Normal Distribution Example 1

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Normal Distribution Example 1

Assume that cholesterol levels for women ages 20-34 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 20-34 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 20-34 has high cholesterol > 240 mg/dL = 0.079. Since this represents population data we can say that approximately 8% of women age 20-34 have cholesterol levels demanding medical attention.

Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal Probabilities

Variable value = 240; mu = 185; sigma = 39; select Upper Tail R Script: pnorm(240, mean=185, sd=39, lower.tail=FALSE)

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b. Compute the probability that a woman age 20-34 has a cholesterol level above 240 mg/dL using the standard normal curve.

We need to compute the z-score for 240. z = (240-185)/39 = 1.410256

The probability under the standard normal curve greater than this z-value is 0.079. Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal

Probabilities

Variable value = 1.410256; select Upper Tail R Script: pnorm(1.410256, lower.tail=FALSE)

1-pnorm(1.410256)

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Cholesterol levels between 200 mg/dL and 240 mg/dL are considered borderline high for women age 20-34.

a. What is the probability that a woman age 20-34 has borderline high cholesterol? Area below 240 = 0.920768

Area below 200 =0.6497388

Area between 200 and 240 = 0.920768 – 0.6497388 = 0.2710292

The probability that a woman age 20-34 has borderline high cholesterol is 0.27.

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.

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b. Compute the probability that a woman age 20-34 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 = (200-185)/39 = 0.3846154 z_240 = (240-185)/39 = 1.410256 Area below z_200 = 0.6497388 Area below z_240 = 0.920768

Area between z_200 and z_240 = 0.920768 - 0.6497388 = 0.2710292

The probability that a woman age 20-34 has borderline high cholesterol is 0.27.

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 20-34 have borderline high cholesterol levels.

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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 = 0.0668072

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. 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 = 0.03593032 Area below 52 = 0.9640697

Area between 34 and 52 = 0.9640697 - 0.03593032 = 0.9281394

The probability that a randomly selected 5 year old boys weighs between 34 and 52 pounds is 0.928 (i.e. 92.8% of 5 year old boys have a weight between 34 and 52 lbs). Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal

Probabilities

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

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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 = 0.8416212

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. Using Rcmdr to get the critical value:

Rcmdr Menu: Distributions > Continuous Distributions > Normal Distribution > Normal Quantiles

Probabilities = 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

Probabilities = 0.8; mu = 43; sigma = 5; select Lower Tail R Script: qnorm(0.8, mean=43, sd=5)