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iVinArrow [24]
3 years ago
13

Please help!! *graphs, median, modes, ranges, and means* thank you!!

Mathematics
1 answer:
kipiarov [429]3 years ago
5 0
<h2><u>Definitions/How to Find It</u></h2>

Mean - The average of all of the numbers. Add up all of the numbers then divide it by the total number of numbers.

Median - The middle number of them all. Put the numbers in order from least to greatest and find the one in the middle.

Mode - The number that appears the most.

Range - The difference between the highest and lowest numbers Subtract the lowest number from the highest number.

<h2><u>Answers</u></h2>

Mean = 56.6

Median = 53

Mode = 41, 43, 63

Range = 56

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Consider a binomial distribution of 200 trials with expected value 80 and standard deviation of about 6.9. Use the criterion tha
zavuch27 [327]

Answer:

120 has a z-score higher than 2.5. So yes, it would be unusual to have more than 120 successes out of 200 trials.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples 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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

\mu = 80, \sigma = 6.9

Use the criterion that it is unusual to have data values more than 2.5 standard deviations above the mean or 2.5 standard deviations below the mean

This means that z-scores higher than 2.5 or lower than -2.5 are considered unusual.

Would it be unusual to have more than 120 successes out of 200 trials

We have to find the Z-score of X = 120.

So

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

Z = \frac{120 - 80}{6.9}

Z = 5.8

120 has a z-score higher than 2.5. So yes, it would be unusual to have more than 120 successes out of 200 trials.

4 0
3 years ago
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Hey User______________

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_____________________

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