Question

The amount of corn chips dispensed into a 16-ounce bag by the dispensing machine has been identified as possessing a normal distribution with a mean of 16.5 ounces and a standard deviation of 0.2 ounce. What chip amount represents the 67th percentile for the bag weight distribution

Answer 1

Answer:

A chip of 16.59 ounces represents the 67th percentile for the bag weight distribution

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

$\mu = 16.5, \sigma = 0.2$

What chip amount represents the 67th percentile for the bag weight distribution

This is X when Z has a pvalue of 0.67, so X when Z = 0.44.

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

$0.44 = \frac{X - 16.5}{0.2}$

$X - 16.5 = 0.44*0.2$

$X = 16.59$

A chip of 16.59 ounces represents the 67th percentile for the bag weight distribution