Solution Sheet---Stat101 Problem Set 41 Ex 4.74 page 281a) X p(X) $1690 0.25562.50 0.25975 0.50 P(X>$1000) = 0.25 b) 625.1050$X 2 Ex 4.75 page 281-282 a)7.55.32)560550(*1.0)555550(*25.0)550550(*3.0)545550(*25.0)540550(*1.0550560*1.0555*25.0550*3.0545*25.0540*1.0222222XXXb)7.5055055022550550XXXXXc)FFYXXYXXY26.10258110223259223259232593 Ex 5.39 page 333a) n=5, p=0.65b) 0, 1, 2, 3, 4, or 5c) X p(X) 0 0.0053 1 0.0488 2 0.1811 3 0.3364X p(X)4 0.31245 0.1160c) Distributionsx-1 0 1 2 3 4 5 6Quantiles100.0% maximum 5.000099.5% 5.000097.5% 5.000090.0% 5.000075.0% quartile 4.000050.0% median 3.000025.0% quartile 3.000010.0% 2.00002.5% 1.00000.5% 0.00000.0% minimum 0.0000MomentsMean 3.2498Std Dev 1.0668242Std Err Mean 0.0106682upper 95% Mean 3.2707119lower 95% Mean 3.2288881N 10000d)07.125.34 a) 1)( dxxf. This implies [1212dxxc], hence c=1/3 b)7984.06375.0)(2.231*25.131*22221222221XXEXEXdxxxEXdxxxEXc)6826.0)1()1()7984.025.17984.025.1(8787.031)7984.025.17984.025.1(24516.02XPdxxXPNWhere NP denotes the probability of normal distribution.d) 5.0)( medianxdxxf . Hence 5.03112medianxdxx 52.1medianx The distribution is skewed to the left. 5 Strategy 1:a) Let W denote “win”, and L denote “lose” Outcomes Probability Value of X WWWW 1/16 $32 LWWW 1/16 $16 WLWW 1/16 $16 WWLW 1/16 $16 WWWL 1/16 $16 LLWW 1/16 $0 LWLW 1/16 $0 LWWL 1/16 $0 WLLW 1/16 $0 WLWL 1/16 $0 WWLL 1/16 $0 LLLW 1/16 -$16 LLWL 1/16 -$16 LWLL 1/16 -$16 WLLL 1/16 -$16 LLLL 1/16 -$32b) Value of X p(X) F(X)-32 1/16 1/16 -16 1/4 5/16 0 3/8 11/16 16 1/4 15/16 32 1/16 1c) 1625632*16116*41)16(*41)32(*161022222XXX Strategy 2 a) Outcomes Probability Value of X W 1/2 1 LW 1/4 1 LLW 1/8 1 LLLW 1/16 1 LLLLW 1/32 1 LLLLL 1/32 -31 b) Value of X p(X) F(x) -31 1/32 1/32 1 31/32 31/32 c) 57.5311*3231)31(*321022XXX Criteria One: Risk As we know, the standard deviation can be used to measure the risk of the action. The standard deviation of Strategy 2 is smaller than that of Strategy 1, soStrategy 2 is better than Strategy 1. Criteria Two: The maximum value of X In Strategy 1, the customer could win as much as $32 in the four consecutive rolls, while in Strategy 2, one could win at most $1. Hence Strategy 1 is more attractive in this