Respuesta :
Answer:
a) 0.6628 = 66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance
b) 0.5141 = 51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance
c) 0.5596 = 55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.
d) 0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry
Step-by-step explanation:
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 = 175[/tex]
(a) What is the probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance?
46.7% of visitors to Rocky Mountain National Park in 2018 entered through the Beaver Meadows. This means that [tex]p = 0.467[/tex]. So
[tex]\mu = E(X) = np = 175*0.467 = 81.725[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.467*0.533} = 6.6[/tex]
This probability, using continuity correction, is [tex]P(X \geq 85 - 0.5) = P(X \geq 84.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 84.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{84.5 - 81.725}{6.6}[/tex]
[tex]Z = 0.42[/tex]
[tex]Z = 0.42[/tex] has a pvalue of 0.6628.
66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance.
(b) What is the probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance?
Using continuity correction, this is [tex]P(80 - 0.5 \leq X < 90 - 0.5) = P(79.5 \leq X \leq 89.5)[/tex], which is the pvalue of Z when X = 89.5 subtracted by the pvalue of Z when X = 79.5. So
X = 89.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{89.5 - 81.725}{6.6}[/tex]
[tex]Z = 1.18[/tex]
[tex]Z = 1.18[/tex] has a pvalue of 0.8810.
X = 79.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{79.5 - 81.725}{6.6}[/tex]
[tex]Z = -0.34[/tex]
[tex]Z = -0.34[/tex] has a pvalue of 0.3669.
0.8810 - 0.3669 = 0.5141
51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance
(c) What is the probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance?
6.3% of visitors entered through the Grand Lake park entrance, which means that [tex]p = 0.063[/tex]
[tex]\mu = E(X) = np = 175*0.063 = 11.025[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.063*0.937} = 3.2141[/tex]
This probability, using continuity correction, is [tex]P(X < 12 - 0.5) = P(X < 11.5)[/tex], which is the pvalue of Z when X = 11.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{11.5 - 11.025}{3.2141}[/tex]
[tex]Z = 0.15[/tex]
[tex]Z = 0.15[/tex] has a pvalue of 0.5596.
55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.
(d) What is the probability that more than 55 visitors have no recorded point of entry?
22.7% of visitors had no recorded point of entry to the park. This means that [tex]p = 0.227[/tex]
[tex]\mu = E(X) = np = 175*0.227 = 39.725[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.227*0.773} = 5.54[/tex]
Using continuity correction, this probability is [tex]P(X \leq 55 + 0.5) = P(X \leq 55.5)[/tex], which is the pvalue of Z when X = 55.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{55.5 - 39.725}{5.54}[/tex]
[tex]Z = 2.85[/tex]
[tex]Z = 2.85[/tex] has a pvalue of 0.9978
0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry
