Mathematics, 10.04.2020 17:53 fish64

Seams Personal advertises on its website that 95% of customer orders are received within four working days. They performed an audit from a random sample of 500 of the 6,000 orders received that month and it shows 470 orders were received on time. Part A: Can we use a normal approximation? Explain. (5 points) Part B: If Seams Personal customers really receive 95% of their orders within four working days, what is the probability that the proportion in the random sample of 500 orders is the same as the proportion found in the audit sample or less? (5 points) (10 points)

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Answer from: billy12008

Using the normal distribution and the central limit theorem, it is found that:

a) Since both $np \geq 10$ and $n(1 - p) \geq 10$ , a normal approximation can be used.b) There is a 0.1515 = 15.15% probability that the proportion in the random sample of 500 orders is the same as the proportion found in the audit sample or less.Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It measures how many standard deviations the measure is from the mean. After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X. By the Central Limit Theorem, for a proportion p in a sample of size n, as long as both $np \geq 10$ and $n(1 - p) \geq 10$ , the mean is $\mu = p$ and the standard error is $s = \sqrt{\frac{p(1 - p)}{n}}$

In this problem:

Sample of 500, hence n = 500

.95% of the orders were received on time, hence

Item a:

np = 500(0.95) = 475

n(1 - p) = 500(0.05) = 25

Since both $np \geq 10$ and $n(1 - p) \geq 10$ , a normal approximation can be used.

Item b:

The mean and the standard error are given by:

$\mu = p = 0.95$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.95(0.05)}{500}} = 0.0097$

The probability that the proportion is of $\frac{470}{500} = 0.94$ or less is the p-value of Z when X = 0.94, hence:

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

By the Central Limit Theorem

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

$Z = \frac{0.94 - 0.95}{0.0097}$

Z = -1.03

Z = -1.03 has a p-value of 0.1515.

0.1515 = 15.15% probability that the proportion in the random sample of 500 orders is the same as the proportion found in the audit sample or less.

To learn more about the normal distribution and the central limit theorem, you can take a look at link

Answer from: maukjackie1974

a. Yes(n=500>=5, n(1-p)=25>=5)

b. 0.15241

Step-by-step explanation:

a. A normal approximation to the binomial can be used $n\geq$ 5 and n(1-p)>=5:

#We calculate our p as follows:

$\hat p$ =x/n=470/500=0.94

n=500

n(1-p)=500(1-0.95)=25

Hence, we can use the normal approximation.

b. This is a normal approximation.

-Given that p=0.95(95%)

-We verify if our distribution can be approximated to a normal:

$np=0.95\times 500=475\\n(1-p)=500(1-0.95)=25\\\\np\geq 5,\ n(1-p)\geq 5$

Hence, we can use the normal approximation of the form:

$P_{bin}(k,n,p)-N(\mu,\sigma^2)\left \{ {{\mu=np=475} \atop {\sigma=\sqrt{np(1-p)}=4.8734}} \right. \\\\\\P_{bin}(k\leq 470)\approx P_{norm}(x\leq 470.5)=P_{norm}(z\leq \frac{470-475}{4.8734})\\\\P_{norm}(z\leq -1.0260)=0.15241$