Historical data shows that the number of goals scored per match at European Football Championships has a mean of 1.93 and a variance of 1.4. A large random sample of n matches is taken. Find the least value of n such that the probability that the average number of goals scored per match exceeds 2 goals is less than 0.24.

1. State the approximate distribution of the sample mean using the Central Limit Theorem.

Let $X$ be the number of goals scored per match.
We are given $\mathrm{E}(X) = 1.93$ and $\mathrm{Var}(X) = 1.4$.
Let $\bar{X}$ be the sample mean number of goals in a sample of size $n$.
Since $n$ is large, by the Central Limit Theorem, $\bar{X}$ is approximately normally distributed:
$\bar{X} \sim \mathrm{N}\left(1.93, \frac{1.4}{n}\right)$

2. Formulate the probability inequality given in the question.

The probability that the average number of goals exceeds 2 is less than 0.24:
$\mathrm{P}(\bar{X} > 2) < 0.24$

3. Standardise the normal variable and equate to the inverse normal value.

Standardising $\bar{X}$ to the standard normal variable $Z$:
$\mathrm{P}\left(Z > \frac{2 - 1.93}{\sqrt{\frac{1.4}{n}}}\right) < 0.24$
$\mathrm{P}\left(Z > \frac{0.07\sqrt{n}}{\sqrt{1.4}}\right) < 0.24$
From standard normal tables or a calculator, the critical value $z$ where $\mathrm{P}(Z > z) = 0.24$ is $z \approx 0.7063$.
For the upper tail probability to be strictly less than $0.24$, the standardised value must be greater than this critical value:
$\frac{0.07\sqrt{n}}{\sqrt{1.4}} > 0.7063$

4. Solve the inequality for $n$.

$0.07\sqrt{n} > 0.7063 \times \sqrt{1.4}$
$0.07\sqrt{n} > 0.8357$
$\sqrt{n} > 11.938$
$n > 142.53$

5. State the final integer value for $n$.

Since the sample size $n$ must be an integer, the least value of $n$ is $143$.