The first-serve percentage of a tennis player in a match is normally distributed with a standard deviation of 4.3%. If a sample of 15 random matches of the player is taken, the mean first-serve percentage is found to be 26.4%. What is the margin of error of the sample mean?

Question

Rambler

Mathematics

18 October, 15:42

The first-serve percentage of a tennis player in a match is normally distributed with a standard deviation of 4.3%. If a sample of 15 random matches of the player is taken, the mean first-serve percentage is found to be 26.4%. What is the margin of error of the sample mean?

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Gabriel Wilcox · Answer 1

Given:

standard deviation = 4.3%

sample size = 15

mean = 26.4%

error = σ/√n

error = 4.3% / √15

error = 4.3% / 3.873

error = 0.0111

0.0111 x 100% = 1.11%