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attashe74 [19]
2 years ago
8

The diameter of small Nerf balls manufactured overseas is expected to be approximately normally distributed with a mean of 5.2 i

nches and a standard deviation of .08 inches. Suppose a random sample of 20 balls is selected. Calculate the mean of the sampling distribution of the sample mean.
Mathematics
1 answer:
ioda2 years ago
7 0

Answer:

5.2 inches

Step-by-step explanation:

The diameter of small Nerf balls follows normally distribution with mean 5.2 inches and standard deviation 0.08 inches.

We have to calculate the mean of sampling distribution when 20 balls are selected randomly.

Here, μ=5.2 and σ=0.08.

We know that mean of sampling distribution for mean is equal to the population mean.

mean of sampling distribution for mean= μxbar=μ= 5.2

Thus, mean of the sampling distribution of the sample mean is 5.2 inches

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Answer:

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Step-by-step explanation:

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A store sells a 33-pound bag of oranges for \$ 3.60$3.60 and a 55-pound bag of oranges for \$ 5.25$5.25. What is the difference
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Answer:

0.01364

Step-by-step explanation:

It is given that,

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Price per pound of 33 pound bag is 3.60/33 = 0.10909 price per pound

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Find the midpoint of the segment between the points (3,17) and (−14,−8)
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Answer:

\large\boxed{\left(-\dfrac{11}{2},\ \dfrac{9}{2}\right)}

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M_{AB}\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)

We have the points (3, 17) and (-14, -8).

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