Answer:
The P-value is 80.23%
Step-by-step explanation:
Given;
number of samples, n = 64
sample mean, X = 1.03
standard deviation, σ = 0.189
[tex]sample \ mean \ (X) = \frac{Sum \ of \ samples}{number \ of \ samples \ -1} \\\\1.03 = \frac{Sum \ of \ samples}{64 -1}\\\\Sum \ of \ samples = 1.03(63)\\\\Sum \ of \ samples = 64.89[/tex]
Population mean (μ) is given as;
[tex]\mu = \frac{Sum \ of \ samples}{number \ of \ samples} \\\\\mu = \frac{64.89}{64} \\\\\mu = 1.01 \\[/tex]
The z-score is given as;
[tex]z = \frac{X -\mu}{\frac{\sigma}{\sqrt{n} } }\\\\z = \frac{1.03 -1.01}{\frac{0.189}{\sqrt{64} } }\\\\z = \frac{0.02}{\frac{0.189}{8} }\\\\z = 0.8466 \\[/tex]
z ≅ 0.85
From the z-table, the P-value at the given z-score is 0.8023 = 80.23%;
