the mean annual income for adult women in one city is $28,520 and the standard deviation of the incomes is $5000. the distribution of incomes is skewed to the right. suppose that we select samples of size 43. determine whether the sampling distribution for is normal (or approximately normal) and give its mean and standard deviation.

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21-10-2022
Mathematics

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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 $5000. the distribution of incomes is skewed to the right. suppose that we select samples of size 43. determine whether the sampling distribution for is normal (or approximately normal) and give its mean and standard deviation.

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rafikiedu08 rafikiedu08 · Answer 1 · 2022-10-25T12:21:37+02:00

The notion of mean is crucial in both mathematics and statistics. The mean is the average or most frequent value among a set of numbers. It is a statistical measure used to determine the median and mode of a probability distribution's central tendency. It's also known as an expected value. Sum of all observations divided by the total observations equals the mean. The term "standard deviation" refers to the degree of dispersion of the data from the mean. Data are grouped around the mean when the standard deviation is low, and are more dispersed when the standard deviation is high.

Given information :

Mean annual income = $28540

Standard Deviation = $5000

Given sample size is n = 43

Sample size doesn't affect the mean, hence the mean remains the same

New mean = $28540

New standard Deviation = Standard Deviation / n^1/2 = 5000/ (43)^1/2

New standard Deviation = 5000/6.5574 = $762.5

Hence, The mean is $28520 and its standard deviation is $762.5

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