solution:-
given that mean = 200 , standard deviation = 60 , n = 100
a. the probability that the sample mean will be within +/- 7 of the population mean
z(193) = (193-200)/(60/sqrt(100)) = -1.17
z(207) = (207-200)/(60/sqrt(100)) = 1.17
=>p(-1.17 < z < 1.17) = p(z < 1.17) - p(z < -1.17)
= 0.8790 - 0.1210
= 0.7580
b. the probability that the sample mean will be within +/-16 of the
population mean
z(184) = (184-200)/(60/sqrt(100)) = -2.67
z(216) = (216-200)/(60/sqrt(100)) = 2.67
P(-2.67 < z < 2.67) = p(z < 2.67) - p(z < -2.67)
= 0.9962 - 0.0038
= 0.9924
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