# Probability with Z-Score

Suppose that we model the distribution of IQ scores in the general population as a normal random variable with mean 100 and standard deviation 15. Find the probability that a randomly selected person's IQ score is between 125 and 130.

I know that an IQ score of 130 is 2 standard deviations away from the mean and 125 is 1.666 standard deviations away. I also know that if X has the standard normal distribution, then σ⋅X+μ has the normal distribution with mean μ and standard deviation σ, for any real μ and any σ>0.

I calculated the z-score for each and they are .9772 and .9515 respectively. However, I am supposed to make sure that at least 6 digits after the decimal point are correct, so my answer of .0257 does not suffice. Is my thinking correct? And if so, how would I get an answer with more decimal places for my answer?

• It may be relevant, and somewhat amusing, to note that actually measuring a six-digit probability would require a sample of $10^{12}$ people, which may be larger than the number of people who have ever lived. – whuber Mar 21 '14 at 16:40

## 2 Answers

From an answer on math.SE

Let $Q(x) = 1 - \Phi(x)$ denote the complementary standard Gaussian distribution function, and $\phi(x)$ the standard Gaussian density function. Many scientific calculators evaluate $Q(x)$ for $x \geq 0$ via a rational function approximation: $$Q(x) \approx \phi(x)(b_1t + b_2t^2 + b_3t^3 + b_4t^4 + b_5t^5) ~~ \mbox{where}~ t = \frac{1}{1 + 0.2316419x},$$ $b_1 = 0.319381530$, $b_2 = 0.356563782$, $b_3 = 1.781477937,$ $b_4 = 1.821255978,$ and $b_5 = 1.330274429.$ The magnitude of the error in the approximation is smaller than $7.5 \times 10^{-8}$ for all $x \geq 0$. This suffices to calculate $\Phi(z)$ to the desired accuracy.

The formula stated above is essentially Formula 26.2.17 in Abramowitz and Stegun.

There are many calculators online that give the probabilities of the normal CDF to more than 6 decimals. One is at danielsoper.com.