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

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

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

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) = (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:

`√(V(X)) = √(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 = (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 = √(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 = √(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 = (X - \mu)/(\sigma)`

`Z = (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.