Respuesta :
Answer:
0.0062 = 0.62% of the 24-ounce bags contain less than the advertised 24 ounces of chips.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean of 24.5 ounces and a standard deviation of 0.2 ounce.
This means that [tex]\mu = 24.5, \sigma = 0.2[/tex]
What proportion of the 24-ounce bags contain less than the advertised 24 ounces of chips?
This is the p-value of Z when X = 24. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{24 - 24.5}{0.2}[/tex]
[tex]Z = -2.5[/tex]
[tex]Z = -2.5[/tex] has a p-value of 0.0062.
0.0062 = 0.62% of the 24-ounce bags contain less than the advertised 24 ounces of chips.
