For a standardized psychology examination intended for psychology majors, the historical data show that scores have a mean of 52
1 answer:
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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:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed samples 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 ,
.
In this problem, we have that:
(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 . So
This probability, using continuity correction, is , which is 1 subtracted by the pvalue of Z when X = 84.5. So
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 , which is the pvalue of Z when X = 89.5 subtracted by the pvalue of Z when X = 79.5. So
X = 89.5
has a pvalue of 0.8810.
X = 79.5
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
This probability, using continuity correction, is , which is the pvalue of Z when X = 11.5. So
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
Using continuity correction, this probability is , which is the pvalue of Z when X = 55.5. So
has a pvalue of 0.9978
0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry