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lesson 2.5 - normal distributions: finding area from values quicknotes …

Question

lesson 2.5 - normal distributions: finding area from values
quicknotes
lesson app 2.5 what cholesterol levels are unhealthy for teen boys?
high levels of cholesterol in the blood increase the risk of heart disease. for teenage boys,
the distribution of blood cholesterol is approximately normal with mean μ = 151.6 milligrams
of cholesterol per deciliter of blood (mg/dl) and standard deviation σ = 25 mg/dl.

  1. about what proportion of teen boys have cholesterol levels less than 100 mg/dl?
  2. cholesterol levels of 200 or higher are considered high for teenagers. what percent of

toen boys have high cholesterol?

  1. cholesterol levels between 170 mg/dl and 200 mg/dl are considered borderline high for

toenagers. what percent of teen boys have borderline high cholesterol levels?

Explanation:

Step1: Calculate the z - score for each value

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x\) is the value, \(\mu = 151.6\) (mean), and \(\sigma=25\) (standard deviation).

For \(x = 100\):

\(z_1=\frac{100 - 151.6}{25}=\frac{- 51.6}{25}=-2.064\)

For \(x = 200\):

\(z_2=\frac{200 - 151.6}{25}=\frac{48.4}{25}=1.936\)

For \(x = 170\):

\(z_3=\frac{170 - 151.6}{25}=\frac{18.4}{25}=0.736\)

Step2: Use the standard normal distribution table (or a calculator with a normal - distribution function)

Part 1:

We want to find \(P(X\lt100)\), which is equivalent to \(P(Z\lt - 2.064)\). Using a standard normal table or a calculator (e.g., in R: pnorm(-2.064)), \(P(Z\lt - 2.064)\approx0.0194\)

Part 2:

We want to find \(P(X\geq200)\), which is equivalent to \(P(Z\geq1.936)\). Since \(P(Z\geq z)=1 - P(Z\lt z)\), and using a standard normal table or a calculator (e.g., in R: 1 - pnorm(1.936)), \(P(Z\geq1.936)\approx1 - 0.9736=0.0264\)

Part 3:

We want to find \(P(170\lt X\lt200)\), which is equivalent to \(P(0.736\lt Z\lt1.936)\). Using the property \(P(a\lt Z\lt b)=P(Z\lt b)-P(Z\lt a)\).
\(P(Z\lt1.936)\approx0.9736\) and \(P(Z\lt0.736)\approx0.7697\)
\(P(0.736\lt Z\lt1.936)=0.9736 - 0.7697 = 0.2039\)

Answer:

  1. Approximately \(0.0194\) (or \(1.94\%\))
  2. Approximately \(0.0264\) (or \(2.64\%\))
  3. Approximately \(0.2039\) (or \(20.39\%\))