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6.12 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.

6.14 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 _________________.

6.30 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

6.33 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

6.50 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)

6.52 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.

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#### Solution Preview

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.

Mean=M = 64.40
Standard deviation =s= 2.40
x= 61.00
z=(x-M )/s -1.4167 =(61-64.4)/2.4
Cumulative Probability corresponding to z= -1.4167 is= 0.0783
Or Probability corresponding to x< 61.00 is Prob(Z)= 0.0783

Thus the probability of percentage of students who are shorter than 61 = 0.0783 or 7.83%

6.14
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 _________________.

a. Obtain the standardized version, z of x.

z=(x-M )/s=(x-206)/44.7

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 -1.2528 and 0.9843.

Mean=M = 206.00 mg/dl
Standard deviation =s= 44.70
x1= 250.00
x2= 150.00
z1=(x1-M )/s= 0.9843 ...

#### Solution Summary

Answers questions on Normal distribution dealing with area under the standard normal curve, standardized version, etc.

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