Solve this statistic question?
A survey of 250 lobster fishermen found that they catch an average of 32.0 pounds of lobster per day with a standard deviation of 4.0 pounds. If a random sample of 36 lobster fishermen is selected, what is the probability that their average catch is less than 31.5 pounds? Assume the distribution of the weights of lobster is normal.
To solve this question, we can use the Central Limit Theorem, which states that the distribution of sample means approximates a normal distribution, regardless of the shape of the original population distribution, as long as the sample size is large enough.
Given that the population mean is 32.0 pounds and the standard deviation is 4.0 pounds, the mean of the sample means will also be 32.0 pounds and the standard deviation of the sample means will be the population standard deviation divided by the square root of the sample size.
First, we need to calculate the standard deviation of the sample means:
Standard deviation (σ) = population standard deviation / √n
= 4.0 / √36
= 4.0 / 6
= 0.6667
Next, we need to calculate the z-score for the sample mean of 31.5 pounds:
z-score (z) = (sample mean - population mean) / standard deviation
= (31.5 - 32.0) / 0.6667
= -0.75 / 0.6667
= -1.125
Using a z-table or a statistical software, we can find the probability associated with the z-score of -1.125.
The probability associated with this z-score is approximately 0.1301.
Therefore, the probability that the average catch of the 36 lobster fishermen is less than 31.5 pounds is approximately 0.1301 or 13.01%.
To solve this statistic question, we will use the concept of sampling distributions and the Central Limit Theorem.
Step 1: Define the problem
We are given that the population mean (μ) is 32.0 pounds with a standard deviation (σ) of 4.0 pounds. We need to find the probability that the average catch of a random sample of 36 lobster fishermen is less than 31.5 pounds.
Step 2: Calculate the standard error
The standard error (SE) represents the variability of the sample means. To calculate it, we use the formula: SE = σ / sqrt(n), where σ is the standard deviation of the population and n is the sample size.
SE = 4.0 / sqrt(36)
SE = 4.0 / 6
SE = 0.67 pounds (rounded to two decimal places)
Step 3: Calculate the z-score
Next, we need to calculate the z-score, which measures the distance between the sample mean and the population mean in terms of standard errors. The formula for calculating the z-score is: z = (x - μ) / SE, where x is the sample mean.
z = (31.5 - 32.0) / 0.67
z = -0.75
Step 4: Find the probability using the z-table
We can then use the z-table (or a statistical calculator) to find the probability associated with the z-score. The z-table gives us the area under the standard normal curve for different z-scores. We need to find the area to the left of the z-score of -0.75.
Looking up the z-score of -0.75 on the z-table, we find that the associated probability is approximately 0.2266.
Step 5: Interpret the result
The probability that the average catch of a random sample of 36 lobster fishermen is less than 31.5 pounds is approximately 0.2266, or 22.66%.
Therefore, there is a 22.66% chance that a random sample of 36 lobster fishermen will have an average catch of less than 31.5 pounds.
If 32 is the average, and the standard deviation is 4, then 31.5 is very very close to the mean.
Use your z-score to find the value to look up : )
z= (x-average)/standard deviation