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The weight of corn chips dispensed into a 24-ounce bag by the dispensing machine has been identified as possessing a normal distribution with a mean of 25 ounces and a standard deviation of 0.8 ounce. What proportion of the 24-ounce bags contain more than the advertised 24 ounces of chips?

Sagot :

The percentage of 24-ounce bags containing more than the advertised 24 ounces of chips will be -1.25.

What is the z score?

The z-score is a numerical assessment of a value's connection to the mean of a set of values, expressed in terms of standards from the mean, that is used in statistics.

Given data;

Mean,μ = 25

Standard deviation,σ=0.8

The Z score is found as;

[tex]\rm Z = \frac{x-\mu}{ \sigma } \\\\ Z = \frac{24-25}{0.8} \\\\ Z= -1.25[/tex]

The P-value for the obtained z score is  .2113.

Hence the percentage of 24-ounce bags containing more than the advertised 24 ounces of chips will be -1.25.

To learn more about the Z score, refer to;

https://brainly.com/question/15016913

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