i)
n=20
p=0.6
using exact binomial distribution,
P(X≥10) = Σ20C x * 0.6^ x * (1-0.6)^(20-x) where x varies from 10 to 20
P(x≥10) = 0.8725
----------------
using normal approximation,
Sample size , n = 20
Probability of an event of interest, p =
0.6
right tailed
X ≥ 10
Mean = np = 12
std dev ,σ=√np(1-p)= 2.1909
P(X ≥ 10 ) = P(Xnormal ≥
9.5 )
Z=(Xnormal - µ ) / σ = ( 9.5 -
12 ) / 2.1909 =
-1.141
=P(Z ≥ -1.141 ) =
0.8731
===============
ii)
Yes, because both probabilities are nearly equal
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