Question

laundry detergent boxes are filled according to an approximately normal distribution with mean 34.15 ounces and standard deviation 0.1 ounces. what is the probability a randomly selected box has less than the 34 ounces of advertised weight?
probability =

Explanation:

Step1: Identify the distribution and parameters

The problem involves a normal distribution with mean $\mu = 34.15$ ounces and standard deviation $\sigma = 0.1$ ounces. We need to find $P(X < 34)$, where $X$ is the weight of a laundry detergent box.

Step2: Calculate the z-score

The z-score formula is $z = \frac{x - \mu}{\sigma}$. Substituting $x = 34$, $\mu = 34.15$, and $\sigma = 0.1$:
$$z = \frac{34 - 34.15}{0.1} = \frac{-0.15}{0.1} = -1.5$$

Step3: Find the probability using the z-table

We need to find $P(Z < -1.5)$, where $Z$ is the standard normal variable. Looking up $z = -1.5$ in the standard normal table, we find that the area to the left of $z = -1.5$ is 0.0668.

Answer:

0.0668