Normal Distribution Example 1

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1 PubH 6414 Worksheet 6a: Normal Distribution 1 of 6 Normal Distribution Example 1 Assume that cholesterol levels for women ages are approximately normally distributed with µ = 185 and σ = 39. Cholesterol levels above 240 mg/dl demand medical attention. a. What is the probability that a woman age has a cholesterol level above 240 mg/dl? This probability is equal to the area under the normal curve for cholesterol > 240. The probability that a woman age has high cholesterol > 240 mg/dl = Since this represents population data we can say that approximately 8% of women age have cholesterol levels demanding medical attention. Variable value = 240; mu = 185; sigma = 39; select Upper Tail R Script: pnorm(240, mean=185, sd=39, lower.tail=false) Draw this area under the curve for cholesterol.

This probability is equal to the area under the normal curve for cholesterol > 240. The probability that a woman age 20-34 has high cholesterol > 240 mg/dl = 0.079.

2 PubH 6414 Worksheet 6a: Normal Distribution 2 of 6 b. Compute the probability that a woman age has a cholesterol level above 240 mg/dl using the standard normal curve. We need to compute the z-score for 240. z = ( )/39 = The probability under the standard normal curve greater than this z-value is Variable value = ; select Upper Tail R Script: pnorm( , lower.tail=false) 1-pnorm( ) Draw this area under the standard normal curve.

We need to compute the z-score for 240. z = (240-185)/39 = 1.

3 PubH 6414 Worksheet 6a: Normal Distribution 3 of 6 Cholesterol levels between 200 mg/dl and 240 mg/dl are considered borderline high for women age a. What is the probability that a woman age has borderline high cholesterol? Area below 240 = Area below 200 = Area between 200 and 240 = = The probability that a woman age has borderline high cholesterol is Variable value = 200, 240; mu=185; sigma=39; select Lower Tail R script: pnorm(240, mean=185, sd=39)-pnorm(200, mean=185, sd=39) Draw this area under the normal curve.

6497388 Area between 200 and 240 = 0.920768 0.6497388 = 0.271

4 PubH 6414 Worksheet 6a: Normal Distribution 4 of 6 b. Compute the probability that a woman age has a borderline high cholesterol using the standard normal distribution. We must calculate the z-scores corresponding to 200 and 240, then find the area between the z-scores. z_200 = ( )/39 = z_240 = ( )/39 = Area below z_200 = Area below z_240 = Area between z_200 and z_240 = = The probability that a woman age has borderline high cholesterol is Variable value = z_200, z_240; select Lower Tail R script: pnorm(z_240)-p(z_240) Draw this area under the standard normal distribution curve. c. What is the interpretation in the context of the original question? Since this is population data we can say that 27% of women ages have borderline high cholesterol levels.

6497388 Area below z_240 = 0.920768 Area between z_200 and z_240 = 0.920768-0.6497388 = 0.271

5 PubH 6414 Worksheet 6a: Normal Distribution 5 of 6 Normal Distribution Example 2: Let X represent the weights of 5-year old boys sampled from a normal distribution with µ = 43 lbs and σ = 5 lbs. a. Find the probability that a randomly selected 5-year old boy weighs less than 35.5 lbs. Area above 35.5 = The probability that a randomly selected 5-year old boy weighs less than 35.5 pounds is Variable value = 35.5; mu=43; sigma=5; select Lower Tail R script: pnorm(35.5, mean=43, sd=5) b. Find the probability that a randomly selected 5-year old boy s weight is greater than 34 lbs and less than 52 lbs. Area below 34 = Area below 52 = Area between 34 and 52 = = The probability that a randomly selected 5 year old boys weighs between 34 and 52 pounds is (i.e. 92.8% of 5 year old boys have a weight between 34 and 52 lbs). Variable value = 34,52; mu=43; sigma=5; select Lower Tail R script: pnorm(52, mean=43, sd=5)-pnorm(34, mean=43, sd=5) Draw this area on the standard normal distribution. z_34 = (34-43)/5 = -1.8 z_52 = (52-43)/5 = 1.8

0668072 The probability that a randomly selected 5-year old boy weighs less than 35.5 pounds is 0.067. Variable value = 35.5; mu=43; sigma=5; select Lower Tail R script: pnorm(35.5, mean=43, sd=5) b.

6 PubH 6414 Worksheet 6a: Normal Distribution 6 of 6 c. Find the weight such that 20% of boys are heavier than the population mean. Using the standard normal distribution: First we need to find the z-value that divides the upper 20% and the lower 80% of the normal distribution (see figure). Critical value = Now rearrange the equation for the z-score and solve for weight. W = z*σ+µ = 0.842*5+43 = 47.2 Twenty percent of boys are heavier than 47.2 pounds. Using Rcmdr to get the critical value: Rcmdr Menu: Distributions > Continuous Distributions > Normal Distribution > Normal Quantiles = 0.8; select Lower Tail R Script: qnorm(0.8) You can also use Rcmdr to get the weight without first using the critical value from the standard normal distribution. Rcmdr Menu: Distributions > Continuous Distributions > Normal Distribution > Normal Quantiles = 0.8; mu = 43; sigma = 5; select Lower Tail R Script: qnorm(0.8, mean=43, sd=5)

8416212 Now rearrange the equation for the z-score and solve for weight. W = z*σ+µ = 0.842*5+43 = 47.2 Twenty percent of boys are heavier than 47.2 pounds.

The Normal Distribution. Alan T. Arnholt Department of Mathematical Sciences Appalachian State University

The Normal Distribution. Alan T. Arnholt Department of Mathematical Sciences Appalachian State University The Normal Distribution Alan T. Arnholt Department of Mathematical Sciences Appalachian State University [email protected] Spring 2006 R Notes 1 Copyright c 2006 Alan T. Arnholt 2 Continuous Random