Ricardo and Tammy practice putting golf balls. Ricardo makes 47% of his putts and Tammy makes 51% of her putts. Suppose that Ricardo attempts 25 putts and Tammy attempts 30 putts. Let R = the proportion of putts Ricardo makes and T = the proportion of putts Tammy makes. What is the probability that Ricardo makes a higher proportion of putts than Tammy?Find the z-table here.

The difference in sample proportions, R − T, has mean 0.47 − 0.51 = −0.04 and standard error √[(0.47)(0.53)/25 + (0.51)(0.49)/30] ≈ 0.135. A difference of 0 corresponds to z = [0 − (−0.04)]/0.135 ≈ 0.30. The probability that R − T is greater than 0 is therefore the area to the right of z = 0.30, about 0.384, so the answer is 0.384.

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