At a new exhibit in the Museum of Science, people are asked to choose between 80

Question

At a new exhibit in the Museum of Science, people are asked to choose between 80 or 110 random draws from a machine. The machine is known to have 90 green balls and 90 red balls. After each draw, the color of the ball is noted and the ball is put back for the next draw. You win a prize if more than 61% of the draws result in a green ball. Use Table 1. a. Calculate the probability of getting more than 61% green balls. (Round “z” value to 2 decimal places, and final answer to 4 decimal places.) n Probability 80 110 b. Would you choose 80 or 110 draws for the game? 80 balls 110 balls

Explanation / Answer

Note that the probability of getting a green ball is 90/(180) = 0.5.

A)

For n = 80:

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.61      
u = mean = p =    0.5      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.055901699      
          
Thus,          
          
z = (x - u) / s =    1.97      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.97   ) =    0.0244 [ANSWER]

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For n = 110:

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.61      
u = mean = p =    0.5      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.047673129      
          
Thus,          
          
z = (x - u) / s =    2.31      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.31   ) =    0.0104 [ANSWER]

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b)

There is greater probability of winning in n = 80.

So, I will choose n = 80. [ANSWER]

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