Answer: (a) 0.344578
(b) 0.211855
Step-by-step explanation: Let X represent Jill's score and Let Y represent Jack's score.
Jill's scores are approximately normally distributed with mean 170 and standard deviation 20 implies :
X ≈(170 , [tex]20^{2}[/tex])
Also , since Jack's scores are approximately normally distributed with mean 160 and standard deviation, it implies
Y ≈ ( 160 , [tex]15^{2}[/tex]).
It is given that their scores are independent which means that the outcome one one will not affect the outcome of the other, we the have:
Y - X ≈ N(-10 ,[tex]20^{2}[/tex]+[tex]15^{2}[/tex] )
Y - X ≈ N(-10 ,625 )
Also , Y + X ≈ N ( 330 , 625 )
(a) We need to find the approximate probability that Jack's score is higher , that is
P ( Y > X)
=P(Y - X >0)
= P ( [tex]\frac{Y-X-(10)}{\sqrt{625} }[/tex] > [tex]\frac{10}{\sqrt{625} }[/tex]
= 1 - Ф([tex]\frac{10}{\sqrt{625} }[/tex])
= 1 - Ф([tex]\frac{10}{25}[/tex])
= 1 - Ф ( 0.4)
= 1 - 0.655422
= 0.344578
P ( Y > X) ≈ 0.345
(b) We need to calculate the approximate probability that their total score is above 350 , that is
P ( X + Y > 350)
= P ( [tex]\frac{x+Y - 330}{\sqrt{625} }[/tex] > [tex]\frac{350 -330}{\sqrt{625} }[/tex])
= 1 - Ф([tex]\frac{20}{25}[/tex])
= 1 - Ф ( 0.8)
= 1 - 0.788145
= 0.211855
P ( X + Y > 350)≈ 0.212
