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Sagot :
Using the normal distribution and the central limit theorem, we have that:
a) A normal model with mean 0.3 and standard deviation of 0.0458 should be used.
b) There is a 0.2327 = 23.27% probability that more than one third of this sample wear contacts.
Normal Probability Distribution
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, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex], as long as [tex]np \geq 10[/tex] and [tex]n(1 - p) \geq 10[/tex].
In this problem:
- 30% of students at a university wear contact lenses, hence p = 0.3.
- We randomly pick 100 students, hence n = 100.
Item a:
[tex]np = 100(0.3) = 30 \geq 10[/tex]
[tex]n(1 - p) = 100(0.7) = 70 \geq 10[/tex]
Hence a normal model is appropriated.
The mean and the standard deviation are given as follows:
[tex]\mu = p = 0.3[/tex]
[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.3(0.7)}{100}} = 0.0458[/tex]
Item b:
The probability is 1 subtracted by the p-value of Z when X = 1/3 = 0.3333, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.3333 - 0.3}{0.0458}[/tex]
[tex]Z = 0.73[/tex]
[tex]Z = 0.73[/tex] has a p-value of 0.7673.
1 - 0.7673 = 0.2327.
0.2327 = 23.27% probability that more than one third of this sample wear contacts.
To learn more about the normal distribution and the central limit theorem, you can take a look at https://brainly.com/question/24663213
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