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muminat
3 years ago
11

A restaurant will select select 1 card from a bowl to win a free lunch. Jimmy put in 10 cards in the bowl. The bowl has a total

of 150 cards in it when they do the drawing. What are the odds of Jimmy winning A. 3/20 B. 1/150. C. 1/15. D. 1/10​
Mathematics
1 answer:
amid [387]3 years ago
5 0
If he put 10 in the bowl and there are 150 cards, the chance for him to win would be 10/150.
now by simplifying it, you get 1/15.
therefore answer is C
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Pani-rosa [81]

Answer:

57.39% probability that between 50 and 60 of them were in favor of leaving the U.K.

Step-by-step explanation:

I am going to use the normal approximmation to the binomial to solve this question.

Binomial probability distribution

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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:

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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.

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:

n = 135, p = 0.38

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What is the probability that between 50 and 60 of them were in favor of leaving the U.K.?

Using continuity correction, this is P(50-0.5 \leq X \leq 60 + 0.5) = P(49.5 \leq X \leq 60.5). So this is the pvalue of Z when X = 60.5 subtracted by the pvalue of Z when X = 49.5. So

X = 60.5

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X = 49.5

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

Z = \frac{49.5 - 51.3}{5.6397}

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0.9484 - 0.3745 = 0.5739

57.39% probability that between 50 and 60 of them were in favor of leaving the U.K.

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