Answer:
the maximum number of containers that the ship can load is 170
Step-by-step explanation:
Given the data in the question;
W ~ N( 600, 60 )
S ~ N( 4n, 0.4√n )
so
p( W > S ) = 0.90
⇒ P( W - S > 0) = 0.9 ------ let this be equation 1
now, since W and S are independent
Mean( W - S ) = Mean( W ) - Mean( S 0 = 600 - 4n
and
SD( W - S ) = √( var(W) + var(S) ) = √( 60² + 0.4²n)
hence;
W - S ~ N( 600 - 4n, √( 60² + 0.4²n) )
now, from equation one, P( W - S > 0) = 0.9
[tex]P( \frac{(W-S)-Mean(W-S)}{SD(W-S)} > \frac{0-(600-4n)}{\sqrt{60^2 + 0.4^2n} }) = 0.90[/tex]
[tex]P( z > \frac{0-(600-4n)}{\sqrt{60^2 + 0.4^2n} }) = 0.90[/tex]
from z- table
[tex]P( z > \frac{0-(600-4n)}{\sqrt{60^2 + 0.4^2n} }) = P( z >-1.282)[/tex]
[tex]\frac{4n - 600}{\sqrt{60^2 + 0.4^2n} } = -1.282[/tex] ------------------ let this be equation 2
now, we square both sides of equation 2
[tex]\frac{(4n - 600)^2}{60^2 + 0.4^2n} } = (-1.282)^2[/tex]
[tex]\frac{(4n - 600)(4n-600)}{3600 + 0.16n} } = 1.643524[/tex]
we cross multiply
16n² + 360000 - 4800n = 1.643524( 3600 + 0.16n )
16n² + 360000 - 4800n = 5916.6864 + 0.26296384n
16n² + 360000 - 5916.6864 - 4800n - 0.26296384n = 0
16n² + 354083.3136 - 4800.26296384n = 0
16n² - 4800.26296384n + 354083.3136 = 0
solving the quadratic equation, we know that;
x = -b±√( b² - 4ac ) / 2a
so we substitute
x = [-(-4800.26296384) ±√( (-4800.26296384)² - (4 × 16 × 354083.3136)] / [2×16]
x = [ 4800.26296384 ±√( 23042524.522 - 22661332.0704 ] / 32
x = [ 4800.26296384 ±√(381192.4516) ] / 32
x = [ 4800.26296384 ± 617.4078 ] / 32
Hence;
x = [ 4800.26296384 - 617.4078 ] / 32 or [ 4800.26296384 + 617.4078 ] / 32
x = 131 or 170
Therefore, the maximum number of containers that the ship can load is 170
