2 A hospital in a large city records the weight of every inf
Solution
population mean,u = 2.9
standard deviation,sigma = 0.45
a.
P(X< 1.9 ) = P(Z< ((1.9-2.9) / 0.45)
= P(Z< -2.22 )
= 1 - P(Z<2.22)
= 1 - 0.9868
= 0.0132
= 1.32%
b.
P(X< 3.9 ) = P(Z< ((3.9-2.9) / 0.45)
= P(Z< 2.22 )
= 0.9868
98.68 percentile
c.
P( 2.2 <X< 3.7 )
= P( ((2.2-2.9) / 0.45) <Z< ((3.7-2.9) / 0.45) )
= P( -1.56 <Z< 1.78 )
= P(Z<1.78) - P(Z<-1.56)
= P(Z<1.78) - (1 - P(Z<1.56))
= 0.9625 - ( 1 - 0.9406)
= 0.9625 - 0.0594
= 0.9031
d.
P( 3.35 <X< 4.4 )
= P( ((3.35-2.9) / 0.45) <Z< ((4.4-2.9) / 0.45) )
= P( 1.00 <Z< 3.33 )
= P(Z<3.33) - P(Z<1)
= 0.9996 - 0.8413
= 0.1583
e.
top 1% corresponds to z-score=2.33
2.9 + 2.33*0.45
= 3.95 kg
f.
bottom 5% correspond to z-score = -1.645
2.9 - 1.645*0.45
= 2.16 kg
g.
P(X< 3.3 ) = P(Z< ((3.3-2.9) / 0.45)
= P(Z< 0.89 )
= 0.8133
20000*0.8133
= 16266

