BSEC Pg 2 - randomly selected guy is over 72 inches 1 Ux=U...

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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 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
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Unformatted text preview: randomly selected guy is over 72 inches. 1. Ux=U Ux Is the mean of x_ U is the mean of the population 2. Ox=O/SQRT of N Ox is the st dev of X_ O is the std dev of the population N is the sample size. 1. The pop. Number {annual salaries of all plumbers} 2. The mean of the population is M=46,700 3. The std dev og the population is 5600 4. The sample size n=42>30 5. The samples mean is X_ is a normal RV (according to the central limit theorem) P(X_<44,00) =P(X_-Ux/Ox_<44,000-Ux_/Ox_) P(Z<44,000-U/O/swrt N) P(Z<44,000-46,700/(5600/sqrt 42) P(Z<-2700/(5600/0.481) P(Z<-2700/864.1) P(Z<-3.12-3.12=-3.1-0.02 0.0009 probability (0.09%) that the means salary of the 42 person sample lies below 44,000 1. The population=all gas prices in cali during the week pop mean, u=3.305 pop std dev...
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