Calculations for the normal distribution

Normal distributions

In Chapter 1.2, the normal distribution and the $68$-$95$-$99.7$ rule were introduced. The empirical rule is helpful for quick estimates, but it only gives a few specific percentages. What if you need a probability that isn’t one of those benchmarks — for example, the percent within half a standard deviation of the mean, or the percent above $2.3$ standard deviations?

Sidenote

Important to know!

Recall that the empirical rule (also known as the $68$-$95$-$99.7$ rule) states the following:

• About $68\%$ of the data lies within one standard deviation (in either direction) of the mean.
• About $95\%$ of the data lies within two standard deviations (in either direction) of the mean.
• About $99.7\%$ of the data (the vast majority) lies within three standard deviations (in either direction) of the mean.

There are two common ways to find these probabilities:

• Use a z-score table (provided on the AP Statistics exam)
• Use software

This text uses software because it’s faster and more accurate, and it doesn’t prevent you from showing clear work on free-response questions.

$X\sim N$ (mean, standard deviation) refers to a normal distribution with the first input as the mean and the second input as the standard deviation.

normalcdf (lower bound, upper bound) gives the probability of falling between two z-scores for the standard normal distribution (mean $=0$ and standard deviation $=1$ for the standard normal).

normalcdf (lower bound, upper bound, mean, standard deviation) gives the probability of falling between two original raw values with the given mean and standard deviation.

invNorm (probability) gives the z-score with the given probability to the left for the normal distribution (mean $=0$ and standard deviation $=1$ for the standard normal).

invNorm (probability, mean, standard deviation) gives the original raw value with the given probability to the left for a normal distribution with the given mean and standard deviation.

Note that if the goal is to find the z-score with the given probability to the right for the normal distribution, we would use invNorm($1-$ probability) for the standard normal and invNorm($1-$ probability, mean, standard deviation) for any non-standard normal distribution.

Practice problem

Example 1:

IQ scores are normally distributed with a mean of $100$ and a standard deviation of $15$.

Since this is not the standard normal distribution (the mean is not $0$ and the standard deviation is not $1$), we must use normalcdf(lower bound, upper bound, mean, standard deviation) and invNorm(probability, mean, standard deviation) to get accurate results. Failing to provide the mean and standard deviation will cause the software to incorrectly assume they are $0$ and $1$.

Part a.

What is the probability that a randomly selected person will score below $70$?

To find $P(X<70)$, we treat “below $70$” as “between a very low value and $70$.” Since an IQ score can’t be below $0$ in this context, we use $0$ as the lower bound:

$\text{normalcdf}(0,70,100,15)=0.0228$

Solution:

You may have expected the answer to be $0.025$ from the $68$-$95$-$99.7$ rule: about $95\%$ of values are within two standard deviations, leaving about $5\%$ outside, and half of that is $2.5\%=0.025$.

The difference happens because $95\%$ is a rounded approximation. Software typically reports a more precise value (often rounded to four decimal places). Also, using $0$ as the lower bound reflects the idea that IQ scores can’t be negative, although that choice has only a small effect because $0$ is many standard deviations below the mean.

Part b.

What is the probability that a randomly selected person will score between $70$ and $110$?

We want $P(70<X<110)$, so we plug both bounds directly into normalcdf:

$\text{normalcdf}(70,110,100,15)=0.7248$

Solution:

Part c.

What is the probability that a randomly selected person will score between $100$ and $150$?

We want $P(100<X<150)$:

$\text{normalcdf}(100,150,100,15)=0.4996$

Solution:

Part d.

What is the probability that a randomly selected person will score above $120$?

We want $P(X>120)$. Since there is no practical maximum IQ score, one clean approach is to find the probability of scoring below $120$ and subtract from $1$:

$\begin{array}{rl}P(X>120)& =1-\text{normalcdf}(0,120,100,15)\\ & =1-0.9088\\ & =0.0912\end{array}$

Alternatively, use a very large number as the upper bound:

$\text{normalcdf}(120,1\text{E}99,100,15)=0.0918$

The slight difference occurs because the first method uses $0$ as the lower bound (reflecting that IQ can’t be negative), while the second method includes the tiny theoretical probability below $0$.

Solution:

Part e.

$20\%$ of people will score less than what score?

We want the 20th percentile, so the probability to the left is $0.2$:

$\text{invNorm}(0.2,100,15)=87.38$

Solution:

Part f.

$30\%$ of people will score more than what score?

We want the cutoff for the top $30\%$, meaning $30\%$ is to the right of the cutoff. Since invNorm uses the probability to the left, we convert:

$\begin{array}{rl}\text{invNorm}(1-0.3,100,15)& =\text{invNorm}(0.7,100,15)\\ & =107.87\end{array}$

Solution:

Part g.

The middle $80\%$ of people will have an IQ between which two scores?

The middle $80\%$ leaves $20\%$ in the tails, split evenly: $10\%$ on the bottom and $10\%$ on the top. We find the 10th and 90th percentiles:

$\begin{array}{rl}\text{invNorm}(0.1,100,15)& =80.78\\ \text{invNorm}(0.9,100,15)& =119.22\end{array}$

Solution: