Respuesta :
Answer:
There is a 22.66% probability that the amount dispensed into a box is less than 12 ounces.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X.
In this problem:
The amount of cereal dispensed into "12-ounce" boxes of Captain Crisp cereal is normally distributed with mean 12.09 ounces and standard deviation 0.12 ounces, so [tex]\mu = 12.09, \sigma = 0.12[/tex].
That is, what is the probability that the amount dispensed into a box is less than 12 ounces?
This is the pvalue of Z when [tex]X = 12[/tex].
So:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{12 - 12.09}{0.12}[/tex]
[tex]Z = -0.75[/tex]
[tex]Z = -0.75[/tex] has a pvalue of 0.2266.
This means that there is a 22.66% probability that the amount dispensed into a box is less than 12 ounces.
