Respuesta :
Using the Central Limit Theorem, we have that:
a) Yes, it can be used, as np > 10 and n(1 - p) > 10.
b) There is a 0.0475 = 4.75% probability that the proportion in the random sample of 100 orders is the same as the proportion found in the audit sample or less.
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 the sample, we have that:
- np = 85 > 10.
- n(1 - p) = 15 > 10.
Hence the normal approximation can be used.
As for part B, if we have p = 0.9 and n = 100, the mean and the standard error are given by:
- [tex]\mu = p = 0.9[/tex].
- [tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.9(0.1)}{100}} = 0.03[/tex]
The probability of a sample proportion of 85% of less is the p-value of Z when X = 0.85, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.85 - 0.9}{0.03}[/tex]
Z = -1.67
Z = -1.67 has a p-value of 0.0475.
There is a 0.0475 = 4.75% probability that the proportion in the random sample of 100 orders is the same as the proportion found in the audit sample or less.
More can be learned about the Central Limit Theorem at https://brainly.com/question/24663213
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