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The mean annual income for adult women in one city is $28,520 and the standard deviation of the incomes is $5,190. The distribution of incomes is skewed to the right. For samples of size 30, which of the following statements best describes the sampling distribution of the mean? The distribution of X is skewed to the right. x is approximately normally distributed. Nothing can be said about the distribution of X. X is normally distributed.

Sagot :

x is approximately normally distributed.

As per the given data,

We need to find Sampling distribution of the given data and state which of the given statements best describes the sampling distribution of the mean.

According to given question,

It is given that,

The mean annual income for adult women in one city is $28,520.

Standard deviation of the incomes is $5,190.

samples of size is 30.

The distribution of incomes is skewed to the right.

Thus, we have given data:

Mean = 28520

Standard Deviation (SD) = 5190

sample n = 30

Now, moving further we get:

here μ = 28520

We can calculate sampling distribution as:

sampling distribution = σ/√n   .

Putting given values in above stated formula, we get:

sampling distribution =   [tex]\frac{5190}{\sqrt{30} }[/tex]

sampling distribution = 947.59.

Hence, x is approximately normally distributed.

For such more questions on Sampling Distribution:

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