solution:-.
given that n = 180 (number of students)
p = 0.5 (probability of correct identification)
standard deviation = sqrt[p(1-p)/200] =
sqrt[(0.5)(0.5)/180]
= 0.0373
(a)p(0.5 < x < 0.6) = p[(0.5 - 0.5) / 0.0373 < Z <
(0.6 - 0.5) / 0.0373]
= p(0 < z < 2.6810)
= p(z < 2.6810) - p(z < 0)
= 0.9963 - 0.5
= 0.4963
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