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.
To find the standard deviation, we can use the information given:
We know that the sunfish length in Grenadier pond is normally distributed with a mean of 15cm. We also know that 91% of these fish have a length less than 17.7cm.
In a normal distribution, we can use the Z-score to determine the area under the curve (probability) corresponding to a given value. The Z-score represents how many standard deviations a given value is from the mean.
In this case, we need to find the Z-score for a length of 17.7cm and determine what Z-score corresponds to the 91st percentile.
The 91st percentile means that 91% of the values in the distribution are below that point. Since the distribution is symmetric, we can find the Z-score that corresponds to the 91st percentile by finding the Z-score that corresponds to the area of (1 - 0.91) = 0.09 in the left tail of the standard normal distribution.
Using a Z-table or statistical software, we can find the Z-score that corresponds to an area of 0.09 in the left tail of the standard normal distribution.
Looking up the Z-score in a Z-table or using statistical software, we find that the Z-score corresponding to an area of 0.09 in the left tail is approximately -1.34.
The Z-score formula is Z = (X - μ) / σ, where Z is the Z-score, X is the observed value, μ is the mean, and σ is the standard deviation.
We can rearrange the formula to solve for σ: σ = (X - μ) / Z.
Plugging the values we know into the formula: σ = (17.7 - 15) / -1.34.
Calculating this: σ ≈ 2.7 / -1.34 ≈ -2.01.
The standard deviation is a measure of the spread of the data, so the negative value doesn't make sense. Therefore, we can take the absolute value of -2.01 to get a positive value.
The standard deviation is approximately 2.01 cm (rounded to one decimal place).