Respuesta :
Answer:
0.121 = 12.1% probability of observing a sample mean of 30.5 grams of fat per pound or less in a random sample of 34 farm-raised trout.
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 normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, 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 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.
Farm-raised trout contain an average of 32 grams of fat per pound, with a standard deviation of 7.5 grams of fat per pound.
This means that [tex]\mu = 32, \sigma = 7.5[/tex]
Sample of 34:
This means that [tex]n = 34, s = \frac{7.5}{\sqrt{34}} = 1.2862[/tex]
Find the probability of observing a sample mean of 30.5 grams of fat per pound or less in a random sample of 34 farm-raised trout.
This is the pvalue of Z when X = 30.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{30.5 - 32}{1.2862}[/tex]
[tex]Z = -1.17[/tex]
[tex]Z = -1.17[/tex] has a pvalue of 0.121
0.121 = 12.1% probability of observing a sample mean of 30.5 grams of fat per pound or less in a random sample of 34 farm-raised trout.
