# 07ffx - Solution Regular Exam Marks Q1 1(a 2 1(b 3(c Mean =...

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Solution Regular Exam Marks Q1 1 (a) Mean = 11.9 \$bn 2 SD = 1.3 \$bn 1 (b) Mean = 11.9 \$bn Covar = 0.28 3 SD = 1.5 \$bn (c ) Tchebyshev P(within 2 SDs of mean) > 0.75 P(outside 2 SDs of mean) < 0.25 P(< mean + 2 * SD) > 0.88 3 Mean + 2 * SD = 10 \$bn Q2 (a) Binomial p= 0.46 n= 10 3 P(x=6) = 0.17 (b) Normal Approx to Binomial n= 100 np = 46 n(1-p) = 54 Since these are >5, we can use the Normal Approximation Mean = 46 SD = 4.98 z= 2.71 Prob from table = 0.5 5 P(>60) = 0 1 (c ) Maybe There is a 17% chance of getting 6/10 satisfied with ineffective training 1 This % is not small enough to prove the training was effective Marks Q3 (a) 1 P(X>0)=P(z>-15/25)=P(z>-0.6)= 1 0.5+0.2257=0.7257 (b) 1 P(x>y)=P(x-y>0) 1 E=-5 1 s.d. = sqrt(25^2+35^2) 43.01 1 P(X-Y>0)=P(z>5/43.01161)=P(z>0.1162)=

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1 0.5-0.0478= 0.45 (c ) Pr(X>R)=0.3 1 Pr(z>(R-20)/35)=0.3 1 (R-20)/35=0.525 1 R= 38.38 Q4 (a) 28,000 29,000 27,000 28,000 25,000 35,000 24,000 28,000 1 mean 28,100 10000 810000 1210000 10000
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07ffx - Solution Regular Exam Marks Q1 1(a 2 1(b 3(c Mean =...

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