Peter K. answered 12/10/19
Math / Statistics / Data Analytics
a) X is normal with mean 103 and variance 18^2, so X~N(103, 324)
b) P(X > 122) = P( (x-µ)/σ > (122 - 103)/18 ) = P(z > 19/18) ≈ 0.1456
c) cutoff is z = -2.053749 by table or calculator; set z = (x-µ)/σ, so -2.053749 = (x - 103)/18 and therefore, by algebra, cutoff value for x is 66.03252, so the highest score with which a student will still be in the bottom 2% is 66.03 rounded to two decimals. Note that there is no such thing as fractional IQ scores.
d) using calculator figure the 75% (Q3) and 25% (Q1) points in this distribution; The IQR is (Q3 - Q1) = 115.1408 - 90.85918 = 24.28162 or 24.28, rounded as required