SOLUTION: The amounts of electricity bills for all households in a city have a skewed probability distribution with mean of $140 and a standard deviation of $30. Find the probability that th
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-> SOLUTION: The amounts of electricity bills for all households in a city have a skewed probability distribution with mean of $140 and a standard deviation of $30. Find the probability that th
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Question 1138232: The amounts of electricity bills for all households in a city have a skewed probability distribution with mean of $140 and a standard deviation of $30. Find the probability that the mean amount of electric bills for a random sample of 75 households selected from the city will be within $6 of the population mean.
this has to do with the central limit theorem that states that, even though the underlying population distribution is not normal, the distribution of sample means approaches normality when the sample size is large enough.
based on that, you should be able to estimate the probability that the mean amount of electric bills for a random sample of 75 households selected from the city will be within 6 dollars of the population mean.
the population mean is 140 and the population standard deviation is 30.
the standard error used in the test is equal to population standard deviation divided by suare root of sample size.
that becomes s = 30 / sqrt(75) = 3.464101615.
that's also part of the central limit theorem.
the sample mean approximates the population mean and the standard deviation of the distribution of sample means gets smaller as the sample size gets larger.
the s-score for the test is z = (x - m) / s.
z is the z-score.
x is the raw score being compared to the mean.
m is the mean.
s is the standard error.
in this test, you will get 2 z-scores and then find the probability that a randomly selected z-score of sample means will be between those values.
your low z-score will be 140 - 6 = 134.
your high z-score will be 140 + 6 = 146.
the low z-score will be equal to (134 - 140) / 3.464101615 = -1.732050808
the high z-score will be equal to (146 - 140) / 3.464101615 = 1.732050808
you would then look up the probability that your z-score is between those values and you will find that the probability is equal to .9167355616.
that's your answer.
this was confirmed visually through use of the following online normal distribution calculator.
the key to successfully completing this analysis is that you use the sample error and not the standard deviation of the population or the standard deviation of the sample if the standard deviation of the population is not available.
but that's another subject that you will probably hit later on.
in either case, you would use the standard error rather than the standard deviation to complete a study such as this.
when you use the online distribution calculator, you use the population mean and the standard error when you are dealing with raw scores and you use the population mean of 0 and standard deviation of 1 when you are dealing with z-scores.
