Answer:
1.) very large sample size
2.) 0.11326
3.) np and n(1-p) should be ≥ 10
4.) 0.010809
Step-by-step explanation:
Given :
p = 0.65 ; n = 500
2.)
P(x ≤ 300)
P(x =x) = nCx * p^x * (1 - p)^(n - x)
Using a binomial probability calculator :
P(x ≤ 300) = 0.11326
To justify why if it can be approximated using a normal distributon :
np and n(1-p) should be ≥ 10
np = 500 * 0.65 = 325
n(1-p) = 500(1 - 0.65) = 175
4.) probability using normal distribution estimate:
Mean, m = np = 500 * 0.65 = 325
Standard deviation, s = sqrt(n*p*(1-p)) = sqrt(500*0.65*(1 - 0.65)) = sqrt(113.75) = 10.665
x ≤ 300
Correction:
x = x + 0.5 = 300 - 0.5 = 300.5
P(x ≤ 300.5)
Zscore = (x - m) / s
Zscore = (300.5 - 325) / 10.665
Zscore = - 24.5 / 10.665
Zscore = - 2.297
P(Z ≤ - 2.297) = 0.010809
