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earnstyle [38]
3 years ago
5

Assuming a normal distribution of 1000 cases, how many cases will be farther away from the mean than + 3 standard deviations?a.A

bout 3 b.At least 500 c.It is impossible to estimate d.327
Mathematics
1 answer:
DanielleElmas [232]3 years ago
8 0

Answer:

P(-3

And then the probability that  will be farther away from the mean than + 3 standard deviations is just

1-0.9973= 0.0027

And the number of cases would be 1000*0.0027=2.7 or approximatley 3 cases. And the bes option is : a.About 3

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

2) Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(\mu,\sigma)  

We are interested on this probability

P(X-3\mu

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

And we can find this probability on this way:

P(-3

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

P(-3

And then the probability that  will be farther away from the mean than + 3 standard deviations is just

1-0.9973= 0.0027

And the number of cases would be 1000*0.0027=2.7 or approximatley 3 cases. And the bes option is : a.About 3

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Answer:

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Step-by-step explanation:

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

A reminder is that the standard deviation is the square root of the variance.

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.

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

In this question:

\mu = 3639, \sigma = \sqrt{145161} = 381, n = 41, s = \frac{381}{\sqrt{41}} = 59.5

Probability that the mean of the sample would differ from the population mean by less than 126 miles

This is the pvalue of Z when X = 3639 + 126 = 3765 subtracted by the pvalue of Z when X = 3639 - 126 = 3513. So

X = 3765

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

By the Central Limit Theorem

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

Z = \frac{3765 - 3639}{59.5}

Z = 2.12

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

Z = \frac{3513 - 3639}{59.5}

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