Respuesta :
Answer:
b. 0.0222
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].
Problems of normally distributed samples can be 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.
In this problem, we have that:
[tex]\mu = 402.7, \sigma = 8.8, n = 40, s = \frac{8.8}{\sqrt{40}} = 1.39[/tex]
What is the probability that the mean weight for a sample of 40 trout exceeds 405.5 grams?
This is 1 subtracted by the pvalue of Z when X = 405.5. So
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{405.5 - 402.7}{1.39}[/tex]
[tex]Z = 2.01[/tex]
[tex]Z = 2.01[/tex] has a pvalue of 0.9778.
So the answer is 1-0.9778 = 0.022
