Mathematics, 08.04.2020 01:12 KayPink5723

Cassandra's Cogs advertises on its website that 90% of customer orders are received within three working days. They performed an audit from a random sample of 100 of the 2,500 orders that month and it shows 85 orders were received on time.

Part A: Can we use a normal approximation? Explain. (5 points)

Part B: If Cassandra's Cogs customers receive 90% of their orders within three working days, what is the probability that the proportion in the random sample of 100 orders is the same as the proportion found in the audit sample or less? (5 points) (10 points)

Answers: 3

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Answer from: scotrebe3626

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 0.0475 = 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

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, the mean is $\mu = p$ and the standard error is $s = \sqrt{\frac{p(1 - p)}{n}}$ , as long as both $np \geq 10$ and $n(1 - p) \geq 10$ .

In this problem:

90% of customer orders are received within three working days, hence p = 0.9

Sample of 100 orders, hence n = 100

The mean and the standard error are given by:

$\mu = p = 0.9$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.9(0.1)}{100}} = 0.03$

Item a:

np = 100(0.9) = 90

n(1 - p) = 100(0.1) = 10

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

Item b:

The probability is the p-value of Z when $X = \frac{85}{100} = 0.85$ , hence:

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

By the Central Limit Theorem

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

$Z = \frac{0.85 - 0.9}{0.03}$

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.

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

Answer from: chloeboo

Part A: yes

Part B: 50%

Step-by-step explanation:

There are two ways we can use a normal approximation for a sample: if the population it is from is normally distributed, or if the sample is sufficiently large (n > 30). In this case, we don't know if the population is normally distributed, but we do know the sample size is larger than 30, so we can use normal approximation.

The sample proportion is normally distributed, with a mean equal to the mean of the population. So there is a 50% chance that sample proportion is less than the population proportion.

Answer from: Quest

7for nonmembers i think

Answer from: Quest

what do u need to know

step-by-step explanation:

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