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sertanlavr [38]
3 years ago
9

While measuring a wire a 5cm wire, a construction worker measured it 5.2 cm by mistake. What is the percent error?

Mathematics
1 answer:
nignag [31]3 years ago
5 0

Answer:

4%

Step-by-step explanation:

Actual length of wire = 5 cm

The measured length of wire = 5.2 cm

We need to find the percent error in the measurement of the length of wire. It can be calculated as follows :

\%=\dfrac{\text{measured length-actiual length}}{\text{actual length}}\times 100\\\\=\dfrac{5.2-5}{5}\times 100\\\\=4\%

So, the required percent error is 4%.

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Simply just divide 5 and 3/5:

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A manager at a local manufacturing company has been monitoring the output of one of the machines used to manufacture chromium sh
denpristay [2]

Answer:

0.682 = 68.2% probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly".

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Normally distributed with a mean of 118 centimeters and a standard deviation of 8 centimeters.

This means that \mu = 118, \sigma = 8

Sample of 16 shells

This means that n = 16, s = \frac{8}{\sqrt{16}} = 2

What is the probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly?"

This is the pvalue of Z when X = 120 subtracted by the pvalue of Z when X = 116.

X = 120

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

By the Central Limit Theorem

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

Z = \frac{120 - 118}{2}

Z = 1

Z = 1 has a pvalue of 0.841

X = 116

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

Z = \frac{116 - 118}{2}

Z = -1

Z = -1 has a pvalue of 0.159

0.841 - 0.159 = 0.682

0.682 = 68.2% probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly".

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