I have a sample question that I need help on.
Suppose 48% of the students in a university are baseball players.
If a sample of 427 students is selected, what is the probability that the sample proportion of baseball players will be greater than 44%?

Question

I have a sample question that I need help on.
Suppose 48% of the students in a university are baseball players. 
If a sample of 427 students is selected, what is the probability that the sample proportion of baseball players will be greater than 44%?

anonymous · Answer

I need an answer quick please its a test.

anonymous · Answer

I came up with 18.18 <z<-18.18

kropot72 · Answer

this can be solved with the normal approximation to the binomial distribution, where:
$$mean=np=427\times 0.48=204.96$$ 
$$\sigma=\sqrt{np(1-p)}=10.324$$
44% of 427 = 187.88
Now you need to calculate the z-score for 187.88 and use a standard normal distribution table.

anonymous · Answer

how would you calculate the zscore for that? i know the formula but not what numbers to plug in

kropot72 · Answer

$$z=\frac{X-\mu}{\sigma}$$
X = 187.88
mu = 206.96
sigma = 10.324

anonymous · Answer

thank you so much @kropot72

kropot72 · Answer

You're welcome :)