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Arte-miy333 [17]
3 years ago
11

If the outliers are not included, what is the mean of the data? 76,79,80,82,50,78,83,79,81,82.

Mathematics
1 answer:
natta225 [31]3 years ago
4 0
50 is the only outlier so take 50 out of the set then add the other 9 data points and divide by 9.<span>
76 + 79 + 80 + 82 + 78 + 83 + 79 + 81 + 82 = 720 / 9 = 80</span>
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A cola-dispensing machine is set to dispense 8 ounces of cola per cup, with a standard deviation of 1.0 ounce. The manufacturer
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Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: ounces per cup dispensed by the cola-dispensing machine.

The population mean is known to be μ= 8 ounces and its standard deviation σ= 1.0 ounce. Assuming the variable has a normal distribution.

A sample of 34 cups was taken:

a. You need to calculate the Z-values corresponding to the top 5% of the distribution and the lower 5% of it. This means you have to look for both Z-values that separates two tails of 5% each from the body of the distribution:

The lower value will be:

Z_{o.o5}= -1.648

You reverse the standardization using the formula Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~N(0;1)

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P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)

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There is a 0.242 probability of the sample means being between the control limits, this means that they will be outside the limits with a probability of 1 - 0.242= 0.758, meaning that the probability of the change of population mean being detected is 0.758.

b. For this item μ= 8.7, the control limits do not change:

P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)

P(Z≤(8.28-8.7)/(1/√34))- P(Z≤7.72-8.7)/(1/√34))

P(Z≤-2.45)- P(Z≤-5.71)=0.007 - 0= 0.007

There is a 0.007 probability of not detecting the mean change, which means that you can detect it with a probability of 0.993.

I hope it helps!

5 0
3 years ago
Read 2 more answers
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Answer:

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