African elephants can eat up to 500 pounds of vegetation in one day. How many pounds of vegetation could an elephant eat in a we
1 answer:
You might be interested in
Answer:
The expected value of the sample proportion is of 0.07.
1. P(p < .02) = 0.0823
2. P(p > .15) = 0.0132
3. P(.05 < p < .09) = 0.4176
Step-by-step explanation:
This question is solved using the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use 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
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean and standard deviation
7% of all registered voters belong to the Green party. 50 voters:
This means that
So, for the normal distribution:
The expected value of the sample proportion is of 0.07.
1. Determine P(p < .02).
This is the pvalue of Z when X = 0.02. So
By the Central Limit Theorem
has a pvalue of 0.0823
So
P(p < .02) = 0.0823
2. Determine P(p > .15).
This is 1 subtracted by the pvalue of Z when X = 0.15. So
has a pvalue of 0.9868
1 - 0.9868 = 0.0132
So
P(p > .15) = 0.0132
3. Determine P(.05 < p < .09).
This is the pvalue of Z when X = 0.09 subtracted by the pvalue of Z when X = 0.05. So
X = 0.09
has a pvalue of 0.7088
X = 0.05
has a pvalue of 0.2912
0.7088 - 0.2912 = 0.4176. So
P(.05 < p < .09) = 0.4176