Respuesta :
Using the normal distribution, it is found that there is a 0.3156 = 31.56% probability that the sample proportion will be less than 14%.
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].
The proportion and the sample size are given, respectively, by:
p = 0.15, n = 294
Hence the mean and the standard error are given, respectively, by:
- [tex]\mu = p = 0.15[/tex]
- [tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.15(0.85)}{294}} = 0.0208[/tex]
The probability is the p-value of Z when X = 0.14, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.14 - 0.15}{0.0208}[/tex]
Z = -0.48
Z = -0.48 has a p-value of 0.3156.
0.3156 = 31.56% probability that the sample proportion will be less than 14%.
More can be learned about the normal distribution at https://brainly.com/question/4079902
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