Question

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.

Answer 1

16.85% of data that falls below 0.274kg.

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

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:

`\mu = 0.297, \sigma = 0.024`

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

`Z = (X - \mu)/(\sigma)`

`Z = (0.274 - 0.297)/(0.024)`

`Z = -0.96`

`Z = -0.96` has a pvalue of 0.1685.

16.85% of data that falls below 0.274kg.