X follows exponential distribution, when $p.d.f.$ of random variable $X$ is $$f(x)=\lambda e^{-\lambda x},\ x \ge 0$$
, where $E(X)=\displaystyle\frac1\lambda$.
Problem
A cellphone, which can be used 50 hours on average after charged, follows exponential distribution.
$$ \mbox{X: the time that the cellphone can be used after charged.}\\ X \sim Exp(0.02) \left(\because E(X)=\frac 1 \lambda = 50,\ \therefore\lambda = 0.02\right) $$
Q$_1$) Find the probability that the cellphone will be used more than 48 hours after charging.
A$_1$) $P(X\gt48)=1-F(48)=e^{\lambda x}\big|_{\lambda=0.02,~x=48}=e^{-0.96}=0.3829$
I don't understand why the answer is 0.3829.
Because average is $50$, $P(X>48)$, and $50$ is larger than $48$, I naively thought the answer is larger than $0.5$.
Why the answer is elicited as 0.3829?
you should not compare it with the mean, you should compare 48 with the median which is $\frac{ln(2)}{0.02}\approxeq 35$ . As $48>35$ the probability is less tha 0.5.