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

Using the normal distribution concept, the chip amount which corresponds to the 67th percentile is 16.59

Using the concept of normal distribution :

The 67th percentile is defined as :

P(Z < 0.67) ; Using a normal distribution table, the corresponding Zscore value is 0.44

Substitute into the Zscore relation and solve for the score, X :

0.44 = (X - 16.5) / 0.2

Cross multiply

0.44 × 0.2 = (X - 16.5)

0.088 = X - 16.5

0.088 + 16.5 = X

X = 16.588

Therefore, the chip amount is 16.59.

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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 and standard deviation , the zscore of a measure X is given by:

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:

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.

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

i think the answer is 90 hope this

on the middle line

step-by-step explanation:

1/5=50