BSEC 10-11 - 257 17 C P(X>10)= =P(X-Ux/Ox > 10-Ux/Ox...

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257 17 C P(X>10)=? =P(X-Ux/Ox > 10-Ux/Ox) =P(Z>10-7/1) =P(Z>3.00)=The marked area =1-the marked area +3.00=+3.0+0.00 1-P(Z<3.00)=1-0.9987=0.0013 There is a 0.0013 probability that the randomly picked guy spends more than 10 hours a week on his pc. A) P(X<5)=0.0228 B) P(5.5<X<9.5)=0.927 C) P(X>10)=0.0013 18 20 13 It’s known that: 1. Let X = height of a randomly selected man X is a normal RV 2. The mean of X is Ux=69.6 inches 3. The standard Deviation of X is Ox=3 inches A) P(X<66)=? P(X<66)=P(X-Ux/Ox < 66-Ux/Ox) =P(Z<66-69-6/3.0) P(Z<-3.6/3) P(Z<-1.20) -1.2=-1.2-0.00 P(Z<-1.2)=0.1151 There is a 0.1151 probability that the height of a randomly selected guy lies below 66
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Unformatted text preview: inches. B) P(66<X<72)=? = P(66-Ux/Ox< X-Ux/Ox < 72-Ux/Ox) = P(66-69.6/3.0 < Z < 72-69.6/3.0) = P(-3.6/3.0 < Z < 2.4/3.0) =P(-1.20 < Z < 0.80) = P(Z < 0.80) – P(Z < -1.20) +0.8=+0.8+0.00 = 0.7881 – 0.1151 = 0.6730 There is a 0.673 probability that the height of the randomly selected guy goes between 66 and 72 inches C) P(X>72)=? P(X-Ux/Ox > 72-Ux/Ox) P(Z>72-69.6/3.0 P(Z>2.4/3.0) P(Z>0.80) =1- the unshaded area = 1-P(Z<0.80) = 1-0.7881 = 0.2119 There is a 0.2119 probability that the height of a randomly selected guy is over 72 inches. A)0.1151 B)0.673 C)0.2119...
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