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briannarose2003
28.02.2020 •
Mathematics
The living space of all homes in a city has a mean of 2100 square feet and a standard deviation of 500 square feet. Let be the mean living space for a random sample of 625 homes from this city. The probability that this mean living space is less than 2069 square feet is. (a) .0478(b) .9394(c) .8788(d) .0606(e) .5239
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Ответ:
(d) .0606
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean
and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation ![s = \frac{\sigma}{\sqrt{n}}](/tpl/images/0528/4269/a95e6.png)
In this problem, we have that:
The probability that this mean living space is less than 2069 square feet is:
This probability is the pvalue of Z when X = 2069. So
By the Central Limit Theorem
So the correct answer is:
(d) .0606
Ответ:
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Step-by-step explanation:
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