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