The Normal Distribution: Example #2

Objective: Compute normal probabilities and quantiles and visualize these values in normal probability and cumulative distributions.

Problem Description:The height of adult males is normally distributed with a mean of 70 inches and a standard deviation of 3 inches. In this example, we will compute probabilities and quantiles from this normal distribution.

The following standard normal distribution is the starting point:



The plot shows the probability of being less than -1 (i.e., the probability of being more than 1 standard deviation below the mean), which is displayed by pressing the probability popup menu Prob and selecting X<= b and then entering -1 as the Upper Limit value in the resulting popup dialog.

The mean and standard deviation can be changed to 70 and 3, respectively, by pressing on the right triangular animate buttons in the mu and sigma controls. However, this would take a long time since the increment is set to 0.1 by default. It is faster to click in the central part of these tools and to type the mean and standard deviation directly in the resulting dialog boxes. The probability is now essentially 0, since -1 is over 23, i.e., (-1 - 70)/3, standard deviations below 70.

What is the probability a randomly selected male is more than 1 standard deviation below the mean? This is given by P(X < 67), since 67 is 1 standard deviation, i.e., 3, below 70. The probability is the same as above since both represent the probability of being more than 1 standard deviation below the mean.

Now calculate the probability a randomly selected male is between 67 and 76 inches, i.e, P(67 < X < 76). We could convert 67 and 76 to a Z scale and calculate P(-1 < Z < 2) from a standard normal density. However, this scaling is not necessary here. We can compute the probabilities directly by selecting a <= x <= b (the equalities do not matter for continuous distributions) and setting a to 67 and b to 76 in the mean and standard deviation dialog boxes. The probability is 0.8186.

Suppose we know P(X < 67) = 0.2525 and the mean is 70. How can we find the standard deviation? P(X < 67) = 0.1587 when the mean is 70 and the standard deviation is 3. We can increase or decrease the standard deviation using the sigma animate tool until we get the desired probability, at least approximately. In this case, sigma = 4.5 gives the correct probability. As seen below, the distribution is flatter since the standard deviation is larger and thus more probability is in the left tail as compared to the above plot when sigma = 3.

Suppose we know P(X < 67) = 0.3085 and the standard deviation is 3. How can we find the mean? As stated above, P(X < 67) = 0.1587 when the mean is 70 and the standard deviation is 3. Reset sigma to 3. We can increase or decrease the mean using the mu animate tool until we get the desired probability, at least approximately. In this case, mu = 68.5 gives the correct probability. As seen below, the distribution has moved to the left since the mean is smaller and thus more probability is in the left tail as compared to the above plot when mu = 70.

Probabilities are calculated from the default radio button f(x), which shows normal density plots. Quantiles are computed from the cumulative distribution function which is displayed by clicking on F(x). The qth quantile, xq, is the value of x such that P(X <= xq) = F( xq) = q. When F(x) is clicked, the median (also the mean for a normal distribution) is shown by default. Notice the median is 70. ( Be sure to reset mu to 70 and sigma to 3.)

What are the quartiles for the height distribution? To get the lower quartile, change q to 0.25 by pressing on the left triangular side of the q control. It may be necessary to click on q and change the increment (Incr) to 0.05. The lower quartile is 67.98. Likewise, change q to 0.75 to get the upper quartile, which is 72.02.

What is the 10th percentile of the distribution? This is the value of x such that P(X < x) = 0.1, i.e. x0.1. Change q to 0.1 and it is seen that x0.1 = 66.1553. For the standard normal distribution, z0.1 = -1.2816. Verify that x0.1 = 70 - 1.2816*3.