2. If $X$ is $N\left(3, \sigma^{2}=4\right)$ find
(a) $P(X< 3)$,
(b) $P(X \leqslant 5)$,
(c) $P(X \leqslant 1)$.

Step 1 ：We are given a normally distributed random variable $X$ with mean 3 and variance 4. We are asked to find the probability of $X$ being less than certain values.

Step 2 ：We can solve this by using the cumulative distribution function (CDF) of the normal distribution. The CDF at a point $x$ gives the probability that the random variable is less than or equal to $x$.

Step 3 ：For (a), we need to find $P(X<3)$. Since the mean of the distribution is 3, the probability that $X$ is less than the mean is 0.5. So, $P(X<3) = \boxed{0.5}$.

Step 4 ：For (b), we need to find $P(X \leqslant 5)$. Using the CDF of the normal distribution, we find that $P(X \leqslant 5) = \boxed{0.8413447460685429}$.

Step 5 ：For (c), we need to find $P(X \leqslant 1)$. Using the CDF of the normal distribution, we find that $P(X \leqslant 1) = \boxed{0.15865525393145707}$.