At a new exhibit in the Museum of Science, people are asked to choose between 71

Question

At a new exhibit in the Museum of Science, people are asked to choose between 71 or 176 random draws from a machine. The machine is known to have 88 green balls and 82 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 59% of the draws result in a green ball.

Calculate the probability of getting more than 59% green balls. (Round your intermediate proportion values and “z” value to 2 decimal places, and final answer to 4 decimal places.)

Calculate the probability of getting more than 59% green balls. (Round your intermediate proportion values and “z” value to 2 decimal places, and final answer to 4 decimal places.)

Explanation / Answer

P(Green) = p = ( 88) / (88 + 82) = 0.5176

a ) n = 71

P(p > 0.59) = P(Z > (0.59- 0.5176)/sqrt(0.5176*(1-0.5176)/71)) = P(Z > 1.22)

= 1 - P(Z < 1.22)

= 1 - 0.8888

= 0.1112

P(p > 0.59) = 0.1112

For n = 176

P(p > 0.59) = P(Z > (0.59 - 0.5176)/sqrt(0.5176*(1-0.5175)/166))

= P(Z > 1.92)

= 1 - P(Z < 1.92)

= 1 - 0.9726

= 0.0274

P(p > 0.59) = 0.0274

b ) I would chose 71 balls , since probability is high

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