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gulaghasi [49]
3 years ago
13

To determine whether they need to hire extra help for the holiday season, the managers of a railroad company want to estimate ho

w long, on average, it takes a crew to unload a freight car at the Lafayette station. They will calculate an estimator as the sample mean of a random sample of unloading times. The managers want the sample mean to have a standard error of 1.51.5 minutes. The corporate office has provided a standard deviation of 99 minutes to use for calculation purposes. How large should the random sample be to ensure the sample mean has the desired standard error?
Mathematics
1 answer:
nadezda [96]3 years ago
3 0

Answer:

36

Step-by-step explanation:

Data provided in the question:

Standard error = 1.5 minutes             ( ∵ number are repetitive 1.51.5 )  

Standard deviation = 9 minutes        ( ∵ number are repetitive 99 )

Now,

Standard error = ( Standard deviation ) ÷ √n

Here,

n is the sample size

Therefore,

on substituting the respective values, we get

1.5 = 9 ÷ √n

or

√n = 9 ÷ 1.5

or

√n = 6

or

n = 6² = 36

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Find the 44th percentile, P44, from the following data! pls help
irina [24]

Answer:

P_{44} = 29.9

Step-by-step Explanation:

The 44th percentile, P_{44}, of the given data can be calculated using the kth formula for percentile given as:

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where,

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k = the percentile = 44

n = total number of the given data values = 23

Plug in the values into the formula to find "i"

i = \frac{44}{100}(23 + 1)

i = \frac{44}{100}(24)

i = \frac{1,056}{100}

i = 10.56

Since "I", 10.56, is not an integer, round the number down, and round it up to the nearest integer, then look for the position each occupy in the ordered data set, and find their average.

Thus,

i_{down} = 10.56 = 10

i_{up} = 10.56 = 11

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P_{44} = \frac{28.7 + 31.1}{2} = \frac{59.8}{2}

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3 years ago
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Answer:

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