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Alexus [3.1K]
3 years ago
13

Applications of Probability Project: Zombie Apocalypse Adapted from "Are You Positive?" Learning Task in the Georgia Frameworks

Zombies are on the loose! A group of researchers wants to be able to detect those who have contracted the zombie virus before they turn into zombies. Those who test positive will be quarantined for two weeks. If they show no signs of being a zombie, they will be freed. They have developed a simple test where they swab the person's ear (mouth would be too dangerous). They gathered 200 people of which 50 were known to be have the zombie virus. Of those who had the virus, 48 tested positive. Of those who did not have the virus, 135 tested negative 1. Based on the data presented above, do you think the test for zombieism is effective at detecting future zombies. Why or why not?
​

Mathematics
1 answer:
Whitepunk [10]3 years ago
5 0

Answer:

Yes, it is a very effective test, having almost a 100% detection rate you wouldn't need much to improve upon except testing multiple times.

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Answer:

(a) The mean and the standard deviation for the numbers of peas with green pods in the groups of 36 is 27 and 2.6 respectively.

(b) The significantly low values are those which are less than or equal to 21.8. And on the other hand, the significantly higher values are those which are greater than or equal to 32.2.

(c) The result of 15 peas with green pods is a result that is significantly​ low value.

Step-by-step explanation:

We are given that hybridization experiments are conducted with peas having the property that for​ offspring, there is a 0.75 probability that a pea has green pods.

Assume that the offspring peas are randomly selected in groups of 36.

The above situation can be represented as a binomial distribution;

where, n = sample of offspring peas = 36

            p = probability that a pea has green pods = 0.75

(a) The mean of the binomial distribution is given by the product of sample size (n) and the probability (p), that is;

                    Mean, \mu  =  n \times p

                                    =  36 \times 0.75 = 27 peas

So, the mean number of peas with green pods in the groups of 36 is 27.

Similarly, the standard deviation of the binomial distribution is given by the formula;

            Standard deviation, \sigma  =  \sqrt{n \times p \times (1-p)}

                                                  =  \sqrt{36 \times 0.75 \times (1-0.75)}

                                                  =  \sqrt{6.75}  =  2.6 peas

So, the standard deviation for the numbers of peas with green pods in the groups of 36 is 2.6.

             

(b) Now, the range rule of thumb states that the usual range of values lies within the 2 standard deviations of the mean, that means;

          \mu - 2 \sigma  =  27 - (2 \times 2.6)

                       =  27 - 5.2 = 21.8

          \mu + 2 \sigma  =  27 + (2 \times 2.6)

                       =  27 + 5.2 = 32.2

This means that the significantly low values are those which are less than or equal to 21.8.

And on the other hand, the significantly higher values are those which are greater than or equal to 32.2.

(c) The result of 15 peas with green pods is a result that is a significantly​ low value because the value of 15 is less than 21.8 which is represented as a significantly low value.

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