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Answer:
By the Central Limit Theorem, both distributions are normal, with mean of 6.4. For samples of size 6, the standard deviation of the the sampling distributions of the sample mean number of damaged avocado fruit is of 0.78, while for samples of 90, it is of 0.2.
Step-by-step explanation:
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.
A recent article states that the population distribution of the number of damaged avocado fruit on moth-infested avocado trees of this variety is symmetric with mean of 6.4 and standard deviation of 1.9.
So normally distributed, with [tex]\mu = 6.4, \sigma = 1.9[/tex]
Samples of 6:
This means that [tex]n = 6, s = \frac{1.9}{\sqrt{6}} = 0.78[/tex]
The mean is the same.
Samples of 90:
This means that [tex]n = 90, s = \frac{1.9}{\sqrt{90}} = 0.2[/tex]
The mean is the same.
By the Central Limit Theorem, both distributions are normal, with mean of 6.4. For samples of size 6, the standard deviation of the the sampling distributions of the sample mean number of damaged avocado fruit is of 0.78, while for samples of 90, it is of 0.2.
From computation, the shapes of the sampling distributions of the sample mean for both sizes of random samples; get closer to normality as the sample size increases with;
Mean = 6.4
Standard deviation for a sample of 6 = 0.776
standard deviation for a sample of 90 = 0.2
We are given;
- Population mean; μ = 6.4
- Population standard deviation; σ = 1.9
Now, according to central limit theorem, since the distribution is normal then the sample mean will be same as Population mean but sample standard deviation is given by;
s = σ/√n
- For random samples of 6 moth-infested trees, the sample standard deviation is; s = 1.9/√6
s = 0.776
- For random samples of 90 moth-infested trees, the sample standard deviation is; s = 1.9/√90
s = 0.2
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