  \n= (-1 x 00..0346) + (3 x 00..1146 ) + (4.5 x 004.26)  \n= 2.6087  \nAns: (A2) [2]  \n[Details are for information. Candidates will not be required to show working.]  \n(i)(b) var(X | Y = 3) = E(X2 | Y = 3)–(E(X | Y = 3))2  \n= (1 x 00..1128) + (9 x 00..0628 )–((1 x 00..1128) + (3 x 00..0628 ))2  \n= 2.3214 –(1.0357)2  \n= 1.2487  \nAns: (A1) [2]  \n[Details are for information. Candidates will not be required to show working.]  \n(ii) Summing columns gives:  \nP(X = 0) = 0.26, P(X = 1) = 0.46, P(X=3) = 0.28 [1] Summing rows gives:  \nP(Y = -1) = 0.11, P(Y = 0) = 0.35, P(Y = 3) = 0.28,  \nP(Y = 4.5) = 0.26 [1]  \n(iii) Test whether P(X = x, Y = y) = P(X = x) P(Y = y) for all pairs [1]  \nShow that this result does not hold for one pair, for example:  \nP(X = 0, Y = -1) = 0.08, not equal to P(X = 0)* P(Y = -1)  \nSoX and Y are not independent. [1]  \nQ4  \nAn actuary is asked to check a linear regression calculation performed by a trainee.  \n􀷠  \nThe trainee reports a least squares slope parameter estimate of 􀜾 = 13.7 and a sample correlation coefficient 􀝎 = − 0.89.  \n(i) Justify why this suggests that the trainee has made an error. [2]  \nIn a different simple linear regression model, a histogram of the residuals is shown ","cbCaiciX3C4HA8p1","https://ap.wps.com/l/cbCaiciX3C4HA8p1","pdf",152753,"English","# Q1\n## (i) Distribution of X\n## (ii) Approximate P(X > 100)\n# Q2\n## (i) E(Xbar)\n## (ii) Var(Xbar)\n## (iii) Compare variances\n## (iv) Impact on actuary precision\n# Q3\n## (i) Conditional expectation and variance options\n## (ii) Marginal distribution functions\n## (iii) Independence test\n# Q4\n## (i) Regression slope vs correlation justification\n## (ii) Residuals and linear model assumptions\n## (iii) Fitted regression line\n## (iv) Variance estimates, predicted mean variance, and 95% CI\n## (v) Compare confidence interval widths","[{\"question\":\"How is the distribution of X determined in Q1?\",\"answer\":\"X is modeled as binomial with parameters (300, 0.4). For approximation, X is then taken as approximately normally distributed with mean 120 and variance 72.\"},{\"question\":\"What is the variance of the sample mean Xbar in Q2?\",\"answer\":\"Var(Xbar) equals sigma2/n, showing that the sample mean’s variance is reduced by a factor of 1/n compared with the variance of an individual observation.\"},{\"question\":\"Why does Q4 suggest the trainee may have made an error?\",\"answer\":\"The regression slope indicates a positive relationship, while the sample correlation coefficient indicates a strong negative relationship. This contradiction suggests an error in the trainee’s calculations or reporting.\"}]","CS1 Specimen Questions and Solutions - July 2020 | PDF",1790488964,18]