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s2008m [1.1K]
2 years ago
15

4- A manufacturing process produces items whose weights are normally distributed. It is known that 22.57% of all the items produ

ced weigh between 100 grams up to the mean and 49.18% weigh from the mean up to 190 grams. Determine the mean and the standard deviation.
Mathematics
1 answer:
galben [10]2 years ago
7 0

Answer:

\\ \mu = 118\;grams\;and\;\sigma=30\;grams

Step-by-step explanation:

We need to use z-scores and a standard normal table to find the values that corresponds to the probabilities given, and then to solve a system of equations to find \\ \mu\;and\;\sigma.

<h3>First Case: items from 100 grams to the mean</h3>

For finding probabilities that corresponds to z-scores, we are going to use here a <u>Standard Normal Table </u><u><em>for cumulative probabilities from the mean </em></u><em>(Standard normal table. Cumulative from the mean (0 to Z), 2020, in Wikipedia) </em>that is, the "probability that a statistic is between 0 (the mean) and Z".

A value of a z-score for the probability P(100<x<mean) = 22.57% = 0.2257 corresponds to a value of z-score = 0.6, that is, the value is 0.6 standard deviations from the mean. Since this value is <em>below the mean</em> ("the items produced weigh between 100 grams up to the mean"), then the z-score is negative.

Then

\\ z = -0.6\;and\;z = \frac{x-\mu}{\sigma}

\\ -0.6 = \frac{100-\mu}{\sigma} (1)

<h3>Second Case: items from the mean up to 190 grams</h3>

We can apply the same procedure as before. A value of a z-score for the probability P(mean<x<190) = 49.18% = 0.4918 corresponds to a value of z-score = 2.4, which is positive since it is after the mean.

Then

\\ z =2.4\;and\; z = \frac{x-\mu}{\sigma}

\\ 2.4 = \frac{190-\mu}{\sigma} (2)

<h3>Solving a system of equations for values of the mean and standard deviation</h3>

Having equations (1) and (2), we can form a system of two equations and two unknowns values:

\\ -0.6 = \frac{100-\mu}{\sigma} (1)

\\ 2.4 = \frac{190-\mu}{\sigma} (2)

Rearranging these two equations:

\\ -0.6*\sigma = 100-\mu (1)

\\ 2.4*\sigma = 190-\mu (2)

To solve this system of equations, we can multiply (1) by -1, and them sum the two resulting equation:

\\ 0.6*\sigma = -100+\mu (1)

\\ 2.4*\sigma = 190-\mu (2)

Summing both equations, we obtain the following equation:

\\ 3.0*\sigma = 90

Then

\\ \sigma = \frac{90}{3.0} = 30

To find the value of the mean, we need to substitute the value obtained for the standard deviation in equation (2):

\\ 2.4*30 = 190-\mu (2)

\\ 2.4*30 - 190 = -\mu

\\ -2.4*30 + 190 = \mu

\\ \mu = 118

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t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626    

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Information given

60, 56, 60, 55, 70, 55, 60, and 55.

We can calculate the mean and deviation with these formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

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Replacing we got:

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n=8 sample size  

\mu_o =60 represent the value that we want to test

\alpha=0.1 represent the significance level

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p_v represent the p value

Hypothesis to test

We want to test if the true mean is less than 60, the system of hypothesis would be:  

Null hypothesis:\mu \geq 60  

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The statistic would be given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info we got:

t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626    

The degrees of freedom are given by:

df=n-1=8-1=7  

The p value would be given by:

p_v =P(t_{(7)}  

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

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