What is the probability? (distribution of sample proportions) A jury is to be selected for a proceeding in which the accused is alleged to have stolen money from a business partner. According to existing data, 40% of potential jurors have been a victim of a major theft at some point in their life that will make them unfairly prejudicial against the accused. If 48 jurors are randomly chosen, calculate the probability that at least half of them will be unfairly prejudicial in this way assuming that the distribution of sample proportions is normal.

You calculated $z=\dfrac{0.5-0.4}{\left(\frac{0.4(1-0.4)}{48}\right)^{1/2}} =1.414213562$

You would have got the same value with $\dfrac{24-48\times 0.4}{\left(48 \times 0.4\times (1-0.4)\right)^{1/2}} $

If you look this up in your normal cumulative distribution function table, it might give you something like $\Pr(Z \le z) = \Phi(z) = 0.9213505$ and so $\Pr(Z \gt z)= 0.07864954$ and this might be what you are looking for as an answer.

Since you are using a normal approximation to a discrete random variable, you might do better with a continuity correction, which could involve starting by calculating $\dfrac{23.5-48\times 0.4}{\left(48 \times 0.4\times (1-0.4)\right)^{1/2}}$ and thus instead coming up with a probability of about $0.1025956$ instead.

As an aside, the actual probability is about $0.1033814$ found using the R code


1 - pbinom(23, 48, 0.4)


so the continuity correction is quite helpful in getting you close.