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Ludmilka [50]
3 years ago
10

A multiple choice test is composed of 10 questions. Each question has four possible answers, only one of the four possible answe

rs is correct. A student randomly selects the answer for each question. Assume that all selections are independent. Let X count the student’s number of correct answers on the test. a) What is the expected number of correct answers? Round your answer to the nearest tenth.
Mathematics
1 answer:
kompoz [17]3 years ago
4 0

Answer:

Let X the random variable of interest "Number of correct anwers on the tet", on this case we now that:

X \sim Binom(n=10, p=0.25)

And the expected value is given by:

E(X) = np =10*0.25 = 2.5

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

Solution to the problem

Let X the random variable of interest "Number of correct anwers on the tet", on this case we now that:

X \sim Binom(n=10, p=0.25)

And the expected value is given by:

E(X) = np =10*0.25 = 2.5

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What function do you know from calculus is such that its first derivative is itself?The above function is a solution of which of
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Answer:

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Step-by-step explanation:

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But when we take the first derivative of  y= e raise to power y

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On simplification

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lny = ln k

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therefore

derivative of e^ky = ke^ky

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