STAT 200 Week 4 Homework Problems.docx - STAT 200 Week 4 Homework Problems Crystal Wold 6.1.2 1 The commuter trains on the Red Line for the Regional

STAT 200 Week 4 Homework Problems.docx - STAT 200 Week 4...

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STAT 200 Week 4 Homework Problems Crystal Wold 6.1.2 1.) The commuter trains on the Red Line for the Regional Transit Authority (RTA) in Cleveland, OH, have a waiting time during peak rush hour periods of eight minutes ("2012 annual report," 2012). a.) State the random variable. x = waiting time during peak hours b.) Find the height of this uniform distribution. 1/(b-a) = 1/(8-0) = 0.125 c.) Find the probability of waiting between four and five minutes. P(4<x<5) = (5-4)*0.125= 0.125 d.) Find the probability of waiting between three and eight minutes. P(3<x<8) = (8-3)*0.125 = 0.625 e.) Find the probability of waiting five minutes exactly. P(x=5) = 0*0.125 = 0 6.3.2 Find the z -score corresponding to the given area. Remember, z is distributed as the standard normal distribution with mean of m = 0 and standard deviation s = 1 . **All calculations done in excel a.) The area to the left of z is 15%. =norm.s.inv(0.15) = -1.0364334 b.) The area to the right of z is 65%. 1-0.65=0.35 =norm.s.inv(0.35) = -0.38532047 c.) The area to the left of z is 10%. =norm.s.inv(0.1) = -1.281552 d.) The area to the right of z is 5%. 1-0.05=0.95 =norm.s.inv(0.95) = 1.6448536 e.) The area between - z and z is 95%. (Hint draw a picture and figure out the area to the left of the - z .) Left of –z = 0.025 (2.5%) Right of z = 0.025 =norm.s.inv(0.025) = -1.959964 and =norm.s.inv(0.975) = 1.959964 f.) The area between - z and z is 99%. Left of -z = 0.005 (0.5%) Right of z = 0.005 =norm.s.inv(0.005) = -2.575829 and =norm.s.inv(0.995) = 2.5758293 6.3.4 According to the WHO MONICA Project the mean blood pressure for people in China is 128 mmHg with a standard deviation of 23 mmHg (Kuulasmaa, Hense & Tolonen, 1998). Assume that blood pressure is normally distributed.
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a.) State the random variable. x = mean blood pressure b.) Find the probability that a person in China has blood pressure of 135 mmHg or more. (135-128)/23=7/23=0.30434378 With the above answer, I used the applet to then find the probability of 0.3804 (38.04%) c.) Find the probability that a person in China has blood pressure of 141 mmHg or less. (141-128)/ 23=13/23=0.56521739 With the above answer, I used the applet to then find the probability of 0.714 (71.4%) d.) Find the probability that a person in China has blood pressure between 120 and 125 mmHg. (120-128)/23=-0.347826 (125-128)/23=-0.130434 With the above answer, I used the applet to then find the probability of 0.0814 (8.14%) e.) Is it unusual for a person in China to have a blood pressure of 135 mmHg? Why or why not? No, as the probability is 38.04% which is above 5%.
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