Respuesta :
Answer:
a) 59.10% probability that 12 or fewer fish were caught.
b) 99.74% probability that 5 or more fish were caught.
c) 58.84% probability that between 5 and 12 fish were caught.
Step-by-step explanation:
I am going to use the normal approximation to the binomial to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]n = 29, p = 0.41[/tex]
So
[tex]\mu = E(X) = np = 29*0.41 = 11.89[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = 2.6486[/tex]
Find the following probabilities.
a) 12 or fewer fish were caught.
Using continuity correction, this is [tex]P(X \leq 12 + 0.5) = P(X \leq 12.5)[/tex], which is the pvalue of Z when X = 12.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{12.5 - 11.89}{2.6486}[/tex]
[tex]Z = 0.23[/tex]
[tex]Z = 0.23[/tex] has a pvalue of 0.5910
59.10% probability that 12 or fewer fish were caught.
b) 5 or more fish were caught.
Using continuity correction, this is [tex]P(X \geq 5 - 0.5) = P(X \geq 4.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 4.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4.5 - 11.89}{2.6486}[/tex]
[tex]Z = -2.79[/tex]
[tex]Z = -2.79[/tex] has a pvalue of 0.0026
1 - 0.0026 = 0.9974
99.74% probability that 5 or more fish were caught.
c) between 5 and 12 fish were caught.
Using continuity correction, this is [tex]P(5 - 0.5 \leq X \leq 12 + 0.5) = P(4.5 \leq X \leq 12.5)[/tex], which is the pvalue of Z when X = 12.5 subtracted by the pvalue of Z when X = 4.5. So.
From a), when X = 12.5, Z has a pvalue of 0.5910
From b), when X = 4.5, Z has a pvalue of 0.0026.
So
0.5910 - 0.0026 = 0.5884
58.84% probability that between 5 and 12 fish were caught.
