Seventy million pounds of trout are grown in the U.S. every year. 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. A random sample of 34 farm-raised trout is selected. The mean fat content for the sample is 30.5 grams per pound. 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.
Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.
To 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, we can use the sampling distribution of the sample mean.
The mean fat content for the population is given as 32 grams per pound, with a standard deviation of 7.5 grams per pound. The sample size is 34.
First, we need to calculate the standard error of the mean (SE):
SE = standard deviation / sqrt(sample size)
SE = 7.5 / sqrt(34)
SE ≈ 1.2869 (rounded to four decimal places)
Next, we need to convert the sample mean of 30.5 grams per pound into a z-score:
z = (sample mean - population mean) / SE
z = (30.5 - 32) / 1.2869
z ≈ -0.6143 (rounded to four decimal places)
Now, we can find the probability using the standard normal distribution table or a statistical software.
P(x ≤ 30.5) = P(z ≤ -0.6143)
Using a standard normal distribution table, we find that the corresponding probability is approximately 0.2686.
Therefore, 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 is approximately 0.269 (rounded to three decimal places).
To 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, we can use the Central Limit Theorem.
The Central Limit Theorem states that the sampling distribution of the sample means approaches a normal distribution, regardless of the shape of the population distribution, as the sample size increases.
To apply the Central Limit Theorem, we need to calculate the sample mean and the standard deviation of the sampling distribution (also known as the standard error).
The sample mean is given as 30.5 grams per pound.
The standard deviation of the sampling distribution (standard error) can be calculated using the formula:
Standard Error = S / sqrt(n)
Where S is the standard deviation of the population (7.5 grams per pound) and n is the sample size (34).
Substituting the values:
Standard Error = 7.5 / sqrt(34)
Calculating this gives us a standard error of approximately 1.2908 grams per pound.
Now, we can calculate the z-score using the formula:
z = (x - μ) / Standard Error
Where x is the sample mean (30.5 grams per pound) and μ is the population mean (32 grams per pound).
Substituting the values:
z = (30.5 - 32) / 1.2908
Calculating this gives us a z-score of approximately -1.3980.
To find the probability of observing a sample mean of 30.5 grams of fat per pound or less, we need to find the area under the normal curve to the left of the z-score (-1.3980).
Using a standard normal distribution table or a calculator, we can find the corresponding probability.
Looking up the z-score -1.3980 in a standard normal distribution table, we find that the probability is approximately 0.0811.
Therefore, 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 is approximately 0.0811, or 8.11%.
,
which is the probability that the mean fat content in farm-raised trout is grams per pound or less. The sample is drawn from a population with mean and standard deviation . We do not know the shape of the population distribution, though, so it would seem that we have too little information about the distribution of to calculate . However, there is an important result called the central limit theorem that applies in this situation. Because the sample size is large enough, regardless of the shape of the population distribution, the distribution of is approximately normal with mean
and standard deviation
.
(Note: we will use the exact value of in the calculations below rather than the approximation of in order to minimize rounding errors.)
Therefore, the variable
follows approximately the standard normal distribution. Thus,
.
The answer is 0.095