ANSWER= 0.9772
STEP-BY-STEP EXPLANATION:
We are assuming that there is an underlying distribution of package weights,even though we don´t speciify the shape of that distribution.
Letting [tex]x_{i}[/tex] denote the ith package weight and S=[tex]x_{1} +....x_{100}[/tex] we are trying to find P(S ≤ 1700)
We know that the distribution of S is approximately normal with mean nцХ and variance по²Х
S≈N( 100 * 15, 100 * 10² ) = N (1500.10000) (note that they gave us the standar deviation aove; variances add, so we need to square this). Finally,
P(S ≤ 1700) = P ([tex]\frac{S-1500}{\sqrt{10000} }[/tex] ≤ [tex]\frac{1700 - 1500}{\sqrt{10000} }[/tex])
≈P (Z≤2) = 0.9772