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You can find this using a table of z scores.
Z = (x - mu) / s
Z = (109 - 100) /15 = 9/15 = 3/5 = .6

You want the percentage of students having a z score above .6.
p = 1 - .7257 = .2743

.2743 * 1800 = 493 students with a score above 109

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Q: How many students of 1800 will have an IQ score above 109 with a mean of 100 and a standard deviation of 15?
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