The random variable $X$ has distribution $N(\mu,\sigma^{2})$.

1. Identify the symmetry relationship between the given limit and the required limit with respect to the mean.

The normal distribution is symmetric about its mean $\mu$.
The distance of $k$ from the mean is $k - \mu$.
The distance of $2\mu - k$ from the mean is $\mu - (2\mu - k) = k - \mu$.
By symmetry, the left tail from $2\mu - k$ is equal in area to the right tail from $k$:
$P(X < 2\mu - k) = P(X > k) = 0.45$

2. Calculate the required probability.

$P(X > 2\mu - k) = 1 - P(X < 2\mu - k)$
$P(X > 2\mu - k) = 1 - 0.45 = 0.55$

1. Find the probability of $X < 1$.

$P(1 < X < 4) = P(X < 4) - P(X < 1)$
Since the mean is $\mu = 4$, $P(X < 4) = 0.5$.
$0.5 - P(X < 1) = 0.2475$
$P(X < 1) = 0.5 - 0.2475 = 0.2525$

2. Standardise the value and solve for $\sigma$.

$P\left(Z < \frac{1 - 4}{\sigma}\right) = 0.2525$
$P\left(Z < \frac{-3}{\sigma}\right) = 0.2525$
Using the inverse normal distribution,
$\frac{-3}{\sigma} \approx -0.66667$
$\sigma = \frac{-3}{-0.66667} = 4.5$

1. Identify the centre of the normal distribution from the mean.

From part (b), the mean is $\mu = 4$. The peak of the normal distribution curve occurs at its mean.
Hence, the maximum point of the curve is located at $x = 4$.

2. Sketch the bell-shaped curve over the specified domain.

The specified interval $[-11, 19]$ is symmetric about the mean $x=4$, since $4 - (-11) = 15$ and $19 - 4 = 15$.
Draw a standard bell-shaped curve over this interval with the following properties:

• Entirely above the horizontal $x$-axis.
• Maximum peak positioned exactly at $x=4$.
• Symmetrical about the vertical line $x=4$.
• The tails of the curve tapering off and asymptotically approaching the $x$-axis at both ends ($x=-11$ and $x=19$).