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Vsevolod [243]
3 years ago
5

In which quadrant is the number –14 – 5i located on the complex plane?

Mathematics
1 answer:
loris [4]3 years ago
5 0

Answer:

The complex number r = -14 - i\,5 belongs to the third quadrant of the complex plane.

Step-by-step explanation:

Let be r = -14 - i\,5. In the complex plane, if \Re (r) < 0 (real component) and \Im (r) < 0 (imaginary component), the number belongs to the third quadrant of the complex plane. The complex number r = -14 - i\,5 belongs to the third quadrant of the complex plane.

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Suppose a large shipment of stereos contained 18% defectives. If a sample of size 306 is selected, what is the probability that
xxMikexx [17]

Answer:

99.36% probability that the sample proportion will differ from the population proportion by less than 6%

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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}

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a sample proportion p in a sample of size n, we have that the sampling distribution of the sample proportions has \mu = p, s = \sqrt{\frac{p(1-p)}{n}}.

In this question:

n = 306, p = 0.18, \mu = 0.18, s = \sqrt{\frac{0.18*0.82}{306}} = 0.0220.

What is the probability that the sample proportion will differ from the population proportion by less than 6%

This is the pvalue of Z when X = 0.18 + 0.06 = 0.24 subtracted by the pvalue of Z when X = 0.18 - 0.06 = 0.12. So

X = 0.24

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

By the Central Limit Theorem

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

Z = \frac{0.24 - 0.18}{0.022}

Z = 2.73

Z = 2.73 has a pvalue of 0.9968

X = 0.12

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

Z = \frac{0.12 - 0.18}{0.022}

Z = -2.73

Z = -2.73 has a pvalue of 0.0032

0.9968 - 0.0032 = 0.9936

99.36% probability that the sample proportion will differ from the population proportion by less than 6%

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