A distribution with unknown mean μ has a variance equal to 1.5. Use central limit theorem to find how large a sample should be taken from the distribution in order that the probability will atleast be 0.95 that the sample mean will be within 0.5 of the population mean.
My try: $P(|\overline{X} -\mu| \leq 0.5) \ge 0.95$
$P\left(\frac{|\overline{X} - μ|}{\frac{\sqrt{1.5}}{\sqrt{n}}} ≤ \frac{0.5}{\frac{\sqrt{1.5}}{\sqrt{n}}}\right) \ge 0.95$
You are almost there...
$$\mathbb{P}\left[\frac{|\overline{X}_n-\mu|}{\sqrt{1.5}}\sqrt{n}\le \frac{0.5}{\sqrt{1.5}}\sqrt{n} \right]\ge 0.95$$
$$\mathbb{P}\left[|Z|\le \frac{0.5}{\sqrt{1.5}}\sqrt{n} \right]\ge 0.95$$
that is
$$\frac{0.5}{\sqrt{1.5}}\sqrt{n} \geq 1.96$$
or
$$n\ge 23.05=24$$
as already noted in the comments by @Misha Lavrov, ceteris paribus, $n=24$ is not great enough to grant a good approximation of the result so, in absence of other satisfied conditions, other ways have to be taken into consideration, for example, Cebishev's inequality, by which
$$\mathbb{P}\left\{ |\overline{X}_n-\mu|<0.5 \right\}\geq 1-\frac{1.5}{0.5^2 n}$$
thus
$$n\geq 120$$
As 120 is very large w.r.t. 24, another way would be to analyze the $n\ge 24$ length' sample with some non-parametric tests to prove its normality