Answer:
a) 0.1180 = 11.80% probability that exactly 30 of them are spayed or neutered.
b) 0.8665 = 86.65% probability that at most 33 of them are spayed or neutered.
c) 0.4129 = 41.29% probability that at least 31 of them are spayed or neutered.
d) 0.5557 = 55.57% probability that between 24 and 30 of them are spayed or neutered.
Step-by-step explanation:
To solve this question, we use the binomial probability distribution, and also it's approximation to the normal distribution.
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 ,
.
62% of owned dogs in the United States are spayed or neutered.
This means that
48 owned dogs are randomly selected
This means that
Mean and standard deviation, for the approximation:
a. Exactly 30 of them are spayed or neutered.
This is P(X = 30), which is not necessary the use of the approximation.
0.1180 = 11.80% probability that exactly 30 of them are spayed or neutered.
b. At most 33 of them are spayed or neutered.
Now we use the approximation. This is, using continuity correction, , which is the pvalue of Z when X = 33.5. So
has a pvalue of 0.8665
0.8665 = 86.65% probability that at most 33 of them are spayed or neutered.
c. At least 31 of them are spayed or neutered.
Using continuity correction, this is , which is 1 subtracted by the pvalue of Z when X = 30.5. So
has a pvalue of 0.5871
1 - 0.5871 = 0.4129
0.4129 = 41.29% probability that at least 31 of them are spayed or neutered.
d. Between 24 and 30 (including 24 and 30) of them are spayed or neutered.
This is, using continuity correction, , which is the pvalue of Z when X = 30.5 subtracted by the pvalue of Z when X = 23.5.
X = 30.5
has a pvalue of 0.5871
X = 23.5
has a pvalue of 0.0314
0.5871 - 0.0314 = 0.5557
0.5557 = 55.57% probability that between 24 and 30 of them are spayed or neutered.