Mathematics, 24.10.2019 03:50 khalibarnes001

(normal approximation) a process yields 10% defective items. suppose 3600 items are randomly selected from the process. use the normal curve approximation (with half-unit correction) to find the probability that the number of defectives exceeds 385? round your z-score to 2 decimals, and your final answer to 4 decimals.

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Answer from: pr4ever

There is a 8.23% probability that the number of defectives exceeds 385.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation \sigma, the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. If we need to find the probability that the measure is larger than X, it is 1 subtracted by this pvalue.

In this problem, we have a binomial probability(since there are only two outcomes for each item, either they are defective or they are not). However, we can find the mean and the standard deviation of these values, and then calculate the zscore.

The mean of a binomial probability distribution with probability $\pi$ and number of trials is given by the following formula.

$\mu = np$

And the standard deviation is given by the following formula:

$\sigma = \sqrt{np(1-p)}$

For this problem, we have 3600 items, so n = 3600 . 10% are defective, so $\pi = 0.10$ .

$\mu = np = 3600(0.10) = 360$

$\sigma = \sqrt{3600(0.10)(0.90)} = 18$

Find the probability that the number of defectives exceeds 385?

This is 1 subtracted by the pvalue of Z when X = 385.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{385-360}{18}$

Z = 1.39

Z = 1.39 has a pvalue of 0.91774.

So, there is a 1-0.91774 = 0.08226 = 0.0823 = 8.23% probability that the number of defectives exceeds 385.

Answer from: Quest

$3 \times 4 = 12 \\ 6 \times 8 = 48$ $48 + 12 = 60$ hope it is correct if not i'm srry i tried to

Answer from: Quest

there is only one way to arrange all the eight $b's$ relative to each other because they are not distinct. we have

$\underline{\ \ }b\underline{\ \ }b\underline{\ \ }b\underline{\ \ }b\underline{\ \ }b\underline{\ \ }b\underline{\ \ }b\underline{\ \ }b\underline{\ \ }$

now there are nine places to place an $a$ (and not more than one $a$ in one place). because we have four $a$ symbols, we need to choose four of these nine places to insert an $a$ . order does not matter because the $a$ are not distinct.

thus there are 1 · c(9,4) ways, or, if we calculate it,

$1\cdot c(9,4) = \dfrac{9! }{4! (9-4)! } = \dfrac{9\cdot8\cdot7\cdot6\cdot5! }{4! 5! } = \dfrac{9\cdot8\cdot7\cdot6}{4\cdot3\cdot2\cdot1} = 126.$

thus there are 126 ways to form such "words."

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(normal approximation) a process yields 10% defective items. suppose 3600 items are randomly selecte...