Answer:
The required probability is 0.533.
Step-by-step explanation:
Consider the provided information.
The actual weight of the chocolate has a uniform distribution ranging from 31 to 32.5 ounces.
Let x is the random variable for the actual weight of chocolate.
According to PDF function.
[tex]P(a\leq x\leq b)=\int\limits^b_a {f(x)} \, dx[/tex]
Where [tex]f(x)=\left\{\begin{matrix}\frac{1}{b-a} & a<x<b\\ 0 & otherwise \end{matrix}\right.[/tex]
It is given that ranging from 31 to 32.5 ounces.
Substitute a=31 and b=32.5 in above function.
[tex]f(x)=\left\{\begin{matrix}\frac{1}{32.5-31} & 31<x<32.5\\ 0 & otherwise \end{matrix}\right.[/tex]
[tex]f(x)=\left\{\begin{matrix}\frac{1}{1.5} & 31<x<32.5\\ 0 & otherwise \end{matrix}\right.[/tex]
We need to find the probability that a box weighs less than 31.8 ounces
Now according to PDF:
[tex]P(x<31.8)=\int\limits^{31.8}_{31} {\frac{1}{1.5} \, dx[/tex]
[tex]P(x<31.8)=\frac{1}{1.5}[31.8-31]\\P(x<31.8)=\frac{0.8}{1.5}\\P(x<31.8)=0.533[/tex]
Hence, the required probability is 0.533.