μ = 100 and standard deviation σ = 5. Find b such that

The standardized score is Z = (X – μ)/σ = (X – 100)/5

Then P(100 ≤ X ≤ b) = P(0 ≤ Z ≤ (b – 100)/5) = 0.3

Use the inverse normal distribution tables to find: (b – 100)/5 = 0.8416, b = 100 + 5·0.8416 = 104.2

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