can you help me solve this

Question

Suppose that about 77% of graduating students attend their graduation. A group of 33 students is randomly chosen, and let X be the number of students who attended their graduation.

1. What is the distribution of X? X ~ ? B U N  (,) 
2. Please show the following answers to 4 decimal places.
3. What is the probability that exactly 25 number of students who attended their graduation in this study? 
4. What is the probability that less than 25 number of students who attended their graduation in this study? 
5. What is the probability that at most 25 number of students who attended their graduation in this study? 
6. What is the probability that between 20 and 26 (including 20 and 26) number of students who attended their graduation in this study? 

Answer 1

Can use normal approximation to binomial distribution here:

mean = np = 33 * 0.77 = 25.41

standard deviation (SD) = sqrt(33 * 0.77 * 0.23) = 2.42

distribution is N(25.41, 2.42)

to answer questions will be using z-stat to translate to standard normal distribution: (x - mean)/SD = (x - 25.41)/2.42

since binomial distribution is discrete will be applying continuity correction (+0.5 or -0.5) to approximate continuous normal distribution.

P(x = 25) = P(24.5 < x < 25.5) = P((24.5 - 25.41)/2.42 < z < (25.5 - 25.41)/2.42) = P(-0.37 < z < 0.04) = P(z < 0.04) - P(z < -0.37) ---> leave it to you to look up those probabilities in standard normal probability table

P(x < 25) = P(x < 24.5) = P(z < -0.37)

P(x <= 25) = P(x < 25.5) = P(z < 0.04)

P(20 <= x <= 26) = P(19.5 < x < 26.5) = P( (19.5 - 25.41)/2.42) < z < (26.5 - 25.41)/2.42) = P(-2.44 < z < 0.45) = P(z < 0.45) - P(z < -2.44)