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Vladimir79 [104]
3 years ago
10

Amilia walk 35 mile. She was so tired that she take a brake for 1 minutes. Then she started to walk 15 mile away from her house.

She was tursthy so she drank 2 cups of water. Then she go to her grandma house and stay there for a week. The next day she took a cab to go to her house. And it took her 1 hour. How long did Amilia walk all the way to her house and her grandma house
Awnser with unknown​
Mathematics
1 answer:
suter [353]3 years ago
8 0

Answer:

50 miles

Step-by-step explanation:

35 + 15

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A random sample of n = 100 observations is selected from a population with mean 20 and standard deviation 15. What is the probab
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Answer:

25.14% probability of observing a mean greater than 21

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.

In this question, we have that:

\mu = 20, \sigma = 15, n = 100, s = \frac{15}{\sqrt{100}} = 1.5

What is the probability of observing a mean greater than 21?

This is 1 subtracted by the pvalue of Z when X = 21. So

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

By the Central Limit Theorem

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

Z = \frac{21 - 20}{1.5}

Z = 0.67

Z = 0.67 has a pvalue of 0.7486

1 - 0.7486 = 0.2514

25.14% probability of observing a mean greater than 21

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