the weights of the fish in a certain lake are normally distributed with a mean of 17 lb and a standard deviation of 12. if 16 fish are randomly selected, what is the probability that the mean weight will be between 14.6 and 20.6 lb?
you can play around with Z table stuff here:
http://davidmlane.com/hyperstat/z_table.html
To find the probability that the mean weight of 16 randomly selected fish will be between 14.6 and 20.6 lb, we can use the Central Limit Theorem.
The Central Limit Theorem states that for a large enough sample size (typically n ≥ 30), the sampling distribution of the sample means will be approximately normally distributed, even if the population distribution is not normal.
In this case, we have a sample size of 16, which satisfies the condition for the Central Limit Theorem, so we can approximate the sampling distribution of the sample means as a normal distribution.
First, we need to calculate the standard error of the mean (SE), which is the standard deviation divided by the square root of the sample size. The standard deviation is given as 12, and the sample size is 16, so:
SE = 12 / √16 = 12 / 4 = 3
Next, we need to standardize the values 14.6 and 20.6 using the formula:
Z-score = (X - mean) / SE
For 14.6:
Z1 = (14.6 - 17) / 3 = -0.8
For 20.6:
Z2 = (20.6 - 17) / 3 = 1.2
Now, we can use the standard normal distribution table or a statistical calculator to find the area between these two Z-scores. Alternatively, we can find the probabilities corresponding to each Z-score separately and subtract them.
Using a standard normal distribution table, we can find that the probability associated with Z = -0.8 is approximately 0.2119, and the probability associated with Z = 1.2 is approximately 0.8849.
To find the probability between these two Z-scores, we subtract the probability associated with Z = -0.8 from the probability associated with Z = 1.2:
Probability = 0.8849 - 0.2119 = 0.6730
Therefore, the probability that the mean weight of 16 randomly selected fish will be between 14.6 and 20.6 lb is approximately 0.6730, or 67.30%.