The 68-95-99.7% rule gives approximate probabilities of a Normal r.v. being within 1, 2, and 3 standard deviations of its mean ... The 68-95-99.7% rule gives approximate probabilities of a Normal r.v. being within 1, 2, and 3 standard deviations of its mean. Derive analogous rules for the following distributions.
a) Unif(0, 1).
b) Expo(1).

For part A, I got $σ = \frac{1}{sqrt(12)}$, $μ = \frac{1}{2}$, X ~ Uni(0, 1)

$P(|U - μ|≤ nσ) = P(|U - \frac{1}{2}|≤ \frac{n}{srt(12)}) = P(\frac{1}{2} - \frac{n}{srt(12)}≤U≤ \frac{1}{2} + \frac{n}{srt(12)})$
$P(|U - μ|≤ nσ) = F_U (\frac{1}{2} + \frac{n}{srt(12)}) - F_U (\frac{1}{2} - \frac{n}{srt(12)})$
Then I plugged in, 1, 2, and 3 and got 57.73%, 100%, and 100%

For part B, I got $σ = \frac{1}{ λ}$, $μ = \frac{1}{ λ} = 1$

$P(|E_1 - μ|≤nσ) = P(|E_1 - 1|≤n) = P(1 - n≤E_1≤1 + n)$
$P(|E_1 - μ|≤nσ) = F_{E_{1}}(1 + n) - F_{E_{1}}(1 - n)$
Then I plugged in, 1, 2, and 3 and got 86.47%, 95.02%, and 98.17%

However I feel like my answers are off I'm just not exactly where my work is off. Please any help would be very much appreciated!
Your answers are correct.
The precise values are as follows.  For the uniform distribution, the probability of being within $1$ standard deviation of the mean is $1/\sqrt{3}$.  The probability of being within $2$ standard deviations is $1$, because $2/\sqrt{3} &gt; 1$.  The same is true of $3$ standard deviations.
For the exponential distribution with mean $1$, the standard deviation is also $1$, therefore being within $1$ standard deviation of the mean implies that $0 \le E_1 \le 2$.  This is simply $1 - e^{-2}$.  To be within two standard deviations, we must have $0 \le E_1 \le 3$, or $1 - e^{-3}$.  And to be within three standard deviations, the probability is $1 - e^{-4}$.
Note that in both cases, we encounter situations where one or more of the endpoints of the interval are truncated due to the support of the random variable not being infinite.
Your answers are correct.
The precise values are as follows.  For the uniform distribution, the probability of being within $1$ standard deviation of the mean is $1/\sqrt{3}$.  The probability of being within $2$ standard deviations is $1$, because $2/\sqrt{3} &gt; 1$.  The same is true of $3$ standard deviations.
For the exponential distribution with mean $1$, the standard deviation is also $1$, therefore being within $1$ standard deviation of the mean implies that $0 \le E_1 \le 2$.  This is simply $1 - e^{-2}$.  To be within two standard deviations, we must have $0 \le E_1 \le 3$, or $1 - e^{-3}$.  And to be within three standard deviations, the probability is $1 - e^{-4}$.
Note that in both cases, we encounter situations where one or more of the endpoints of the interval are truncated due to the support of the random variable not being infinite.