## Normal: Example #1

Objective: Compute probabilities and quantiles for a normal random variable representing LDL cholesterol levels and visualize these levels using the Normal Applet.

Problem Description: High blood cholesterol increases the risk of atheroscleros, the thickening of the arteries that can reduce blood flow to the heart, brain, kidneys, etc. This increases the risk of heart attack, stroke, kidney failure, etc. The cholesterol level for adult males of a specific racial group is approximately normally distributed with a mean of 4.8 mmol/L and a standard deviation of 0.6 mmol/L.

The LDL cholesterol level is normally distributed with parameters µ = 4.8 and sigma = 0.6. The Normal Applet below displays the normal curve for these parameters.

A person has moderate risk if his cholesterol level is more than 1 but less than 2 standard deviations above the mean, i.e., between 5.4 (= 4.8 + 1*0.6) and 6.0 (= 4.8 + 2*0.6) mmol/L. What proportion of the population has moderate risk according to this criterion? To compute this probability: 1) select a<=x<=b from the Prob menu; 2) and type 5.4 in the Lower Limit and 6.0 in the Upper Limit field; and 3) click OK. The resulting probability is: P( 5.4 <= X <= 6.0) = 0.136 as shown in the bottom panel of the applet. Thus, 13.6% of the population have moderate risk.

A person has high risk if his cholesterol level is more than 2 standard deviations above the mean, i.e., greater than 6.0 mmol/L. What proportion of the population has high risk? Select x>=a from the Prob menu and type 6.0 in the resulting Dialog and click OK. The probability is: P(X >= 6.0) = 0.0228 as shown in the display.

A person within 1 standard deviation of the mean has normal cholesterol risk. Select a<=x<=b from the Prob menu and enter 4.2 ( = 4.8 -1*0.6) and 5.4 as limits. Then P(4.2 <:= X <= 5.4) = 0.6826 as shown in the botton panel. Thus, more than two-thirds of the population have normal risk.

Finally, a person has low risk if his cholesterol level is 1 standard deviation or more below the mean. Select x<=a from the Prob menu and type 4.2 in the resulting Dialog. Then P(X <= 4.2) = 0.1587, i.e., about 16% of the population have low risk.

Suppose we want to compute the quartiles of the LDL cholesterol distribution. Click on F(x) to display the cumulative distribution function (CDF). The upper quartile can be found by clicking on the right arrow button of the q control until q = 0.75. The upper quartile is seen to be 5.2047. Likewise, by clicking on the left arrow button, the lower quartile is seen to be 4.3952. Therefore, the interquartile range (IQR) is 0.9095.

What is the 90th percentile of the distribution, i.e., the cholesterol level that exceeds 90% of the population? This is the value of x such that P(X<=x) = 0.9, i.e., x0.9. Change q to 0.9 and observe that x0.9 = 5.569 mmol/L. For the standard normal distribution, z0.9 = 1.282. (This can be easily checked in the introductory section.) Verify that x0.9 = 4.8 + 1.282*0.6 = 5.569. Put another way, the 90th percentile is 1.282 standard deviation above the mean and this is true for any normal distribution. More generally, xq = µ + zq*sd. Why?