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Savatey [412]
3 years ago
6

On a map, the distance from Happy Hill Park to Rainbow Valley Park is 4/1/2 inches. The scale is 1/2 inch: 3 miles. What is the

actual distance between the two parks (in miles)?
Mathematics
1 answer:
7nadin3 [17]3 years ago
7 0

That would be 39miles.

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pete owes his brother $35 from five weeks of borrowing money if pete contiuess to borrow money how much money will he owe after
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I think the answer is 56

Step-by-step explanation:

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Consider the question of whether the home team wins more than half of its games in the National Basketball Association. Suppose
spayn [35]

Answer:

0.0037 = 0.37% probability that the home team would win 65% or more of its games in a simple random sample of 80 games

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 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 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}}

The home team therefore wins 50% of its games

This means that p = 0.5

Determine the probability that the home team would win 65% or more of its games in a simple random sample of 80 games

Sample of 80 means that n = 80 and, by the Central Limit Theorem:

\mu = p = 0.65

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5*0.5}{80}} = 0.0559

This probability is 1 subtracted by the pvalue of Z when X = 0.65. So

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

By the Central Limit Theorem

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

Z = \frac{0.65 - 0.5}{0.0559}

Z = 2.68

Z = 2.68 has a pvalue of 0.9963

1 - 0.9963 = 0.0037

0.0037 = 0.37% probability that the home team would win 65% or more of its games in a simple random sample of 80 games

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