The mean and standard deviation of the x sampling distribution for each of the following sample sizes will be [tex]N(100,10/\sqrt{41} ), N(100,10/\sqrt{130} )[/tex].
Suppose X has a distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma,[/tex]then,
the distribution of the mean of samples of size n, drawn from the values of X, is normally distributed with a mean equal to the mean of X, and a standard deviation equal to [tex]\sigma/\sqrt{n}[/tex]
Symbolically, we write it as:
[tex]X \sim N(\mu, \sigma) \implies \overline{X} \sim N(\mu,\sigma/\sqrt{n} )[/tex]
A random sample is selected from a population with a mean = of 100 and a standard deviation = of 10.
The mean and standard deviation of the x sampling distribution for each of the following sample sizes.
If N=41
[tex]X \sim N(100, 10) \implies \overline{X} \sim N(100,10/\sqrt{41} )[/tex]
If N=130
[tex]X \sim N(100, 10) \implies \overline{X} \sim N(100,10/\sqrt{130} )[/tex]
Learn more about standard normal distribution here:
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