Question

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?

Answer 1

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

=$\frac{4.3}{\sqrt{15} } =1.110%$

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%

Answer 2

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%