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Cans of drink are packed in boxes, each containing 4 cans. The weights of these cans are normally distributed with mean 510 g and standard deviation 14 g. The weights of the boxes, when empty, are independently normally distributed with mean 200 g and standard deviation 8 g.

a) Find the probability that the total weight of a full box of cans is between 2200 g and 2300 g.

b) Two cans of drink are chosen at random. Find the probability that they differ in weight by more than 20 g.

پاسخ تشریحی :
نمایش پاسخ

a) $W\~N\left( {2240,{\text{ }}848} \right)$

$\frac{{2200 - 2240}}{{\sqrt {848} }}\,\,\left( { =  - 1.374} \right)$

$\Phi \left( {'' - 1.374''} \right) = 1 - \Phi \left( {''1.374''} \right){\text{ }}\left( { = 0.0847} \right)$

$\frac{{2300 - 2240}}{{^{\sqrt {848} }}}\,\,\,\left( { = 2.060} \right)$

$\Phi \left( {''2.060''} \right){\text{ }}\left( { = 0.9803} \right)$

$\Phi \left( {''2.060''} \right) - \left( {1 - \Phi \left( {''1.374''} \right)} \right)$

$ = 0.896$  (3 sfs)

b) ${X_1} - {X_2}\~N\left( {0,{\text{ }}392} \right)$

$\frac{{20 - 0}}{{\sqrt {392} }}\,\,\,\left( { = 1.010} \right)$

$(\Phi \left( {''1.010'' = 0.8438} \right)$

$P\left( {X \gt 20} \right) = 1 - \Phi \left( {''1.010''} \right){\text{ }}\left( { = 0.1562} \right)$

$2 \times P\left( {X \gt 20} \right)$

$ = 0.312$ (3 sfs)

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