Question

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.

Answer 1

Answer:

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.