Respuesta :
Answer:
The standard deviation of this remaining weight is 0.56.
Step-by-step explanation:
The information provided is:
- The weight of large bags of one brand of potato chips (Z) is a random variable with a mean of E (Z) = 18 ounces and a standard deviation of SD (Z) = 0.7 ounces.
- The amount of chips a professional caterer pours into a certain size of serving bowl (X) can be described as a random variable with a mean of E (X) = 4 ounces and a standard deviation of SD (X) = 0.3 ounces.
- The mean amount left in the bag after the caterer has filled two bowls (Y) is E (Y) = 10 ounces.
- The two variables are independent.
For independent random variables the variance of the sum is the sum of the individual variances.
Compute the standard deviation of this remaining weight as follows:
[tex]SD(Z)=\sqrt{V (X_{1}+V(X_{2})+V(Y)}\\\\0.7=\sqrt{0.3^{2}+0.3^{2}+V (Y)}\\\\0.7^{2}=0.18+V (Y)\\\\V (Y) = 0.31\\\\SD(Y) = \sqrt{0.31}\\\\SD (Y)=0.556778\\\\SD (Y) = 0.56[/tex]
Thus, the standard deviation of this remaining weight is 0.56.
