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3 years ago

9

Jill's bowling scores are approximately normally distributed withmean 170 and standard deviation 20, while Jack's scores areappr

oximately normally distributed with mean 160 and standarddeviation 15. If Jack and Jill each bowl one game, then assumingthat their scores are independent random variables, approximate theprobability that(a) Jack's score is higher.(b) the total of their scores is above 350.

1 answer:

Lady bird [3.3K]3 years ago

8 0

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 , $20^{2}$ )

Also , since Jack's scores are approximately normally distributed with mean 160 and standard deviation, it implies

Y ≈ ( 160 , $15^{2}$ ).

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 , $20^{2}$ + $15^{2}$ )

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 ( $\frac{Y-X-(10)}{\sqrt{625} }$ > $\frac{10}{\sqrt{625} }$

= 1 - Ф( $\frac{10}{\sqrt{625} }$ )

= 1 - Ф( $\frac{10}{25}$ )

= 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 ( $\frac{x+Y - 330}{\sqrt{625} }$ > $\frac{350 -330}{\sqrt{625} }$ )

= 1 - Ф( $\frac{20}{25}$ )

= 1 - Ф ( 0.8)

= 1 - 0.788145

= 0.211855

P ( X + Y > 350)≈ 0.212

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