determine that $P(Y-X<0.28) = 0.7881$. Calculate $i$.

Question

Last year, the daily price of corn per bushel,X, was normally distributed with a mean of $5$ and a standard deviation of $0.75$. This year, assume the daily price of corn bushel, Y , has the same distribution as last year, but is i% greater than the last year due inflation. An analysis determined that $P(Y -X < 0.28) = 0.7881$. Calculate $i$.

This is what I have:

$Y-X\sim N(5i-5,.75^2+(.75i^2))$

CLT:

$P\bigg(\frac{ .28-(5i-5))}{\sqrt{.75^2+(.75i)^2}}<.28\bigg)=.7881$

Using Z score table $P(Z<Z_1)=.81$

$P\bigg(\frac{ .28-(5i-5))}{\sqrt{.75^2+(.75i)^2}}=.81\bigg)$

Solve for $i$

And I get $i=.83$ However the answer is 5% where did I go wrong?

Answer 1

The phrasing of the problem is, I think, extremely hard to follow. Reverse engineering the official solution ($i=.05$) leads me to the following interpretation:

For any given $i$: choose a value for $X$ from the given distribution. Compute the probability that $(1+i)X-X=iX$ is less than $.28$ Solve for the value of $i$ that makes that probability $.7881$

For instance, had we guessed the value $i=.03$ we'd now want to find the probability that $X<\frac {.28}{.03}$. As that comes to $X<9.\overline 3$ we see that the probability is effectively $1$, so $i=.03$ is not correct. Indeed, numerically we can solve this problem to get $i=.05000103$ which, indeed, is effectively $.05$

Answer 2

Given that $Y$ is $i\%$ greater than $X$ $\implies Y=X\left(1+\frac{i}{100}\right)\implies Y-X=\frac{i}{100}X$.

Therefore, $Y-X\sim\mathcal{N}\left(\frac{5i}{100},\left(\frac{0.75i}{100}\right)^2\right)$. Let $\Phi$ be the cdf of standard normal distribution. Then,

\begin{align} P(Y−X<0.28)=0.7881 \implies &\Phi\left(\frac{0.28-\frac{5i}{100}}{\frac{0.75i}{100}}\right)=0.7881\\ \implies& \frac{0.28-0.05i}{0.075i}=0.7998\implies i = 5.0001\%. \end{align}