A tube if Listerine tartar control toothpaste contains 4.2 ounces. As people use

Question

A tube if Listerine tartar control toothpaste contains 4.2 ounces. As people use the toothpaste, the amount remaining in any tube is random. Assume the amount of toothpaste left in the tube follows a uniform distribution. From this information, we can determine the following information about the amount remaining in a toothpaste tube without invading anyone’s privacy.



a. How much toothpaste would you expect to be remaining in the tube?
b. What is the standard deviation of the amount remaining in the tube?
c. What is the likelihood there is less than 3.0 ounces remaining in the tube?
d. What is the probability there is more than 1.5 ounces remaining in the tube?

Explanation / Answer

X - the amount of toothpaste left in the tube X: U(a, b) f(x) = { 0, x b } F(x) = { 0, x b } a.) How much toothpaste would you expect to be remaining in the tube? M(X) = )xdx/(b - a) (from a to b) = (a + b) / 2 b.) What is the standard deviation of the amount remaining in the tube? D(X) = M[X -M(X)]^2 = )(x - (a + b)/2)^2 dx/(b - a) (from a to b) = (b - a)^2 / 12 c.) What is the likelihood there is less than 3.0 ounces remaining in the tube? Probability is p(X < 3.0) = )f(x)dx (from a to 3.0) = F(3.0) - F(a) but,(strictly) a likelihood function is a conditional probability function considered as a function of its second argument with its first argument held fixed, thus: y -> P(Z|Y=y) d.)What is the probability there is more than 1.5 ounces remaining in the tube? " p(X > 1.5) = 1 - P(x = 1.5) = 1 - )f(x)dx (from a to 1.5) = 1 - F(1.5) - F(a)

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