The fill weight of a certain brand of adult cereal is normally distributed with a mean of 910 grams and a standard deviation of 5 grams. If we select one box of cereal at random from this population, what is the probability that it will weigh less than 900 grams?

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Answer:

2.28% probability that it will weigh less than 900 grams

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 = 910, \sigma = 5[/tex]

What is the probability that it will weigh less than 900 grams?

This probability is the pvalue of Z when X = 900. So

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

[tex]Z = \frac{900 - 910}{5}[/tex]

[tex]Z = -2[/tex]

[tex]Z = -2[/tex] has a pvalue of 0.0228.

2.28% probability that it will weigh less than 900 grams

Answer:

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Step-by-step explanation:

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