The weights of the fish in a certain lake are normally distributed with a mean of 15lbs. and standard deviation of 6lbs. If a sample of 5 fish are taken from the lake, find the probability their means weights are between 12.6 lbs. and 18.6 lbs.
z = (x -x bar)/ std dev/sqrt(n)
z = (12.6-15)/ 6/sqrt(5)
z = -0.89
.1867
z = (x -x bar)/ (std dev/sqrt(n))
z = (18.6-15)/ 6/sqrt(5)
z= 1.34
.9099
.9909-.1867 = .8042
what is the sqrt?
Well, I must say it's quite fishy that we're talking about fish weights! Let's dive in and calculate the probability, shall we?
We know that the mean weight of the fish is 15 lbs, with a standard deviation of 6 lbs. Since we're dealing with a sample of 5 fish, we'll need to calculate the standard deviation of the sample mean. This is done by dividing the standard deviation of the population (6 lbs) by the square root of the sample size (5).
So, the standard deviation of the sample mean is 6 lbs / √5 ≈ 2.68 lbs.
Now, we can convert the weights of 12.6 lbs and 18.6 lbs to z-scores. The z-score can be calculated using the formula:
z = (x - μ) / σ,
where x is the given weight, μ is the mean weight, and σ is the standard deviation of the sample mean.
For 12.6 lbs:
z1 = (12.6 - 15) / 2.68 ≈ -1.12,
For 18.6 lbs:
z2 = (18.6 - 15) / 2.68 ≈ 1.12.
Now, we need to find the probability that the sample mean weights fall between these z-scores. We can find this probability by looking up the z-values in the standard normal distribution table.
Using the table, we find that the probability corresponding to z = -1.12 is approximately 0.1314, and the probability corresponding to z = 1.12 is also approximately 0.1314.
To find the probability of the sample mean weights falling between 12.6 lbs and 18.6 lbs, we subtract the probability corresponding to z1 from the probability corresponding to z2:
P(12.6 < x < 18.6) ≈ 0.1314 - 0.1314 = 0.
Oh dear, it seems like the probability is zero! That means it's highly unlikely for the sample mean weights to fall between 12.6 lbs and 18.6 lbs. Perhaps the fish in this lake have gone on a weight-loss program!
Keep in mind though, this calculation assumes that the fish weights are normally distributed. If there are any other factors at play, like different species of fish or environmental conditions, the distribution may not be exactly normal.
To find the probability that the mean weights of a sample of 5 fish are between 12.6 lbs. and 18.6 lbs., we can use the Central Limit Theorem.
The Central Limit Theorem states that for a large enough sample size (n > 30), the sample mean will be approximately normally distributed regardless of the distribution of the population.
In this case, the sample size is 5 which is smaller than 30. However, since the population distribution is normal, we can still use the normal distribution to approximate the sample means.
To calculate the probability, we need to standardize the values using Z-scores and then use the Z-table or a calculator to find the corresponding probabilities.
The Z-score formula is:
Z = (X - μ) / (σ / sqrt(n))
Where:
Z = Z-score
X = Value we want to standardize (in this case, the sample mean)
μ = Population mean
σ = Population standard deviation
n = Sample size
To find the Z-scores for 12.6 lbs. and 18.6 lbs., we use the formula with the given values:
Z1 = (12.6 - 15) / (6 / sqrt(5))
Z2 = (18.6 - 15) / (6 / sqrt(5))
Calculating the Z-scores:
Z1 = -0.73
Z2 = 0.73
Now we can use the Z-table or a calculator to find the probabilities associated with these Z-scores.
P(12.6 lbs. < X < 18.6 lbs.) = P(-0.73 < Z < 0.73)
Using the Z-table or calculator, we can find that the probability is approximately 0.552.
Therefore, the probability that the mean weights of a sample of 5 fish from the lake are between 12.6 lbs. and 18.6 lbs. is approximately 0.552 or 55.2%.
To solve this problem, we need to use the properties of a normal distribution. The mean weight of the fish in the lake is 15 lbs, and the standard deviation is 6 lbs. We know that the sample size is 5, so we are interested in finding the probability that the mean weight of 5 randomly selected fish falls between 12.6 lbs and 18.6 lbs.
To find this probability, we need to standardize the values using the formula for the standard error of the mean, which is the standard deviation divided by the square root of the sample size. In this case, the standard error would be 6 lbs / sqrt(5) ≈ 2.683 lbs.
Next, we can use the standard normal distribution table or a calculator to find the probabilities for the standardized values. We need to find the probability that the standardized mean falls between (12.6 - 15) / 2.683 and (18.6 - 15) / 2.683.
Calculating the standardized values, we get (12.6 - 15) / 2.683 ≈ -1.0037 and (18.6 - 15) / 2.683 ≈ 0.7486.
Using the standard normal distribution table or a calculator, we can find the probability associated with the range -1.0037 to 0.7486. Subtracting the cumulative probability associated with -1.0037 from the cumulative probability associated with 0.7486 will give us the probability that the mean weight of the 5 fish falls between 12.6 lbs and 18.6 lbs.
Note: The standard normal distribution table provides probabilities for standardized values between -∞ and a given value. To find the cumulative probability associated with a standardized value, you can look up the corresponding value in the table or use a calculator.
I hope this explanation helps in solving the problem!