Mathematics, 17.05.2021 14:00 alisonlebron15

10 points 4. Suppose you are dealing with a binomial distribution. Calculate the following probabilities:
a) With 10 trials and a 41 chance of success, the probability of exactly seven successes.
b) With 150 trials and a 39 chance of success, the probability of at least twenty-two successes.

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Answer from: jlo15

a) 0.048 = 4.8% probability of exactly seven successes.

b) 100% probability of at least twenty-two successes.

Step-by-step explanation:

The first question we use the binomial distribution, while for the second we use the approximation to the normal.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Question a:

Here, we have n = 10, p = 0.41 .

This probability is P(X = 7). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 7) = C_{10,7}.(0.41)^{7}.(0.59)^{3} = 0.048$

0.048 = 4.8% probability of exactly seven successes.

b) With 150 trials and a 39 chance of success, the probability of at least twenty-two successes.

Here we have n = 150, p = 0.39

Mean and standard deviation:

$\mu = 150*0.39 = 58.5$

$\sigma = \sqrt{150*0.39*0.61} = 5.9737$

This probability is, using continuity correction, $P(X \geq 22 - 0.5) = P(X \geq 21.5)$ , which is 1 subtracted by the p-value of Z when X = 21.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{21.5 - 58.5}{5.9737}$

Z = -6.19

Z = -6.19 has a p-value of 0

1 - 0 = 1

100% probability of at least twenty-two successes.

Answer from: Quest

answer: option a.

step-by-step explanation:

you know that the quadrilateral opqr ( a four sided figure) is inscribed in circle. this means that all its vertices are on the circle.

then, by definition, the opposite angles of the quadrilateral are supplementary, this means that the sum of their measures is 180 degrees.

you can observe in the figure that the angles ∠opq and the angle ∠orq are opposite, therefore:

$m\angle opq+m\angle orq=180\°\\\\m\angle opq+ (2x + 16)\°=180\°$

then, $m\angle opq+ (2x + 16)\°=180\°$ can be used to d to calculate the measure of ∠opq

Quadrilateral opqr is inscribed in circle n, as shown below. which of the following could be used to

Answer from: Quest

bull

step-by-step explanation:

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