Question

Statistics: Using the integral and z-score find the probability of a normal scenario with a mean of 1 and a standard deviation of 1/2 between 0 to 5.

Answer 1

z1 = (0 - 1)/0.5 = -2
z2 = (5-1)/0.5 = 8

cum probability for z2 = 8 is 100%

subtract small amount of normal curve below z1 = -2
from 100% to get probability to be between 0 and 5, z = -2 to 8

Answer 2

Thanks for starting this off, @douglaswinslowcooper! I agree that the area will be equivalent to the area between z = -2 and z = 8. I would like to now set up an integral for this area.

Answer 3

\[P(a < z < b) = \frac{ 1 }{ \sqrt{2\pi} } \int\limits_{a} ^{b} e^{-\frac{1}{2}z^2}~dz\]
\[P(-2 < z < 8) = \frac{ 1 }{ \sqrt{2\pi} } \int\limits_{-2} ^{8} e^{-\frac{1}{2}z^2}~dz\]
\[u = \frac{ z }{ \sqrt{2} }\]
\[u ^{2} = \frac{ z^{2} }{ 2 }\]
\[du = \frac{ 1 }{ \sqrt{2} }~dz\]
\[\frac{ 1 }{\sqrt{2\pi} } \int\limits \limits_{-2} ^{8} e^{- \frac{ 1 }{ 2 } z ^{2}}~dz = \frac{ 1 }{\sqrt{\pi} } \int\limits \limits_{-\sqrt{2}} ^{4\sqrt{2}} e^{-u^{2}}~du\]
\[e^{-u^2} = \sum_{n = 0}^{\infty} \frac{ (-1)^{n} u^{2n} }{ n! }\]
\[\frac{ 1 }{\sqrt{\pi} } \int\limits \limits_{-\sqrt{2}} ^{4\sqrt{2}} e^{-u^{2}}~du = \frac{ 1 }{\sqrt{\pi} } \int\limits \limits_{-\sqrt{2}} ^{4\sqrt{2}}(\sum_{n = 0}^{\infty} \frac{ (-1)^{n} u^{2n} }{ n! })~du\]
\[\frac{ 1 }{\sqrt{\pi} } \int\limits \limits_{-\sqrt{2}} ^{4\sqrt{2}}(\sum_{n = 0}^{\infty} \frac{ (-1)^{n} u^{2n} }{ n! })~du = \frac{ 1 }{\sqrt{\pi} } (\sum_{n = 0}^{\infty} \frac{ (-1)^{n} (4\sqrt{2})^{2n+1} }{ n!(2n+1) } - \sum_{n = 0}^{\infty} \frac{ (-1)^{n} (-\sqrt{2})^{2n+1} }{ n!(2n+1) })\]

Answer 4

Both of these series converge by the ratio test, however, there does not seem to be a nice way to find their sum. Ideally I would like to write the solution in "closed form" (using a finite number of terms), however, I have read that it is impossible to do so for this problem.

What are some of the preferred ways of expressing the area under the bell curve using integrals, NOT decimal approximations? Is it enough to simply write the definite integral? Or is it preferred to write it as an infinite series, as I have attempted? Or there other ways to express this?

Answer 5

I don't know. I think the question does not ask you to do this. "Use the integral" I think means "use the cumulative distribution function," which is widely tabulated, and is the way I would answer the question.