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?
2 answers:
Answer:
Step-by-step explanation:
Given that X the first-serve percentage of a tennis player in a match is normal
Std deviation =4.3%
Sample size n = 15
Std error of sample = std deviation/square root of n
=
Margin of error =Z critical value x std error
If 95% confidence is considered then we have
Z critical value=1.96
Hence margin of error =1.96(1.11)%
=2.1756%
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%
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