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Studentka2010 [4]
3 years ago
13

Find mean median and mode of 6,1,3,8,5,11,1,5

Mathematics
2 answers:
gizmo_the_mogwai [7]3 years ago
6 0

1, 1, 3, 5, 5, 6, 8, 11

Mean: 1 + 1 + 3 + 5 + 5 + 6 + 8 + 11 = 40/8 = 5

Median: 5

Mode: 5, 1

Y_Kistochka [10]3 years ago
5 0

Okay so the mean is 5


Median is (5+5)/2 = 5


and Mode is 1,5




Hope this helps, best of luck




~Animaljamissofab ♥

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Answer:

0.9319 = 93.19% probability that at least 88 out of 153 registered voters will vote in the presidential election.

Step-by-step explanation:

Binomial probability distribution

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The expected value of the binomial distribution is:

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In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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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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This means that n = 153

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\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{153*0.63*0.37} = 5.97

Consider the probability that at least 88 out of 153 registered voters will vote in the presidential election.

Using continuity correction, this is: P(X \geq 88 - 0.5) = P(X \geq 87.5), which is 1 subtracted by the p-value of Z when X = 87.5.

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

Z = \frac{87.5 - 96.39}{5.97}

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1 - 0.0681 = 0.9319

0.9319 = 93.19% probability that at least 88 out of 153 registered voters will vote in the presidential election.

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