Respuesta :
Considering the problem described, we have that:
a) The normal distribution can be used, as both [tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex].
b) There is a 0.0132 = 1.32% probability that proportion in the random sample of 200 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.
- The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with [tex]\mu = np, \sigma = \sqrt{np(1-p)}[/tex], as long as [tex]np \geq 10[/tex] and [tex]n(1 - p) \geq 10[/tex].
The parameters of the binomial distribution are:
n = 2000, p = 0.92.
Hence:
- np = 200 x 0.92 = 184.
- n(1-p) = 200 x 0.08 = 16.
Since both [tex]np \geq 10, n(1 - p) \geq 10[/tex], a normal approximation can be used. The mean and the standard deviation of the approximation are given as follows:
- [tex]\mu = np = 200 \times 0.92 = 184[/tex].
- [tex]\sigma = \sqrt{np(1 - p)} = \sqrt{200 \times 0.92 \times 0.08} = 3.8367[/tex]
The probability that proportion in the random sample of 200 orders is the same as the proportion found in the audit sample or less, using continuity correction, is P(X < 175.5), which is the p-value of Z when X = 175.5, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{175.5 - 184}{3.8367}[/tex]
Z = -2.22.
Z = -2.22 has a p-value of 0.0132.
0.0132 = 1.32% probability that proportion in the random sample of 200 orders is the same as the proportion found in the audit sample or less.
More can be learned about the normal distribution at https://brainly.com/question/24537145
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