U2x2ux20 sd2x2sdx2 4 pz 1 1 pz 1 true false 5 pz 0 5

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U2x=2*Ux=20$ SD(2x)=2*SDx=2$ 4. P(Z > 1) = 1 – P(Z < -1) True False 5. P(Z = 0) = .5 True False 6. Suppose X has a normal distribution with mean 70 and standard deviation 10. What is the chance that X is at most 75? Z=(75-70)/10=0.5 P(x<75)=P(z<0.5)=0.6915 1
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7. Suppose Bob took two exams; one was a math exam whose mean was 70 and standard deviation was 10, and the other was a statistics exam whose mean was 80 and standard deviation was 5. Bob scored 90 on both exams. On which exam did he do better, compared to the rest of the class? Justify your answer. Ans: Z1=(90-70)/10=2 for math exam Z2=(90-80)/5=2 for stat exam So he did same on both exams. He is scored at the 97.72% percentile in his class. So he did better than 97.72% of his classmates. 8. Suppose X has a normal distribution with mean -10 and standard deviation 2. What value of X is 3 standard deviations above the mean and what is its corresponding value of Z? Z=3 according to the definition X=u+Z*σ= -10+3*2= -4 9. Suppose X has a normal distribution whose mean is 0 and standard deviation is 2. Find the probability that X is between -3 and -1. Z1=-3/2=-1.5 Z2=-1/2=-0.5 P(-3<x<-1)=P(-1.5<z<-0.5)=0.3085-0.0668= 0.2417 10. The 70 th percentile for Z is equal to 0.70. True False A certain brand of tire has a life expectancy of 50,000 miles with a standard deviation of 5,000 miles. 11. What’s the chance that your tires go at least 40,000 miles before they wear out? P(X>40,000)=P(z>-2)=1-P(z<-2)=1-0.0228=0.9772 12. 10% of the tires last more than how many miles? Go to table find p=0.9 then z=1.28 So X=u+z*σ=50,000+1.28*5000=56400 miles 2
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