Discuss the following:
Q1: a. The area under the standard normal curve with parameters m = 64.4 and o= 2.4 that lies to the left of 61 is 0.0783. Use this information to estimate the percentage of female's students who are shorter than 61 inches.
Q2: According to the National Health and Nutrition Examination Survey, the serum (noncellular portion of blood) total cholesterol level of U.S. females 20 years old and older is normally distributed with a mean of 206 mg/dl (milligrams per deciliter) and a standard deviation of 44.7 mg/dl. Let x denote serum total cholesterol level for U.S. females 20 years old and older.
a. Obtain the standardized version, z of x.
b. The percentage of U.S. females 20 years old or older who have a serum total cholesterol level between 150mg/dl and 250 mg/dl is equal to the area under the standard normal curve between _______________ and _____________.
c. The percentage of U.S. females 20 years old or older who have a serum total cholesterol level below 220 mg /dl is equal to the area under the standard normal curve that lies to the _____________ of _________________.
Q3: Find that area under the standard normal curve that lies to the right of
a. -1.07
b. 0.6
c. 0.
d. 4.2
Q4: Determine that area under the standard normal curve that lies
a. either to the left of -2.12 or to the right of 1.67.
b. either to the left of 0.63 or to the right of 1.54
Q5: Opisthotrochopodus n.sp. is a poly chaete worm that inhabits deep sea hydrothermal vents along the Mid-Atlantic Ridge. According to an article by Van Dover et al. in Marine Ecology Progress Series (Vol. 181, pp. 201-214) the lengths of female polychaete worms are normally distributed with mean 6.1 mm and standard deviation 1.3 mm. Let x denote the length of a randomly selected female polychaete worm.
Determine:
a. P(X < 3).
b. P(5< X<7)
Q6: The A.C. Nielsen Company reports in the Nielsen Report on Television that the mean weekly television viewing time for children aged 2-11 years is 24.50 hours. Assume that the weekly television viewing times of such children are normally distributed with a standard deviation of 6.23 hours and apply the 68.26-95.44-99.97 rule to fill in the blanks.
a. 68.26% of all such children watch between _______ and __________ hours of TV per week.
b. 95.44% of all such children watch between _______ and __________ hours of TV per week.
c. 99.74% of all such children watch between ________ and _________ hours of TV per week.