This is a binomial distribution question with
n = 100
p = 0.56
q = 1 - p = 0.43999999999999995
This binomial distribution can be approximated as Normal distribution since
np > 5 and nq > 5
Since we know that
a) x = 60
P(x > 60.0)=?
The z-score at x = 60.0 is,
z = 0.8058
This implies that
P(x > 60.0) = P(z > 0.8058) = 1 - 0.7898209109675304
P(x > 60.0) = 0.2102
b)
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