assume a probability distribution with a mean of %10 and standard deviation of %1.5. In wanting to solve the probability being lower than %5, the normal distribution is written down and integrated as follows:

$\int_{-∞}^{0.05} e^{(-\frac{1}{2} ((x - 0.1)/0.015)^2)}/(\sqrt{2 π} 0.015) dx = 0.00042906$

which gives %0.04

if you calculate it through the z score(assuming a sample size of 10) it would be:


looking at a z score table would then give a probability of 2.798*10^-26

why do the two approaches give different answeres?

  • 3
    $\begingroup$ Where does the sqrt(10) come from? Clearly, $$z\equiv \frac{x-\mu}{\sigma}=-10/3$$, and $N(z)=0.0042906$ $\endgroup$ Commented Aug 23, 2022 at 9:20
  • $\begingroup$ "standard deviation of %1.5" is unusually small. Are you sure of this value? I think 15% (or more) would be a more usual value for stock returns. No surprise then that the probability you found is quite low. Any outcome outside of $\mu \pm 2 \sigma$ is rare. $\endgroup$
    – nbbo2
    Commented Aug 23, 2022 at 9:57

1 Answer 1


Seems like you calculated the z-score incorrectly. There’s no need for the sqrt(10) in the denominator since we just want the standard deviation there.


Not the answer you're looking for? Browse other questions tagged or ask your own question.