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Ludmilka [50]
3 years ago
10

If you have 2,000 milliliters of water how many liters do you have

Mathematics
2 answers:
klemol [59]3 years ago
8 0

1 liter = 1000 milliliters

2000/1000 = 2

 you have 2 liters

quester [9]3 years ago
3 0
Two litres.
1000ml=1l
2000ml=2l
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vekshin1

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0.0179 = 1.79% probability that the mean of this sample is less than 15.99 ounces of water.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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.

In this question, we have that:

\mu = 16.15, \sigma = 0.45, n = 35, s = \frac{0.45}{\sqrt{35}} = 0.0761

What is the probability that the mean of this sample is less than 15.99 ounces of water?

This is the pvalue of Z when X = 15.99. So

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

By the Central Limit Theorem

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

Z = \frac{15.99 - 16.15}{0.0761}

Z = -2.1

Z = -2.1 has a pvalue of 0.0179

0.0179 = 1.79% probability that the mean of this sample is less than 15.99 ounces of water.

3 0
3 years ago
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