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kari74 [83]
3 years ago
13

A machine fills containers with a particular product. Assume the filling weights are normally distributed with a variance of 0.1

6 ounces2. If 85% of the containers hold greater than 12 ounces, find the machine's mean filling weight (in ounces).
Mathematics
1 answer:
Ghella [55]3 years ago
8 0

Answer:

z=-0.674

And if we solve for \mu we got

\mu=12 +0.674*0.4=12.270

So then the mean is 'mu = 12.270 for this case.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(\mu,\sqrt{0.16}= 0.4)  

For this part we want to find a value a, such that we satisfy this condition:

P(X>12)=0.85   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.85 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.85

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=-0.674

And if we solve for \mu we got

\mu=12 +0.674*0.4=12.270

So then the mean is 'mu = 12.270 for this case.

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Answer:

There is not sufficient evidence to support the claim.

Step-by-step explanation:

The claim to be tested is:

The mean respiration rate (in breaths per minute) of students in a large statistics class is less than 32.

To test this claim the hypothesis can be defined as follows:

<em>H₀</em>: The mean respiration rate of students is 32, i.e. <em>μ</em> = 32.

<em>Hₐ</em>: The mean respiration rate of students is less than 32, i.e. <em>μ</em> < 32.

The sample mean  respiration rate of students is 31.3.

According to the claim the sample mean is less than 32.

The sample mean value is not unusual if the claim is true, and the sample mean value is also not unusual if the claim is false.

Thus, there is not sufficient evidence to support the claim.

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The 12th term of the arithmetic sequence is 10.5. The 18th term of this sequence is 13.5. Find the common difference and the fir
xxMikexx [17]

Answer:

common difference is 0.5

first term is 5

Step-by-step explanation:

<em>use </em><em>the </em><em>formula</em><em> for</em><em> </em><em>the</em><em> </em><em>nth </em><em>term </em><em>of </em><em>an </em><em>ap </em><em>Tn=</em><em>a+</em><em>(</em><em>n-1)</em><em>d</em>

<em>T12=</em><em>1</em><em>0</em><em>.</em><em>5</em>

<em>T18=</em><em>1</em><em>3</em><em>.</em><em>5</em>

<em>therefore</em><em> </em><em>come </em><em>up </em><em>with </em><em>two </em><em>equations</em>

<em>T12=</em><em>a+</em><em>(</em><em>1</em><em>2</em><em>-</em><em>1</em><em>)</em><em>d</em>

<em>1</em><em>0</em><em>.</em><em>5</em><em>=</em><em>a+</em><em>1</em><em>1</em><em>d</em><em>(</em><em>1</em><em>s</em><em>t</em><em> </em><em>equation</em><em>)</em>

<em>T18</em><em>=</em><em>a+</em><em>(</em><em>1</em><em>8</em><em>-</em><em>1</em><em>)</em><em>d</em>

<em>1</em><em>3</em><em>.</em><em>5</em><em>=</em><em>a+</em><em>1</em><em>7</em><em>d</em><em>(</em><em>2</em><em>n</em><em>d</em><em> </em><em>equation</em><em>)</em>

<em>then </em><em>solve </em><em>both </em><em>as </em><em>a </em><em>simultaneous</em><em> </em><em>equation</em>

<em>a+</em><em>1</em><em>1</em><em>d</em><em>=</em><em>1</em><em>0</em><em>.</em><em>5</em>

<em>a+</em><em>1</em><em>7</em><em>d</em><em>=</em><em>1</em><em>3</em><em>.</em><em>5</em>

<em> </em><em> </em><em> </em><em> </em><em>-6d/</em><em>-</em><em>6</em><em>=</em><em>-</em><em>3</em><em>/</em><em>-</em><em>6</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>d=</em><em>0</em><em>.</em><em>5</em>

<em>use </em><em>one </em><em>of</em><em> the</em><em> </em><em>equations</em><em> </em><em>to </em><em>find </em><em>the </em><em>first</em><em> </em><em>term</em>

<em>a+</em><em>1</em><em>1</em><em>(</em><em>0</em><em>.</em><em>5</em><em>)</em><em>=</em><em>1</em><em>0</em><em>.</em><em>5</em>

<em>a+</em><em>5</em><em>.</em><em>5</em><em>=</em><em>1</em><em>0</em><em>.</em><em>5</em>

<em>a=</em><em>1</em><em>0</em><em>.</em><em>5</em><em>-</em><em>5</em><em>.</em><em>5</em>

<em>a=</em><em>5</em>

<em>I </em><em>hope </em><em>this </em><em>helps</em>

<em>please </em><em>mark</em><em> </em><em>as </em><em>brainliest</em>

7 0
3 years ago
(-5c - 3) - 2 = -10c + 20 <br> How would I find c?
Ksivusya [100]
<span>(-5c - 3) - 2 = -10c + 20
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In short, Your Answer would be 5

Hope this helps!</span>
4 0
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