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lulu8167
11.04.2020 •
Mathematics
According to a survey, 63% of the Scottish population has visited woodland in the previous year. Skeptical about this claim, you decide to take a simple random sample of 650 people in this population, and you ask them if they had visited woodland in the previous year. You find that 60% of the sample replied "yes" to your question.
Assuming that the original survey's 63% claim is correct, what is the approximate probability that less than 60% of the sample would report that they had visited woodland in the previous year?
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Ответ:
5.71% probability that less than 60% of the sample would report that they had visited woodland in the previous year
Step-by-step explanation:
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.
For a sampling propotion p in a sample of size n, we have that![\mu = p, \sigma = \sqrt{\frac{p(1-p)}{n}}](/tpl/images/0594/9923/77fd8.png)
In this problem, we have that:
What is the approximate probability that less than 60% of the sample would report that they had visited woodland in the previous year?
This is the pvalue of Z when X = 0.6. So
5.71% probability that less than 60% of the sample would report that they had visited woodland in the previous year
Ответ:
4x5=20(base)
20x6=120 cubic inches is ur answer