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pychu [463]
3 years ago
12

At a new exhibit in the Museum of Science, people are asked to choose between 73 or 175 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. [You may find it useful to reference the z table.] a. Calculate the probability of getting more than 70% green balls. (Round your intermediate proportion values and “z” value to 2 decimal places, and final answer to 4 decimal places.) b. Would you choose 73 or 175 draws for the game?

Mathematics
1 answer:
Stolb23 [73]3 years ago
6 0

Answer:

Probability of winning first situation:  0.0793

Probability of winning second situation:  0.0146

You should go with the first option

Step-by-step explanation:

See attached photo for answers.   For this situation you need to identify p-hat, p, q and n.

P-hat, p, and q all stay the same for both situations, only n changes.

p-hat is the proportion of green balls we want, which is 0.7

p = 98/159 = 0.62

q = 0.38 because q = 1 - p, which in this case is q = 1 - 0.62 = 0.38

n = 73 for the first situation, and n = 175 for the second situation

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