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A supplier of kerosene has a 150-gallon tank that is
Chapter 4, Problem 13E(choose chapter or problem)
A supplier of kerosene has a 150-gallon tank that is filled at the beginning of each week. His weekly demand shows a relative frequency behavior that increases steadily up to 100 gallons and then levels off between 100 and 150 gallons. If Y denotes weekly demand in hundreds of gallons, the relative frequency of demand can be modeled by
\(f(y)=\left\{\begin{array}{ll} y, & 0 \leq y \leq 1, \\ 1, & 1<y \leq 1.5, \\ 0, & \text { elsewhere. } \end{array}\right. \)
a Find \(F(y)\).
b Find \(P(0 \leq Y \leq .5)\).
c Find \(P(.5 \leq Y \leq 1.2)\).
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QUESTION:
A supplier of kerosene has a 150-gallon tank that is filled at the beginning of each week. His weekly demand shows a relative frequency behavior that increases steadily up to 100 gallons and then levels off between 100 and 150 gallons. If Y denotes weekly demand in hundreds of gallons, the relative frequency of demand can be modeled by
\(f(y)=\left\{\begin{array}{ll} y, & 0 \leq y \leq 1, \\ 1, & 1<y \leq 1.5, \\ 0, & \text { elsewhere. } \end{array}\right. \)
a Find \(F(y)\).
b Find \(P(0 \leq Y \leq .5)\).
c Find \(P(.5 \leq Y \leq 1.2)\).
Step 1 of 3
Let Y denote weekly demand in hundreds of gallons, the relative frequency of demand can be modeled by
\(f(y) =\begin{cases} y, & 0\leq y\leq 1\\ 1, & 1< y\leq 1.5\\ 0, & elsewhere \end{cases}\)
Our goal is:
a) We need to find \(F(y)\).
b) We need to find \(P(0\leq y\leq 0.5)\).
c) We need to find \(P(0.5\leq y\leq 1.2)\).
a)
Now we have to find \(F(y)\).
We have been asked to calculate the distribution function.
The calculation yields the following results,
For \(0\leq y\leq 1\),
Here we take the limit 0 to y.
\(F(y)=\int_{0}^{y}tdt\)
\(F(y)=\frac {y^2}{2}\)
Then,
For \(1<y\leq 1.5\),
Here we take the limit 0 to 1 and 1 to y.
\(F(y)=\int_{0}^{1}t\ dt+\int_{1}^{y}dt\)
\(F(y)=\frac {1}{2}+y-1\)
\( y=1-\frac{1}{2}\)
The density function is given below
\(f(y) =\begin{cases} 0, & y< 1\\ \frac{y^2}{2}, & 0\leq y\leq 1\\ y-\frac{1}{2}, & 1<y \leq 1.5\\ 1, & y>1.5 \end{cases}\)
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