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liberstina [14]
3 years ago
11

James was trying to sleep one night but there was too much noise around him. His clock ticked every

Mathematics
1 answer:
g100num [7]3 years ago
4 0

Answer:

<em>It will occur zero times between midnight and one o'clock.</em>

Step-by-step explanation:

<u>Least Common Multiple (LCM)</u>

Three events keep James from sleeping: his clock ticking every 20 seconds, a tap dripping every 15 seconds, and his dog snoring every 27 seconds.

All three events happened together at midnight. They will happen together again the first time the numbers 20, 15, and 27 have a common multiple. This is the LCM.

List the prime factors of each number:

20: 2,2,5

15: 3,5

27: 3,3,3

Now multiply all the factors the maximum number of times they appear:

LCM=2*2*3*3*3*5=540

(a) All the events will happen together again after 540 minutes.

(b) Since 540 minutes = 9 hours, this event won't happen again until 9 am. Thus, it will occur zero times between midnight and one o'clock.

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Suppose 42% of the population has myopia. If a random sample of size 442 is selected, what is the probability that the proportio
sineoko [7]

Answer:

0.7994 = 79.94% probability that the proportion of persons with myopia will differ from the population proportion by less than 3%.

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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 proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Suppose 42% of the population has myopia.

This means that p = 0.42

Random sample of size 442 is selected

This means that n = 442

Mean and standard deviation:

\mu = p = 0.42

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.42*0.58}{442}} = 0.0235

What is the probability that the proportion of persons with myopia will differ from the population proportion by less than 3%?

Proportion between 0.42 + 0.03 = 0.45 and 0.42 - 0.03 = 0.39, which is the p-value of Z when X = 0.45 subtracted by the p-value of Z when X = 0.39.

X = 0.45

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

By the Central Limit Theorem

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

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Z = 1.28

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

Z = \frac{0.39 - 0.42}{0.0235}

Z = -1.28

Z = -1.28 has a p-value of 0.1003

0.8997 - 0.1003 = 0.7994

0.7994 = 79.94% probability that the proportion of persons with myopia will differ from the population proportion by less than 3%.

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Answer:

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2 years ago
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