Computing and manipulating CDFs and PDFs without calculator

Question

The strength X of a certain material is such that its distribution is found by $X=e^Y$, where Y~N(10,1). Find the cdc and pdf of X and compute P(10,000

I've approached this first by finding the cdf:

$ F(x)=P(X \leq x)=P(e^Y \leq x)=P(Y \leq ln x)$

So F (ln x) or the cdf is simply the integral from negative infinity to positive ln x of the pdf of the normal distribution N(10,1).

To get pdf I used the chain rule to get f(ln x) * 1/x. This gives me a complex equation. I know that I would use this equation to do an integral from 10,000 to 20,000. How would you do this without using a computer?

Answer 1

$$P(10000 \le X \le 20000) = P(\ln(10000) \le Y \le \ln(20000)) \approx P(-0.7897 \le (Y-10) \le -0.0965),$$ where $Y-10 \sim N(0,1)$. Indeed, you need a computer/table to compute these kinds of probabilities, since there is no closed form for these kinds of integrals.

Answer 2

As Angryavian said in their answer, there is no closed form for the normal cumulative distribution. However at some point in time people had to compute these things by hand (or refer to a table hand-computed by someone else). Picking up where Angryavian left off, you want $$ I=\frac{1}{\sqrt{2\pi}}\int_{-0.7897}^{-0.0965} e^{-x^2/2}dx =\frac{1}{\sqrt{2\pi}}\int_{0.0965}^{0.7897} e^{-x^2/2}dx.$$ The integration range is vaguely in the central region of the plot of the integrand (though $0.7897$ is pushing it) so we might get away with approximating the integrand by its Taylor series at the origin. So you can take $$ I\approx \frac{1}{\sqrt{2\pi}} \int_a^b (1-x^2/2+x^4/8)dx =\frac{1}{\sqrt{2\pi}} ((b-a) - (b^3-a^3)/6 + (b^5-a^5)/40).$$ Plugging in $b=0.7897$ and $a=0.0965,$ I get $$ I \approx 0.24692 $$ whereas the exact answer is $$ I\approx0.24671.$$