Respuesta :
Answer:
67.30% probability that the mean weight will be between 16.6 and 22.6 lb
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], 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]
In this problem, we have that:
[tex]\mu = 19, \sigma = 6, n = 4, s = \frac{6}{\sqrt{4}} = 3[/tex]
If 4 fish are randomly selected, what is the probability that the mean weight will be between 16.6 and 22.6 lb
This is the pvalue of Z when X = 22.6 subtracted by the pvalue of Z when X = 16.6.
X = 22.6
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{22.6 - 19}{3}[/tex]
[tex]Z = 1.2[/tex]
[tex]Z = 1.2[/tex] has a pvalue of 0.8849
X = 16.6
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{16.6 - 19}{3}[/tex]
[tex]Z = -0.8[/tex]
[tex]Z = -0.8[/tex] has a pvalue of 0.2119
0.8849 - 0.2119 = 0.6730
67.30% probability that the mean weight will be between 16.6 and 22.6 lb
