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Irina18 [472]
3 years ago
13

For a specific type of machine part being produced, the diameter is normally distributed, with a mean of 15.000 cm and a standar

d deviation of 0.030 cm. Machine parts with a diameter more than 2 standard deviations away from the mean are rejected. If 15,000 machine parts are manufactured, how many of these will be rejected?
a. 560
b. 680
c. 750
d. 970
Mathematics
2 answers:
Savatey [412]3 years ago
8 0

Answer:

The answer is 750

Step-by-step explanation:

Verizon [17]3 years ago
6 0
This problem involves Statistics. When you are talking about the number of deviations, you directly refer to what we called the Empirical Rule.

The empirical rule states that 95% of the values will lie within 2 standard deviations of the mean. So 100% - 95% = 5% of the parts will be rejected. So
(0.05)*(15,000) = 750

Therefore, 750 machine parts are rejected. Hope this helps! :D
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Answer:

A: 15%

Step-by-step explanation:

First test:

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Second Test:

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You can plug -x - 1 for y in the first equation

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Divide

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Step-by-step explanation:

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27. The average hourly wage of workers at a fast food restaurant is $7.25/hr
morpeh [17]

Answer:

0.0668 = 6.68% probability that the worker earns more than $8.00

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.

The average hourly wage of workers at a fast food restaurant is $7.25/hr with a standard deviation of $0.50.

This means that \mu = 7.25, \sigma = 0.5

If a worker at this fast food restaurant is selected at random, what is the probability that the worker earns more than $8.00?

This is 1 subtracted by the pvalue of Z when X = 8. So

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

Z = \frac{8 - 7.25}{0.5}

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% probability that the worker earns more than $8.00

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