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23. There are frogs and koi in a pond, and their numbers are independent. Let [tex]\(X\)[/tex] represent the number of frogs in any given week, and let [tex]\(Y\)[/tex] represent the number of koi in any given week.

Given:
- [tex]\(X\)[/tex] has a mean of 28 and a standard deviation of 2.7.
- [tex]\(Y\)[/tex] has a mean of 15 and a standard deviation of 1.6.

Which answer choice correctly calculates and interprets the standard deviation of the difference, [tex]\(D = X - Y\)[/tex]?

A. [tex]\(\sigma_D = 1.05\)[/tex]; this pond can expect the difference of frogs and koi to vary by approximately 1.05 from the mean.
B. [tex]\(\sigma_D = 1.1\)[/tex]; this pond can expect the difference of frogs and koi to vary by approximately 1.1 from the mean.
C. [tex]\(\sigma_D = 3.1\)[/tex]; this pond can expect the difference of frogs and koi to vary by approximately 3.1 from the mean.
D. [tex]\(\sigma_D = 13\)[/tex]; this pond can expect the difference of frogs and koi to vary by approximately 1.05 from the mean.

Sagot :

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]