The length of sunfish in Grenadier pond are normally distributed with a mean of 15cm. Find the standar deviation to one decimal place. If 91% of these fish have a length less than 17.7 cm.
I get about 2.01 for standard deviation. Use your tables.
http://davidmlane.com/hyperstat/z_table.html
To find the standard deviation for the length of sunfish in Grenadier pond, we can use the concept of standard deviation in a normal distribution.
Given:
Mean (μ) = 15 cm
Length (x) of fish where 91% have a length less than = 17.7 cm
To find the standard deviation (σ), we'll use the Z-score formula.
Z = (x - μ) / σ
Since 91% of the fish have a length less than 17.7 cm, this corresponds to a Z-score of approximately 1.34. We can find this value using a standard normal distribution table or a calculator.
Plugging in the values, we have:
1.34 = (17.7 - 15) / σ
To find the standard deviation (σ), we can solve for it:
σ = (17.7 - 15) / 1.34
Performing the calculation, we get:
σ ≈ 1.687 cm (rounded to one decimal place)
Therefore, the standard deviation for the length of sunfish in Grenadier pond is approximately 1.7 cm.
To find the standard deviation, we need to use the information given about the mean and the percentage of fish with a certain length.
Step 1: Convert the given percentage to a z-score.
The z-score represents the number of standard deviations above or below the mean a particular value is. We can use the z-score formula:
z = (x - μ) / σ
where:
z = z-score
x = given value
μ = mean
σ = standard deviation
In this case, we know that 91% of the fish have a length less than 17.7 cm. We can find the z-score for this value using the formula:
z = (17.7 - 15) / σ
Step 2: Calculate the z-score using the standard normal distribution table or a calculator.
Using the standard normal distribution table or a calculator, we can find the z-score corresponding to a given percentage. In this case, we need to find the z-score that corresponds to a 91% probability. The z-score for this probability is approximately 1.34.
Step 3: Solve for the standard deviation (σ).
Now that we have the value of the z-score, we can solve for the standard deviation using the rearranged formula:
σ = (x - μ) / z
Plugging in the known values:
1.34 = (17.7 - 15) / σ
To isolate σ, we multiply both sides of the equation by σ:
1.34 * σ = 17.7 - 15
Finally, we can solve for σ:
1.34 * σ = 2.7
σ = 2.7 / 1.34
σ ≈ 2.01
Therefore, the standard deviation of the length of sunfish in Grenadier pond is approximately 2.01 cm, rounded to one decimal place.