# Solved:The actual amount of coffee (in grams) in a 230-gram jar filled by a certain machine

#### ByDr. Raju Chaudhari

Sep 20, 2020

The actual amount of coffee (in grams) in a 230-gram jar filled by a certain machine is a random variable whose probability density is given by

 \begin{aligned} f(x)=\left\{ \begin{array}{ll} 0 & \hbox{x \leq 227.5;}\\ \frac{1}{5} , & \hbox{227.5 < x < 232.5;} \\ 0, & \hbox{x\geq 232.5.} \end{array} \right. \end{aligned}

Find the probabilities that a 230-gram jar filled by this machine will contain

(a) at most 228.65 grams of coffee;
(b) anywhere from 229.34 to 231.66 grams of coffee;
(c) at least 229.85 grams of coffee.

### Solution

(a) The probabilities that a 230-gram jar filled by this machine will contain
at most 228.65 grams of coffee is

 \begin{aligned} P(X\leq 228.65) &= \int_{227.5}^{228.65} f(x) \; dx\\ &= \frac{1}{5}\int_{227.5}^{228.65} \; dx\\ &= \frac{1}{5}\big[x\big]_{227.5}^{228.65} \\ &= \frac{1}{5}\big[ 228.65 -227.5]\\ &= 0.23 \end{aligned}

(b) The probabilities that a 230-gram jar filled by this machine will contain anywhere from 229.34 to 231.66 grams of coffee is

 \begin{aligned} P(229.34< X< 231.66) &= \int_{229.34}^{231.66} f(x) \; dx\\ &= \frac{1}{5}\int_{229.34}^{231.66} \; dx\\ &= \frac{1}{5}\big[x\big]_{229.34}^{231.66} \\ &= \frac{1}{5}\big[ 231.66 -229.34]\\ &= 0.464 \end{aligned}

(c) The probabilities that a 230-gram jar filled by this machine will contain
at least 229.85 grams of coffee is

 \begin{aligned} P(X\geq 229.85) &= \int_{229.85}^{232.5} f(x) \; dx\\ &= \frac{1}{5}\int_{229.85}^{232.5} \; dx\\ &= \frac{1}{5}\big[x\big]_{229.85}^{232.5} \\ &= \frac{1}{5}\big[ 232.5 -229.85]\\ &= 0.53 \end{aligned}