Answer to Question #137793 in Statistics and Probability for John Kafs

Question #137793
Customers at TAB are charged for the amount of salad the take. Sampling suggests that the amount of salad taken is uniformly distributed between 5 ounces and 15 ounces. Let
Χ = Salad plate filling weight.
i. Find the probability density function of Χ
ii. What is the probability that a customer will take between 12 and 15 ounces of salad? iii. What is the probability that a customer will take fewer than 5 ounces of salad.
iv. Find Ε ( Χ) and Var ( Χ)
1
Expert's answer
2020-10-13T19:11:08-0400

solution


i) PDF of X


For a uniform distribution,


f(x)=1baf(x)=\frac{1}{b-a}

For X between 5 and 15 ounces:


f(x)=1155=110f(x) =\frac{1}{15-5} =\frac{1}{10}



Answer: f(x)=110f(x)=\frac{1}{10}


ii) Probability that 12<X<1512<X<15


f(12<X<15)=(1512)110=310f(12<X<15)=(15-12)*\frac{1}{10} = \frac{3}{10}



answer: f(12<x<15)=310f(12<x<15)=\frac{3}{10}

iii) probability that X<5X<5


The domain of X is the range of values where the PDF is valid. The domain of X=[5,15]X=[5,15]

The PDF is not valid for values of X<5X<5


answer: f(X<5)=0f(X<5) = 0


iv) E(X) and Var(X)



E(X)=b+a2E(X) = \frac{b+a}{2}

=15+52=10= \frac{15+5}{2}= 10




Var(X)=(ba)212Var(X)= \frac{(b-a)^2}{12}

=(155)212=10012=8.3333= \frac{(15-5)^2}{12}=\frac{100}{12}=8.3333


answer: E(X)=10E(X) = 10 and Var(X)=8.3333Var(X)=8.3333



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