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Degger [83]
3 years ago
10

Select the correct answer. A simulated game of chess is programmed between two computers. The game is supposed to be biased in f

avor of player B winning 4 out of 5 times. Which is the most suspicious set of outcomes of 30 games played between the two computers? A. AABABBBABBBBBBBBABBBABBBBBBBBA B. ABBABBAABBBBBABBBBBBBBBBBBBABB C. ABBABABABABBABABABABABABABABAA D. ABBABBBABBBBBBBBABBBABBBBBBBBB
Mathematics
2 answers:
pishuonlain [190]3 years ago
7 0

Answer:

C is the most suspicious set.

Step-by-step explanation:

We would expect that B wins 4/5 * 30 =  24 games.

In set  A,  B  wins 23 games.

In B, B wins 24 games.

In C, B wins  15 games.

In D, B wins 25 games.

So  result C is very suspicious.

wariber [46]3 years ago
3 0

Answer: The answer is C

Step-by-step explanation: I just took the test myself and got a 100%

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Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

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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.

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.

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The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 18.6, \sigma = 5.9

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

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33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768

This probability is 1 subtracted by the pvalue of Z when X = 20.4. So

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

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1 - 0.9961 = 0.0039

0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

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