The mean and standard deviation of the distribution of R−S is;
Option B; Mean 3 and standard deviation 5
We are given details of two random variables R and S.
RandomVariableR;
- Mean; μ_R = 10
- Standard deviation; σ_R = 4
RandomVariable S;
- Mean; μ_S = 7
- Standard Deviation; σ_S = 3
We want to carry out R - S for both mean and standard deviation
1) For Mean; R - S would simply be done by subtracting normally as done in arithmetic operations.
Thus; μ_R - μ_S = 10 - 3
μ_R - μ_S = 7
2) For standard deviation;
We know that standard deviation is not independent as it is the square root of variance.
Therefore, if we want to add or subtract standard deviations, it means that the variation will increase.
Thus, the difference in standard deviation is;
σ_R - σ_S = √((σ_R)² + (σ_S)²)
σ_R - σ_S = √((4)² + (3)²)
σ_R - σ_S = √25
σ_R - σ_S = 5
In conclusion, the distribution of R - S gives;
Mean = 3 and Standard deviation = 5
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Answer: B
Mean of 3 and Standard deviation of 5
Explanation:
Performing operations on means of independent variables is similar to performing arithmetic operations on natural numbers.
Subtracting the mean of S from R is 10 - 7 = 3
For the difference in standard deviation. It should be noted that when combining variables, the variation within the distribution. So therefore, adding or subtraction of variables, the variation will increase and will yield same answer.
To obtain the difference in standard deviation, the variances of the random variables has to be added and then the square root will be taken to return back to standard deviation.
SD(R-S)=sqrt(R²+S²)
SD=sqrt(4²+3³)
SD=sqrt(25)
SD=5
Answer:
Down below
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