Hi, Re reply: P(z ≤ 1.7)= NORMSDIST(1.7)(Excel)  

0rP(z ≤ 1.7)=  normalcdf(-10,1.7)(TI Calculator) See Graph Below

 z = (117-100)/10 = 1.7.  P(z>1.7)  = 1 - P(z ≤ 1.7) = 1 - .9554 = .0446 0r 4.46% 

B) find the probability that of 49 randomly selected people, their average IQ is less than 105.

 z = 

P(z ≤ 3.5) = .9554  0r  95.54%



z = 1.7

Note: P(z ≤ 1.7) is the Area under the Normal Curve to the 'left' of Blue Line

P(z > 1.7)is the Area under the Normal Curve to the 'right' of Blue Line



 

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