Question

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?

Answer 1

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 910, \sigma = 5$

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

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{900 - 910}{5}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228.

2.28% probability that it will weigh less than 900 grams

Answer 2

Answer:

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Step-by-step explanation: