Using Z-score= ( X - mean) /standard deviation
With help of standard normal distribution table we find prob
Let X1 be indicated IQ score such that
P( X<X1) = 0.7622
We get Z =0.713 such that P(Z <0.713 ) =0.7622
Z=0.713 =(X1 - mean) /s.d. = (X1 - 100) /15
X1 = 15(0.713) + 100 = 110.7
Option D is correct
Solution file is attached go through iit
Thanks
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