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 196 random draws from a machine. The machine is known to have 98 green balls and 61 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 70% of the draws result in a green ball. Use Table 1.

Calculate the probability of getting more than 70% green balls. (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

At a new exhibit in the Museum of Science, people are asked to choose between 71 or 196 random draws from a machine. The machine is known to have 98 green balls and 61 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 70% of the draws result in a green ball. Use Table 1.

Explanation / Answer

Solution:

a. Calculate the probability of getting more than 70% green balls.
When 71 draws from a machine

Proportion rate of green ball( P ) = 98/ (98+61) = 0.6164
Standard Deviation= sqrt(P*Q /n) = sqrt(0.6164*0.3836/71) = 0.0577
Normal Distribution = Z= X- u / sd ~ N(0,1)
P(X > 0.70) = (0.70-0.6164)/0.0577
= 0.0836/0.0577 = 1.45
= P(Z >1.45) From Standard Normal Table
= 0.0735

When 196 draws from a machine
Proportion ( P ) = 0.6164
Standard Deviation = sqrt(P*Q /n) = sqrt(0.6164*0.3836/196) = 0.0347
Normal Distribution = Z= X- u / sd ~ N(0,1)

P(X > 0.70) = (0.70-0.6164)/0.0347
= 0.0836/0.0347 = 2.408
= P ( Z > 2.408) From Standard Normal Table
= 0.0080

n Probability
71 0.0735

196 0.0080

71 balls , Since the Probability of Success rate is more

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