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1 year ago

You toss a fair coin 10000 times. what are the odds of obtaining more than 5100 tails, approximately?

1 answer:

3 0

This can be solved by using the normal approximation to the binomial distribution.
mean = np = 10.000 * 0.5 = 5,000
The standard deviation is given by:
$S.D.= \sqrt{npq} = \sqrt{5000\times0.5} =50$
$z=\frac{5100-5000}{50}=2$
The probability of obtaining more than 5100 tails is 0.0228 and the probability of obtaining fewer than 5100 tails is 0.9772.
The odds of obtaining more than 5100 tails is therefore:
0.0228:0.9772 = 1:42.86.

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Which of the following statement is true about k-NN algorithm?

BaLLatris [955]

The answer is D. All of the above.

The computational complexity of K-NN increases as the size of the training data set increase and the algorithm gets significantly slower as the number of examples and independent variables increase.

Also, K-NN is a non-parametric machine learning algorithm and as such makes no assumption about the functional form of the problem at hand.

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2 years ago

Which expression is equivalent to x Superscript negative five-thirds? StartFraction 1 Over RootIndex 5 StartRoot x cubed EndRoot

Anastasy [175]

Option B : $\frac{1}{\sqrt[3]{x^{5} } }$ is the expression equivalent to $x^{-\frac{5}{3}$

Explanation:

The given expression is $x^{-\frac{5}{3}$

Rewriting the expression $x^{-\frac{5}{3}$ using the exponent rule, $a^{-b}=\frac{1}{a^{b}}$

Hence, we get,

$\frac{1}{x^{\frac{5}{3} } }$

Simplifying, we get,

$\frac{1}{\left(x^{5}\right)^{\frac{1}{3}}}$

Applying the rule, $a^{\frac{1}{n}}=\sqrt[n]{a}$

Thus, we have,

$\frac{1}{\sqrt[3]{x^{5} } }$

Now, we shall determine from the options that which expression is equivalent to $x^{-\frac{5}{3}$

Option A: $\frac{1}{\sqrt[5]{x^{3} } }$

The expression $\frac{1}{\sqrt[5]{x^{3} } }$ is not equivalent to simplified expression $\frac{1}{\sqrt[3]{x^{5} } }$

Thus, the expression $\frac{1}{\sqrt[5]{x^{3} } }$ is not equivalent to $x^{-\frac{5}{3}$

Hence, Option A is not the correct answer.

Option B: $\frac{1}{\sqrt[3]{x^{5} } }$

The expression $\frac{1}{\sqrt[3]{x^{5} } }$ is equivalent to the simplified expression $\frac{1}{\sqrt[3]{x^{5} } }$

Thus, the expression $\frac{1}{\sqrt[3]{x^{5} } }$ is equivalent to $x^{-\frac{5}{3}$

Hence, Option B is the correct answer.

Option C: $-\sqrt[3]{x^5}$

The expression $-\sqrt[3]{x^5}$ is not equivalent to the simplified expression $\frac{1}{\sqrt[3]{x^{5} } }$

Thus, the expression $-\sqrt[3]{x^5}$ is not equivalent to $x^{-\frac{5}{3}$

Hence, Option C is not the correct answer.

Option D: $-\sqrt[5]{x^3}$

The expression $-\sqrt[5]{x^3}$ is not equivalent to the simplified expression $\frac{1}{\sqrt[3]{x^{5} } }$

Thus, the expression $-\sqrt[5]{x^3}$ is not equivalent to $x^{-\frac{5}{3}$

Hence, Option D is not the correct answer.

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2 years ago

Read 2 more answers

Is the right arm or left arm stronger? Twenty five right handed people were studied. They were asked to do as many bicep curls a

melisa1 [442]

Answer:

The p-value should be higher than 0.05

Step-by-step explanation:

solution is found below

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1 year ago

Suppose that 20% of the adult women in the United States dye or highlight their hair. We would like to know the probability that

Rasek [7]

Answer:

71.08% probability that pˆ takes a value between 0.17 and 0.23.

Step-by-step explanation:

We use the binomial approxiation to the normal to solve this question.

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:

$p = 0.2, n = 200$ . So

$\mu = E(X) = np = 200*0.2 = 40$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.2*0.8} = 5.66$

In other words, find probability that pˆ takes a value between 0.17 and 0.23.

This probability is the pvalue of Z when X = 200*0.23 = 46 subtracted by the pvalue of Z when X = 200*0.17 = 34. So

X = 46

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

$Z = \frac{46 - 40}{5.66}$

$Z = 1.06$

$Z = 1.06$ has a pvalue of 0.8554

X = 34

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

$Z = \frac{34 - 40}{5.66}$

$Z = -1.06$

$Z = -1.06$ has a pvalue of 0.1446

0.8554 - 0.1446 = 0.7108

71.08% probability that pˆ takes a value between 0.17 and 0.23.

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2 years ago

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Answer: Hello! The answer to your question is B, the intersection of the lines drawn to bisect each vertex of the triangle. Hope this helped! Please pick my answer as the Brainliest!

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