A factory produces rectangular white marble floor tiles. The length of the floor tiles in centimetres can be regarded as a normal random variable with mean 18.9 and standard deviation 0.3. The breadth of the tiles in centimetres can be regarded as an independent normal random variable with mean 8.9 and standard deviation 0.1. Find the mean and standard deviation of the total perimeter of ten randomly chosen tiles.

1. Define the random variable and state the exact binomial distribution with its mean and variance.

Let $X$ be the number of red-tinted tiles in a packaging of 500 tiles.
$X \sim \text{B}(500, 0.6)$
Mean, $\mu = np = 500(0.6) = 300$
Variance, $\sigma^2 = np(1-p) = 300(0.4) = 120$

2. State the normal approximation and set up the probability inequality with continuity correction.

Since $n$ is large and both $np = 300 > 5$ and $n(1-p) = 200 > 5$, $X$ can be suitably approximated by a normal distribution:
$Y \sim \text{N}(300, 120)$
We require the claim to be correct at least 95% of the time, so $P(X \ge k) \ge 0.95$.
Applying continuity correction for discrete to continuous approximation:
$P(Y \ge k - 0.5) \ge 0.95$

3. Standardise the normal variable and use the inverse normal distribution to find the range of values for k.

Standardising the random variable:
$P\left(Z \ge \frac{k - 0.5 - 300}{\sqrt{120}}\right) \ge 0.95$
From the standard normal distribution, the critical $z$-value for an upper tail area of $0.95$ (or lower tail of $0.05$) is approximately $-1.6449$.
Thus, we require:
$\frac{k - 300.5}{\sqrt{120}} \le -1.6449$
$k - 300.5 \le -1.6449\sqrt{120}$
$k \le 300.5 - 18.018$
$k \le 282.48$

4. Conclude the final greatest integer value for k.

Since $k$ represents a number of tiles, it must be an integer.
The greatest possible integer value satisfying $k \le 282.48$ is $k = 282$.