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frosja888 [35]
4 years ago
14

Use the Central Limit Theorem to find the mean and standard error of the mean of the indicated sampling distribution. The monthl

y rents for studio apartments in a certain city have a mean of $ 1 comma 060 and a standard deviation of $ 190. Random samples of size 30 are drawn from the population and the mean of each sample is determined. Round the answers to the nearest hundredth.
Mathematics
1 answer:
malfutka [58]4 years ago
4 0

Answer:

Mean $1060

Standard error $34.69

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error, s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Population:

Mean $1,060 and standard deviation $190.

Sampling distriution of samples of size 30:

Mean $1060

Standard deviation s = \frac{190}{\sqrt{30}} = 34.69

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Answer:

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Step-by-step explanation:

Given

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Required

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First, we calculate the mean

\overline x = \frac{\sum x}{n}

\overline x = \frac{9/5 + 9/5 + 2 + 9/5}{4}

\overline x = \frac{7.4}{4}

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s^2 = \frac{(9/5 - 1.85)^2+(9/5 - 1.85)^2+(2 - 1.85)^2+(9/5 - 1.85)^2}{4 - 1}

s^2 = \frac{(-0.05)^2+(-0.05)^2+(0.15)^2+(-0.05)^2}{4 - 1}

s^2 = \frac{0.0025+0.0025+0.0225+0.0025}{3}

s^2 = \frac{0.03}{3}

s^2 = 0.01

<em>Hence, the variance is 0.01</em>

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