A vending machine is designed to dispense a mean of 7.6 oz of coffee into an 8-oz cup. If the standard deviation of the amount of coffee dispensed is 0.3 oz and the amount is normally distributed, find the percent of times the machine will dispense less than 7.57 oz.

We have been given that a vending machine is designed to dispense a mean of 7.6 oz of coffee into an 8-oz cup. The standard deviation of the amount of coffee dispensed is 0.3 oz and the amount is normally distributed.

We are asked to find the percent of times the machine will dispense less than 7.57 oz.

First of all, we will find the z-score corresponding to 7.57 using z-score formula.

[tex]z=\frac{x-\mu}{\sigma}[/tex], where,

z = z-score,

x = Random sample score,

[tex]\mu[/tex]= Mean,

[tex]\sigma[/tex] = Standard deviation.

[tex]z=\frac{7.57-7.6}{0.3}[/tex]

[tex]z=\frac{-0.03}{0.3}[/tex]

[tex]z=-0.1[/tex]

Now we will use normal distribution table to find the probability of z-score less than -0.1 that is [tex]P(z<-0.1)[/tex]

Using normal distribution table we will get:

[tex]P(z<-0.1)=0.46017[/tex]

Let us convert [tex]0.46017[/tex] into percentage as:

[tex]0.46017\times 100\%=46.017\%\approx 46.02\%[/tex]

Therefore, 46.02% of times the machine will dispense less than 7.57 oz.

A vending machine is designed to dispense a mean of 7.6 oz of coffee into an​ 8-oz cup. If the standard deviation of the amount of coffee dispensed is 0.3 oz and the amount is normally​ distributed, find the percent of times the machine will dispense less than 7.57 oz.

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A vending machine is designed to dispense a mean of 7.6 oz of coffee into an 8-oz cup. If the standard deviation of the amount of coffee dispensed is 0.3 oz and the amount is normally distributed, find the percent of times the machine will dispense less than 7.57 oz.