The equation for this polynomial graphed will be P(x) = a(x + 2)(x + 1)(x – 3)³.
<h3>What is polynomial?</h3>A polynomial expression is an algebraic expression with variables and coefficients.
The graph of the polynomial is given below.
From the graph, the roots of the polynomial will be -2, -1, and 3.
Then the factor of the polynomial will be
(x + 2), (x + 1), and (x – 3)
Then the polynomial will be
P(x) = a(x + 2)(x + 1)(x – 3)³
More about the polynomial link is given below.
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Answer:
Using the probability mass function we can find the probability required
We need to check the conditions in order to use the normal approximation.
We assume independence in the events
So we see that we satisfy the conditions and then we can apply the approximation.
If we appply the approximation the new mean and standard deviation are:
So then we can approximate the normal with
But for this case since we have a continuous distribution with this approximation we can't find since the are below a point is 0, w can find
or
but not
using the normal approximation.
Step-by-step explanation:
Previous concepts
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".
Let X the random variable of interest, on this case we now that:
The probability mass function for the Binomial distribution is given as:
Where (nCx) means combinatory and it's given by this formula:
Solution to the problem
For this case we have the following distirbuion for X the number of undergraduates
Using the probability mass function we can find the probability required
We need to check the conditions in order to use the normal approximation.
We assume independence in the events
So we see that we satisfy the conditions and then we can apply the approximation.
If we appply the approximation the new mean and standard deviation are:
So then we can approximate the normal with
But for this case since we have a continuous distribution with this approximation we can't find since the are below a point is 0, w can find
or
but not
using the normal approximation.