A.
n = 298
= 0.1
center = 0.1
Std. Deviation = 0.0174 (

Since,
np = 29.8> 5 and
n(1-p) = 268.2 > 5
Thus assumptions to consider sampling distribution of
approximately normal are fulfilled.
2.
P(>0.12) =
0.1252

Let z be the standard normal variable. so,
Z = (-p)/
P(>0.12) =
p(Z>(0.12-0.1)/0.0174)
= P(Z>1.1494)
= 0.1252
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