SOLUTION: Bowling is being played by Ana and Bea. Assume that both are experienced bowlers and that Bea's and Ana's bowling scores are regularly distributed, with Ana's having a mean of 175

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Question 1197441: Bowling is being played by Ana and Bea. Assume that both are experienced bowlers and that Bea's and Ana's bowling scores are regularly distributed, with Ana's having a mean of 175 and a standard deviation of 10, and Bea's having a mean of 170 and a standard deviation of 5. determine the likelihood that

(A) At least one of them registers a game-high 185 points.

(B) Ana and Bea both receive scores of between 168 and 173.
Found 2 solutions by ewatrrr, ikleyn:
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!

Hi  

(A) At least one of them registers a game-high 185 points.
Using TI or similarly an inexpensive calculator like an Casio fx-115 ES plus
Bea: µ = 170 and σ = 5    Ana: µ = 175 and σ = 10

 P(Bea OR Anna have 185)  = .0009 + .024 = .0249

(B) Ana and Bea both receive scores of between 168 and 173.
   B(both between 168 and 173) = (.38)(.18)

Wish You the Best in your Studies.


Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.
Bowling is being played by Ana and Bea. Assume that both are experienced bowlers and that Bea's and Ana's bowling scores
are regularly distributed, with Ana's having a mean of 175 and a standard deviation of 10,
and Bea's having a mean of 170 and a standard deviation of 5. determine the probability that
(A) At least one of them registers a game-high 185 points.
(B) Ana and Bea both receive scores of between 168 and 173.
~~~~~~~~~~~~~~~~~~~


            The solution in the post by @ewatrrr is incorrect for question (A).
            So, I came to bring you a correct solution to this question.


                        Part (A) 


P(Ana above 185) = P_normal (µ = 175; σ = 10; above 185) = 0.1587 (use your calculator or any tool which you like).


P(Bea above 185) = P_normal](µ = 170; σ =  5; above 185) = 0.0013.



P(Ana OR Bea above 185) = P(Ana above 185) + P(Bea above 185) - P(Ana above 185)*P(Bea above 185) = 

                               = 0.1587 + 0.0013 - 0.1587 *0.0013 = 0.1598.   ANSWER to question A)

Solved (correctly).



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