# 6.1.2 1.) The commuter trains on the Red Line for the Regional Transit Authority (RTA) in Cleveland

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=The waiting time of the trains on the Red Line b.) Find the height of this uniform distribution. F(x)= 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. Probability of waiting 5 minutes exactly is 0 since continuous distribution at a particular point is zero. 6.3.2 Find the z-score corresponding to the given area. Remember, z is distributed as the standard normal distribution with mean of and standard deviation . a.) The area to the left of z is 15%. -1.04+(-1.03+1.04)*(8/23)=-1.0365 b.) The area to the right of z is 65%. -0.39+(-0.38+0.39)*(17/37)=-0.3854 c.) The area to the left of z is 10%. -1.29+(-1.28+1.29)*(5/6)=-1.2817 d.) The area to the right of z is 5%. 1.65+(1.64-1.65)*(1/2)=1.645 e.) The area between and z is 95%. (Hint draw a picture and figure out the area to the left of the .) P(-z<Z<z)=0.95 f.) The area between and z is 99%. P(-z<Z<z)=0.99 -2.58+(-2.57+2.58)*(1/2)=-2.575 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. a.) State the random variable. X= Blood pressure for people in China Mean=128 Standard Deviation=23 b.) Find the probability that a person in China has blood pressure of 135 mmHg or more. P(z > (135-128)/23)= P(z>0.3043) = 0.3804 c.) Find the probability that a person in China has blood pressure of 141 mmHg or less. P(z≤(141-128)/23)= P(z≤0.5652)= 0.7140

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