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Stock price relative change can be modeled with random variable $R-1$ where random variable $R$ follows log-normal distribution. Meaning $\log(R)$ (log as natural logarithm follows normal distribution. Random variables $\log(R)$ estimated values for mean and standard deviation are $\mu=5.1$ and $\sigma = 3$

What is probability for relative stock price dropping more than 4% from it's current value within one day.

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$$\ln(x)\sim N(5.1,9)$$

As given by WolframAlpha, the CDF is

$$F_X(x)=\Phi(\frac{\ln(x)-5.1}{3})$$

$\Phi$ is the CDF of the Standard Normal.

Now one plugs in values to this formula. Your $x$-value is 4, so plugging this in will give you your answer. Remember to subtract this value from $1$ as you are considering when the price is dropping more than $4$% and the CDF gives you less than $4$%.

I am only considering the absolute value of $x$ (even thought you said it was a drop), because $\ln(x)$ does not take on negative values.

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Our relative stock price change is modeled with random variable $R-1$, we can set our random variable to $x=(R-1)$.

We want to find out probability for: $$P(X \le -0.04)$$ Since we want drop that is larger than $-0.04$. Larger drop means larger negative percentile which is $P(x\le -0.04)$ (same as smaller).

$$P((R-1) \le -0.04)$$ $$P(R\le -0.04 + 1)$$ $$P(R \le 0.96)|\log()$$ $$P(\log(R)\le \log(0.96))$$ log is natural logirthm in this case.

We know that $log(R)$ follows normal distribution so by normalizing value of $log(0.96)$ we can get our probability

$$Z=\frac{x-\mu}{\sigma}$$ plugging in values gives us: $$Z=\frac{\log(0.96-5.1)}{3}≈-1.71360$$

Z is standardized normal. By looking at value for z from standardized normal table. $$P(X\le -0.04)≈\phi(-1.71)$$ $$P(X\le -0.04)≈0.0436≈4.36\%$$

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