Question

The weights of a small box of healthy oats are normally distributed with a mean of 8.9 oz and a standard deviation of 0.1 oz. Find the probability that a randomly chosen box of healthy oats weighs more than 8.8 oz. Express the probability as a decimal

Answer 1

Using the normal distribution, it is found that there is a 0.8413 probability that a randomly chosen box of healthy oats weighs more than 8.8 oz.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.

In this problem:

• The mean is of 8.9 oz, hence $\mu = 8.9$.

• The standard deviation is of 0.1 oz, hence $\sigma = 0.1$.

The probability that a randomly chosen box of healthy oats weighs more than 8.8 oz is 1 subtracted by the p-value of Z when X = 8.8, hence:

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

$Z = \frac{8.8 - 8.9}{0.1}$

$Z = -1$

$Z = -1$ has a p-value of 0.1587.

1 - 0.1587 = 0.8413.

0.8413 probability that a randomly chosen box of healthy oats weighs more than 8.8 oz.

To learn more about the normal distribution, you can take a look at brainly.com/question/24663213