The mean weight of a brand of cereal is 0.297 kg with a standard deviation of .024kg. Assuming a normal distribution, find the percentage of data that falls below 0.274kg.

Respuesta :

Answer:

16.85% of data that falls below 0.274kg.

Step-by-step explanation:

Problems of normally distributed samples are 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]

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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 0.297, \sigma = 0.024[/tex]

Assuming a normal distribution, find the percentage of data that falls below 0.274kg.

This is the pvalue of Z when X = 0.274. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{0.274 - 0.297}{0.024}[/tex]

[tex]Z = -0.96[/tex]

[tex]Z = -0.96[/tex] has a pvalue of 0.1685.

16.85% of data that falls below 0.274kg.