Question 1007909
 IQs are normally distributed with mean 100 and standard deviation 16. 
a)	Less than 110	
P(x < 110) = ?
z(110) = (110-100)/16 = 5/8
P(x < 110) = P(z < 5/8) = normalcdf(-100,5/8) = 0.7340
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b)	Between 110 and 130
z(130) = (130-100)/16 = 30/16 = 15/8
P(110 < x < 130) = P(5/8 < z < 15/8) = normalcdf(5/8,15,8) = 0.2356
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I'll leave the rest to you.
Cheers,
Stan H.
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P(110 c)	More than 130 
P(130 that is how i had it set up... am i just taking (iq score - mean) and dividing it by 16 to get the percentage? what do i with less, between or more though?