To answer this question, we'll use the properties of the normal distribution.
1. First, we need to calculate the z-score for the height of 39 inches. The z-score is a measure of how many standard deviations an element is from the mean.
The formula to calculate a z-score is:
\[
z = \frac{X - \mu}{\sigma}
\]
where \(X\) is the value we are interested in (39 inches), \(\mu\) is the mean height (38 inches), and \(\sigma\) is the standard deviation (2 inches).
2. We substitute in the values:
\[
z = \frac{39 - 38}{2} = \frac{1}{2} = 0.5
\]
3. The z-score of 0.5 tells us that 39 inches is half a standard deviation above the mean. Now we need to find out what percentage of the boys are taller than this. We do this by looking up the z-score in a standard normal distribution table or using a calculator that provides the cumulative distribution function (CDF) for the normal distribution.
The CDF gives us the probability that a value is less than or equal to a certain value in a standard normal distribution. For a z-score of 0.5, the CDF tells us the percentage of data points that are less than or equal to 39 inches.
4. Since we want to know the probability that a boy is taller than 39 inches, we need to subtract this CDF value from 1.
Looking up the z-score of 0.5 on the standard normal distribution table or using a calculator, we can find the CDF value, which is approximately 0.6915. This represents the probability of a boy being 39 inches or shorter.
5. Finally, to find the probability of a boy being taller than 39 inches:
\[
\text{Probability taller than 39 inches} = 1 - \text{CDF}(0.5)
\]
\[
\text{Probability taller than 39 inches} = 1 - 0.6915 = 0.3085
\]
So, there is roughly a 30.85% probability that a three-year-old boy will be taller than 39 inches, given the normal distribution of heights with a mean of 38 inches and a standard deviation of 2 inches.