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3 years ago

Write a paragraph on testing probability

Biology

1 answer:

VashaNatasha [74]3 years ago

5 0

Explanation:

Probabilities are described as ratios of favorable event outcome to the total number of event outcomes.

This is written as...

$P (E) =\frac{n(E)}{n(S)} \\$

where...

E= the number of times the event occurs

S= the number of trials

In biology experiments, hypotheses are formed based on research questions, and tested with the use of variables to provide a particular outcome. Statistics allows for testing data for consistency with the hypothesis, while statistical probability testing can be used in experiments to determine a range of outcomes, from genetic inheritance, evolutionary rates to theoretical experimental results.

In these statistical models, probability distributions are functions that give probabilities for certain event outcomes within an experiment (a set of trials). These may be either continuous, taking a value within a range of two numbers; or discrete, which may be either of two specified values. Discrete probability distributions list each value that a random variable may possibly take on.

Further Explanation:

For example, two types of probability distributions are employed in experimental biology:

Binomial distributions, which are discrete distributions, provide probability of a certain number of successful events for x a random variable, in a specific number of trials, n; here, the probability of success of an individual trial is constant at P and only one of two outcomes are possible- this is sampling with replacement.

where...

b(x; n, P)-the probability that an experiment of n trials results in x successes

nCx- the number of combinations of n things at r time

$b(x; n, P) = [ nCx ]* P^{x} * (1-P)^{n-x}\\$

This is often used in determining potential outcomes before data collection.

A type of continuous distribution, the student's t-test, compares standard deviations and means from two sets of samples or groups to check for significant differences between them.

$t= \frac{(x_{1} - x_{2}) }{\sqrt{(\frac{(S_{1}) ^{2} }{n1} }+ (\frac{(S_{2}) ^{2} }{n2 }}$

where...

x1 and s1 are the mean and standard deviation of sample 1 respectively
x2 and s2 are the mean and standard deviation of sample 1 respectively
n1 and n2 are sample sizes in samples 1 and 2 respectively

The null and alternate hypotheses typically theorize the likelihood and significance of certain event outcome probabilities. Critical values of t, along with degrees of freedom are used to determine a range of probable outcomes; probability or p- values along with this range, are used to determine whether either hypothesis is rejected or accepted.

For instance, significant differences between an experimental control and a specific treatment group would show that these occurrences are not due to sampling errors or random chance...

Learn more about calculating probability at brainly.com/question/4021035

Learn more about calculating event probability at brainly.com/question/6649771

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<h3>What is Carbon 14?</h3>

Carbon 14, also known as radiocarbon, is a radioactive carbon isotope.

Isotopes are the atoms of the same element -carbon- that vary in neutrons and, hence, in their massic number. They are alternative forms of the same element.

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The term half-life is a reference. It means that an organism that has been dead for 5730 years has half the C14 amount or concentration than the same organism had when it was alive.

Knowing the half-life of an element is useful to determine the age of the dead matter.

C14 is used in radiocarbon dating techniques or methods to estimate the age of fossils. This is a reliable technique used for dating organic samples that are less than 50,000 years old.

Available data:

The half-life of C14 is 5730 years
Bones activity of 0.10 Bq per gram of carbon

To answer this question we can make use of the following equation

Ln (C14T₁/C14 T₀) = - λ T₁

Where,

C14 T₀ ⇒ Amount of carbon in a living body. We know, by bibliography, that living organism activity is 0.23 Bq per gram of carbon. So, C14 T₀ = 0.23 Bq/g

C14T₁ ⇒ Amount of carbon in the dead body. C14T₁ = 0.1 Bq/g

λ ⇒ radioactive decay constant = (Ln2)/T₀,₅

T₀,₅ ⇒ The half-life of carbon 14 = 5730 years

T₀ = Time when the organism was alive

T₁ = Age of bones

Let us first calculate the radioactive decay constant.

λ = (Ln2)/T₀,₅

λ = 0.693/5730

λ = 0.0001209

Now, let us calculate the first term in the equation

Ln (C14T₁/C14 T₀) = Ln (0.1/0.23) = Ln 0.4347 = - 0.833

Finally, let us replace the terms, clear the equation, and calculate the value of T₁.

Ln (C14T₁/C14 T₀) = - λ T₁

- 0.833 = - 0.0001209 x T₁

T₁ = - 0.833 / - 0.0001209

T₁ = 6889.99 ≅ 6890 years

The bones are approximately 6890 years.

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