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.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
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 ,
.
Question a:
Here, we have .
This probability is P(X = 7). So
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
Mean and standard deviation:
This probability is, using continuity correction, , which is 1 subtracted by the p-value of Z when X = 21.5. So
has a p-value of 0
1 - 0 = 1
100% probability of at least twenty-two successes.