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zzz [600]
3 years ago
5

A piñata is a container filled with candy and is broke by hitting it with a stick. The probability that Amanda will break a piña

ta on her first his is 0.58. She will continue to hit the piñata until it breaks. If she does not break the piñata on a particular hit, the piñata is weakened and the probability that she will break it on the next hit is 0.14 greater than the probability on the previous hit. For example, if the piñata does not break on the first hit, the probability that it will break on the second hit is 0.72 (found by 0.58 + 0.14 = 0.72). Calculate the probability that Amanda does not break the piñata on the first hit and does break the piñata on the second hit
Mathematics
1 answer:
Jet001 [13]3 years ago
7 0

Answer:

0.3024

Step-by-step explanation:

Given the following :

Probability of breaking pinata on first hit = 0.58;

P(first hit break) = 0.58

P(does not break on first hit) = 1 - p(first hit break) = 1 - 0.58 = 0.42

P(2nd hit break) = p(first hit break) + 0.14 = 0.58 + 0.14 = 0.72

Hence, Calculate the probability that Amanda does not break the piñata on the first hit and does break the piñata on the second hit

P(does not break on first hit) * p(2nd hit break)

0.42 * 0.72 = 0.3024

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Here are the steps:

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Assume that hybridization experiments are conducted with peas having the property that for​ offspring, there is a 0.75 probabili
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(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)

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(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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