The probability of obtaining a sample mean this large or larger is 0.0228.
Given the mean is 3.2, standard deviation is 0.8 and the sample size is 64.
We calculate the probability of a mean of 3.4 as follows:
#First determine the z-value:
[tex]z=\frac{\bar{X}-\mu}{\sigma / \sqrt{n}}$=\frac{3.4-3.2}{0.8 / \sqrt{64}}$=2.000$[/tex]
#We then determine the corresponding probability on the z tables:
[tex]Z(X \geq 3.4)=1-P(X < 3.4)$=1-0.97725$=0.0228$[/tex]
Hence, the probability of obtaining a sample mean this large or larger is 0.0228.
The sample mean is regularly distributed for samples of any size derived from a population with a normal distribution, with mean X= and standard deviation X=/n, where n is the sample size.
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