The weights of the fish in a certain lake are normally distributed with a mean of 11 lb and a
standard deviation of 6. If 4 fish are randomly selected, what is the probability that the mean
weight will be between 8.6 and 14.6 lb?
A) 0.0968 B) 0.4032 C) 0.3270 D) 0.6730
i still don't get how to do it?
To solve this problem, we need to use the properties of the normal distribution and the concept of the sampling distribution.
The weights of the fish in the lake are normally distributed with a mean of 11 lb and a standard deviation of 6 lb. When we randomly select a sample of fish and calculate the mean weight of that sample, this is called the sample mean. The mean of the sample means is the same as the population mean, which is 11 lb. The standard deviation of the sample means, also known as the standard error, can be calculated by dividing the standard deviation of the population by the square root of the sample size.
In this case, we are randomly selecting 4 fish, so the standard deviation of the sample means is 6 / √4 = 3 lb.
Now, we need to compute the z-scores for the lower and upper limits of the desired range of mean weights. The z-score of a value represents the number of standard deviations it is away from the mean. We can calculate the z-score using the formula:
z = (x - μ) / σ
Where:
- x is the value of interest (in this case, the lower and upper limits of the desired range)
- μ is the mean of the population (11 lb)
- σ is the standard deviation of the sample means (3 lb)
For the lower limit (8.6 lb):
z1 = (8.6 - 11) / 3 = -0.8
For the upper limit (14.6 lb):
z2 = (14.6 - 11) / 3 = 1.2
Next, we need to find the probability associated with each z-score using the standard normal distribution table (also known as the z-table). The z-table provides the cumulative probability up to a given z-score. We want to find the probability between these two z-scores, so we need to calculate the area between them.
Using the z-table, we find that the area to the left of z1 (-0.8) is 0.2119, and the area to the left of z2 (1.2) is 0.8849. To find the area between these two z-scores, we subtract the smaller area from the larger area:
P(z1 < z < z2) = P(z < z2) - P(z < z1)
= 0.8849 - 0.2119
= 0.6730
Therefore, the probability that the mean weight will be between 8.6 and 14.6 lb for a sample of 4 fish is 0.6730. The correct answer is (D) 0.6730.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
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 proportions related to the Z scores.