Suppose a population of 10 spiders doubles in size every month. How
2 answers:
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Answer:
a) By the Central Limit Theorem, approximately normally distributed, with mean 26 and standard error 0.44.
b) s = 0.44
c) 0.84% of the sample means will be greater than 27.05 seconds
d) 98.46% of the sample means will be greater than 25.05 seconds
e) 97.62% of the sample means will be greater than 25.05 but less than 27.05 seconds
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(also called standard error)
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
a. What can we say about the shape of the distribution of the sample mean time?
By the Central Limit Theorem, approximately normally distributed, with mean 26 and standard error 0.44.
b. What is the standard error of the mean time? (Round your answer to 2 decimal places)
c. What percent of the sample means will be greater than 27.05 seconds?
This is 1 subtracted by the pvalue of Z when X = 27.05. So
By the Central Limit Theorem
has a pvalue of 0.9916
1 - 0.9916 = 0.0084
0.84% of the sample means will be greater than 27.05 seconds
d. What percent of the sample means will be greater than 25.05 seconds?
This is 1 subtracted by the pvalue of Z when X = 25.05. So
has a pvalue of 0.0154
1 - 0.0154 = 0.9846
98.46% of the sample means will be greater than 25.05 seconds
e. What percent of the sample means will be greater than 25.05 but less than 27.05 seconds?"
This is the pvalue of Z when X = 27.05 subtracted by the pvalue of Z when X = 25.05.
X = 27.05
has a pvalue of 0.9916
X = 25.05
has a pvalue of 0.0154
0.9916 - 0.0154 = 0.9762
97.62% of the sample means will be greater than 25.05 but less than 27.05 seconds