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Sagot :
Answer:
a) Approximately normal variable.
b) 0.05.
c) 0.0123
d) Approximately 0% probability that more than 10% of her sample will be Snorlaxes
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
5% chance that a Snorlax will appear. Tanya will search Mill Creek park until she captures 312 random pokemon.
This means that [tex]p = 0.05, n = 312[/tex]
a. What type of random variable can we use to approximate the sample proportion?
By the Central Limit Theorem, an approximately normal variable.
b. What is the mean of the sample proportion?
[tex]\mu = p = 0.05[/tex]
The mean is 0.05.
c. What is the standard deviation of the sample proportion?
[tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.05*0.95}{312}} = 0.0123[/tex]
d. What is the approximate probability that more than 10% of her sample will be Snorlaxes?
This is 1 subtracted by the pvalue of Z when X = 0.1. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.1 - 0.05}{0.0123}[/tex]
[tex]Z = 4.05[/tex]
[tex]Z = 4.05[/tex] has a pvalue of approximately 1
1 - 1 = 0
Approximately 0% probability that more than 10% of her sample will be Snorlaxes
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