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Sagot :
Using the normal distribution and the central limit theorem, it is found that there is a 0.0409 = 4.09% probability that, from a simple random sample of 300 adults in the county, less than 50% would say they believe that gardening should be part of the school curriculum.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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 sample proportions for a proportion p in a sample of size n has [tex]\mu = p, s = \sqrt{\frac{p(1 - p)}{n}}[/tex]
In this problem:
- The proportion is of 55%, hence [tex]p = 0.55[/tex]
- The sample has 300 adults, hence [tex]n = 300[/tex]
Then, the mean and the standard error are given by:
[tex]\mu = p = 0.55[/tex]
[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.55(0.45)}{300}} = 0.0287[/tex]
The probability is the p-value of Z when X = 0.5, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.5 - 0.55}{0.0287}[/tex]
[tex]Z = -1.74[/tex]
[tex]Z = -1.74[/tex] has a p-value of 0.0409.
0.0409 = 4.09% probability that, from a simple random sample of 300 adults in the county, less than 50% would say they believe that gardening should be part of the school curriculum.
A similar problem is given at https://brainly.com/question/25800303
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