Respuesta :
Using the normal distribution, the probability that the sample proportion will be less than 0.1768 is of 0.9222 = 92.22%.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, 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].
For this problem, the proportion and the sample size are given by:
p = 0.12, n = 66.
Hence the mean and the standard error are:
- [tex]\mu = p = 0.12[/tex].
- [tex]s = \sqrt{\frac{0.12(0.88)}{66}} = 0.04[/tex]
The probability that the sample proportion will be less than 0.1768 is the p-value of Z when X = 0.1768, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.1768 - 0.12}{0.04}[/tex]
Z = 1.42
Z = 1.42 has a p-value of 0.9222.
More can be learned about the normal distribution at https://brainly.com/question/4079902
#SPJ1
