To calculate the standard deviation of the difference [tex]\( D = X - Y \)[/tex], where [tex]\( X \)[/tex] and [tex]\( Y \)[/tex] are independent random variables, we follow several steps. Here is a detailed breakdown:
1. Understand the Given Information:
- The random variable [tex]\( X \)[/tex] represents the number of frogs and has a mean [tex]\( \mu_X = 28 \)[/tex] and a standard deviation [tex]\( \sigma_X = 2.7 \)[/tex].
- The random variable [tex]\( Y \)[/tex] represents the number of koi and has a mean [tex]\( \mu_Y = 15 \)[/tex] and a standard deviation [tex]\( \sigma_Y = 1.6 \)[/tex].
2. Calculate the Mean of [tex]\( D \)[/tex]:
The mean of the difference [tex]\( D = X - Y \)[/tex] is given by:
[tex]\[
\mu_D = \mu_X - \mu_Y
\][/tex]
However, for this problem, we are not required to report [tex]\(\mu_D\)[/tex], so we focus on the standard deviation.
3. Calculate the Variance of [tex]\( D \)[/tex]:
Since [tex]\( X \)[/tex] and [tex]\( Y \)[/tex] are independent, their variances add when calculating the variance of the difference [tex]\( D \)[/tex]:
[tex]\[
\sigma_D^2 = \sigma_X^2 + \sigma_Y^2
\][/tex]
Substituting the given values:
[tex]\[
\sigma_D^2 = (2.7)^2 + (1.6)^2
\][/tex]
[tex]\[
\sigma_D^2 = 7.29 + 2.56
\][/tex]
[tex]\[
\sigma_D^2 = 9.85
\][/tex]
4. Calculate the Standard Deviation of [tex]\( D \)[/tex]:
To find the standard deviation, we take the square root of the variance:
[tex]\[
\sigma_D = \sqrt{9.85}
\][/tex]
[tex]\[
\sigma_D \approx 3.1384709652950433
\][/tex]
5. Interpret the Result:
The standard deviation [tex]\(\sigma_D \approx 3.14\)[/tex] means that the difference between the number of frogs and koi can be expected to vary by approximately 3.14 from the mean difference.
Thus, the correct answer is:
[tex]\[
\sigma_D = 3.1 \quad \text{(rounded to 1 decimal place)}
\][/tex]
Interpretation: This pond can expect the difference of frogs and koi to vary by approximately 3.1 from the mean.
Therefore, the correct answer choice is:
[tex]\[
\dot{\sigma}_0 = 3.1; \, \text{this pond can expect the difference of frogs and koi to vary by approximately 3.1 from the mean.}
\][/tex]
