Let \(X\) and \(Y\) be independent and identically distributed normal random variables. If \[ P(X+2Y\le3)=P(2X-Y\ge4), \] then \(E(X)\) is
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\(X+2Y\) and \(2X-Y\) are both normal with the same variance \(5\sigma^2\). Equal probabilities mean equal standardized cutoffs, so set \(\dfrac{3-3\mu}{\sqrt5}=\dfrac{\mu-4}{\sqrt5}\) and solve for \(\mu\).
Step 1: Note what stays the same across the two combinations. Call the common mean $\mu=E(X)=E(Y)$ and the common variance $\sigma^2$. Build $U=X+2Y$ and $V=2X-Y$. Because $X$ and $Y$ are independent normal variables, any linear combination of them is again normal, so both $U$ and $V$ are normal random variables.
Step 2: Compute variance first, since it turns out to be the shortcut. \[ Var(U)=1^2\cdot\sigma^2+2^2\cdot\sigma^2=5\sigma^2 \] \[ Var(V)=2^2\cdot\sigma^2+(-1)^2\cdot\sigma^2=5\sigma^2 \] Both variances come out equal to $5\sigma^2$. This match happens because $1^2+2^2=2^2+1^2$, and it is the reason the standardizing step below becomes so clean.
Step 3: Standardize both probability statements. Since $U\sim N(3\mu,5\sigma^2)$ and $V\sim N(\mu,5\sigma^2)$, write each probability as a statement about a standard normal variable $Z$: \[ P(U\le3)=P\left(Z\le\frac{3-3\mu}{\sigma\sqrt5}\right) \] \[ P(V\ge4)=P\left(Z\ge\frac{4-\mu}{\sigma\sqrt5}\right)=P\left(Z\le\frac{\mu-4}{\sigma\sqrt5}\right) \] The second line uses the mirror symmetry of the standard normal curve around $0$: the area to the right of a point equals the area to the left of its negative.
Step 4: Match the two z-values. Because the given condition says these two probabilities are equal, and because the standard normal cdf never repeats a value, the two z-values behind them must themselves be equal: \[ \frac{3-3\mu}{\sigma\sqrt5}=\frac{\mu-4}{\sigma\sqrt5} \]
Step 5: Clear the denominator and solve. The $\sigma\sqrt5$ on each side cancels directly, so \[ 3-3\mu=\mu-4 \] Collect the $\mu$ terms on one side and the plain numbers on the other: \[ 3+4=4\mu \] \[ \mu=\frac{7}{4} \]
Step 6: Rule out the distractors. $7/3$ appears if you swap the roles of the coefficients $3$ and $4$ while collecting terms. $3/7$ and $4/7$ appear if the equation is inverted by mistake. None of these track the algebra correctly.
Step 7: State the result. The common mean of $X$ and $Y$ works out to $7/4$, so \[ \boxed{E(X)=\dfrac{7}{4}} \]
