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lisabon 2012 [21]
3 years ago
7

There are 32 student,nine of those are women,what percent are men,round to the nearest tenth?

Mathematics
1 answer:
12345 [234]3 years ago
8 0
First you have to find out how many guys you have-1.) 32-9=23, then you divide the number of men by the number of students-2.) 23÷32=0.718, so the answer is 0.7
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Using the <u>normal distribution and the central limit theorem</u>, it is found that there is a 0.0166 = 1.66% probability of a sample proportion of 0.59 or less.

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

  • It 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, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sampling proportions of a proportion p in a sample of size n has mean \mu = p and standard error s = \sqrt{\frac{p(1 - p)}{n}}

In this problem:

  • 1,190 adults were asked, hence n = 1190
  • In fact 62% of all adults favor balancing the budget over cutting taxes, hence p = 0.62.

The mean and the standard error are given by:

\mu = p = 0.62

s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.62(0.38)}{1190}} = 0.0141

The probability of a sample proportion of 0.59 or less is the <u>p-value of Z when X = 0.59</u>, hence:

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{0.59 - 0.62}{0.0141}

Z = -2.13

Z = -2.13 has a p-value of 0.0166.

0.0166 = 1.66% probability of a sample proportion of 0.59 or less.

You can learn more about the <u>normal distribution and the central limit theorem</u> at brainly.com/question/24663213

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