An island has 12 fur seal rookeries (breeding places). To estimate the fur seal pup population in Rookery A, 6334 fur seal pup
1 answer:
Answer:
37080
Step-by-step explanation:
To solve this problem we need to use proportions:
We don't know the total population in Rookery A but we know that 6334 of the total were tagged. Later, from a sample of 1200 pups, 205 were previously tagged.
The proportions we get are:
where p is the total population.
Now we will solve the proportion for p:
Therefore the total number of fur seal pups is 37077 ≈ 37080
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Answer:
3.67% probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.
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 normally distributed samples are 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.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question, we have that:
Find the probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.
This is the pvalue of Z when X = 0.809.
By the Central Limit Theorem
has a pvalue of 0.0367
3.67% probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.