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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?

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    $\begingroup$ 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. $\endgroup$ – whuber Mar 21 '14 at 16:40
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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.

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

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