Mathematics, 24.03.2020 18:37 couscous5307

Two types of coins are produced at a factory: a fair coin and a biased one that comes up heads 60 percent of the time. We have one of these coins, but do not know whether it is a fair coin or a biased one. In order to ascertain which type of coin we have, we shall perform the following statistical test: We shall toss the coin 1000 times. If the coin lands on heads 550 or more times, then we shall conclude that it is a biased coin, whereas if it lands on heads less than 550 times, then we shall conclude that it is a fair coin. Use a normal approximation to compute the following probabilities (ignore continuity correction). (i) If the coin is actually fair, what is the probability that we shall reach a false conclusion? [7] (ii) If the coin is actually unfair, what is the probability that we shall reach a false conclusion? [8] Your

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Answer from: cassandrabeliles

i) 0.1% probability that if the coin is actually fair, we reach a false conclusion.

ii) 0.05% probability that if the coin is actually unfair, we reach a false conclusion

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:

Fair coin:

Comes up heads 50% of the time, so p = 0.5

1000 trials, so n = 1000

E(X) = np = 1000*0.5 = 500

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1000*0.5*0.5} = 15.81$

If the coin lands on heads 550 or more times, then we shall conclude that it is a biased coin.

(i) If the coin is actually fair, what is the probability that we shall reach a false conclusion?

This is the probability that the number of heads is 550 or more, so this is 1 subtracted by the pvalue of Z when X = 549.

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

$Z = \frac{549 - 500}{15.81}$

Z = 3.1

Z = 3.1 has a pvalue of 0.9990

1 - 0.9990 = 0.001

0.1% probability that if the coin is actually fair, we reach a false conclusion.

(ii) If the coin is actually unfair, what is the probability that we shall reach a false conclusion?

Comes up heads 60% of the time, so p = 0.6

1000 trials, so n = 1000

E(X) = np = 1000*0.6 = 600

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1000*0.6*0.4} = 15.49$

If the coin lands on less than 550 times(that is, 549 or less), then we shall conclude that it is a biased coin.

So this is the pvalue of Z when X = 549.

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

$Z = \frac{549 - 600}{15.49}$

Z = -3.29

Z = -3.29 has a pvalue of 0.0005

0.05% probability that if the coin is actually unfair, we reach a false conclusion

Answer from: Quest

The first one, i believe, would be correct. all the others have one thing in common, and that's that they have repeating numbers. in the second one, x and y have repeating 6's and 3's, and in the third, x and y have repeating 6's as well. in the last one, there are repeating 3's. the first one has no repeating numbers.
Which table shows a proportional relationship between x and y