I've this problem, to compute the probabilities of a normal distribution
Fifty-seven percent of students at Calabria Business School support making Introduction to Philanthropy a required course. The school plans to survey 100 students to gauge opinion on this issue. What is the probability that no more than half of those surveyed will say they support requiring the course?
My issue, I don't see how to get the mean. I've Pop.Sample_size = 100 P(Z<0.5) We know 0.57 support the required course. Any clues are welcome
Since the discussion in the comments has gotten a bit involved, I thought I'd post something here.
As discussed in the comments, the problem is missing critical information. If the number of students, $S$, is low then we need to know what $S$ is. For instance, if $S=100$ then the answer is $0$ since we know that $57$ students support the choice.
For a general $S$, we'd have $.57\times S$ supporters, leaving $.43\times S$ non-supporters. In that case the answer we want would be $$P_S=\frac {\displaystyle \sum_{i=0}^{50}\binom {.57\times S}i\times \binom {.43\times S}{100-i}}{\binom {S}{100}}$$
Where the binomial symbols are set to $0$ when the denominator is $<0$.
For instance,$P_{100}=.07143$ and $P_{1000}= .0836$.
Now, if $S$ is very large we can safely assume that each student independently has a $.57$ probability of being a supporter. In that case the exact answer is given by the binomial distribution: $$\sum_{i=0}^{50}\binom {100}i\times .57^i\times .43^{100-i}=.095$$
Note that even with $S=1000$ we're quite far from this value. Even $P_{5000}=.09124$ which is still a bit off of the limiting value.
We could get an approximate answer by using the normal approximation to the answer. The mean for the normal is $57$ and $\sigma = \sqrt {100\times .57\times .43}=4.95$ Using that data, we compute the probability that a random value drawn from the distribution is $≤50.5$ and get $.0946$ which is pretty close to the limiting value. Note that the $50.5$ includes the so-called continuity correction.