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djverab [1.8K]
3 years ago
14

A machine used to fill​ gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 137 ounces and

a standard deviation of 0.30 ounces. You randomly select 45 cans and carefully measure the contents. The sample mean of the cans is 136.9 ounces. Does the machine need to be​ reset? Explain your reasoning.
Mathematics
2 answers:
ElenaW [278]3 years ago
5 0
 <span>Standard error of mean = Sigma / sqrt n 
= 0.30 / sqrt 45 
= 0.30 / 6.71 
= 0.045 
Xbar - Mu = 138.9 - 139 = - 0.1 
- 0.1 / 0.045 = - 2.22 

YES, because the sample mean lies beyond 2 Standard deviations on the left side of mu.

If this was the appropriate answer mark as the brainliest!
-procklown</span>
tensa zangetsu [6.8K]3 years ago
3 0

Answer:

Step-by-step explanation:

Let X be the amount of paint dispensed by a machine used to fill gallon=sized paint cans.

X is Normal(137, 0.30)

Sample size = 45

Sample std error = 0.30/sqrt 45 =0.0447

x bar = 136.9

Create hypotheses as:

H0: sample mean = 137

Ha: sample mean not equal to 137

(Two tailed test)

Test statistic = (136.9-137)/0.0447

=-2.237

p value = 0.0257

Since p >0.01, at 99% level we accept null hypothesis

Machine need not be reset.

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Suppose you have a bag containing 2 black marbles and 3 red marbles. You reach into the bag, select a marble, see what color it
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Answer:

\dfrac{9}{25}

Step-by-step explanation:

Given that the bag contains black and red marbles.

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Number of red marbles in the bag = 3

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Let us have a look at the formula for probability of an event E, which can be observed as:

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P(\text{First red marble}) = \dfrac{\text{Number of red marbles}}{\text{Total number of marbles}} = \dfrac{3}{5 }

Now, the marble chosen at first is replaced.

Therefore, the count remains the same.

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Now, the <em>required probability</em> can be found as:

P(\text{First red marble})\times P(\text{Second red marble}) = \dfrac{3}{5}\times \dfrac{3}{5} = \bold{\dfrac{9}{25} }

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3 years ago
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0.08x+0.03 (29000-x)=
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