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A study on the latest fad diet claimed that the amounts of weight lost by all people on this diet had a mean of 22.4 pounds and a standard deviation of 5.8 pounds. Step 1 of 2 : If a sampling distribution is created using samples of the amounts of weight lost by 73 people on this diet, what would be the mean of the sampling distribution of sample means? Round to two decimal places, if necessary.

Respuesta :

Answer:

sampling distribution = 22.4

sampling deviation = 0.68

Step-by-step explanation:

Given that :

Mean , μ = 22.4

Standard deviation, σ = 5.8

n = 73

Therefore, the sampling distribution of the mean and the standard deviation is

[tex]$\mu_{\overline x} = 22.4$[/tex]

So, [tex]$\sigma_{\overline x}=\frac{\sigma}{\sqrt n}$[/tex]

            [tex]$=\frac{5.8}{\sqrt {73}}$[/tex]

          = 0.68

So the sampling distribution of the mean is 22.4

The standard deviation is 0.68

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