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MaRussiya [10]
4 years ago
15

Which assumption will work best for a large data set with a normal distribution? a.Most of the data points are close to the mini

mum value of the set. b. Most of the data points are close to the maximum value of the set. c. Most of the data points are close to the mean value of the set d. The data points are distributed uniformly over the range of the set.
Mathematics
2 answers:
masha68 [24]4 years ago
3 0
I think the correct answer would be option C. The best assumption that will work for a large data set having a normal distribution would be that most of the data points are close to the mean value of the set. Having this assumption, we can use freely the common measures of central tendency as statistical tools.
11Alexandr11 [23.1K]4 years ago
3 0

Answer:

Option C) Most of the data points are close to the mean value of the set

Step-by-step explanation:

We are given the following information in the question:

Properties of Normal Distribution

  • It is a probability distribution which  is symmetric about the mean that means data near the mean.
  • In graph form, normal distribution will appear as a bell curve.
  • In a normal distribution, Mean = Mode = Median

We have to choose the best assumption that will work for a large data set with normal distribution.

Option C) Most of the data points are close to the mean value of the set

It is one of the properties of the normal distribution. Most of the data point cannot be close to the minimum or maximum value and nor the data can be distributed uniformly over the range of the set.

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What is the answer to 7(4p+2p+3r) help me pls =(
Alex Ar [27]

Answer:

42p + 21r

Step-by-step explanation:

First look at the parentheses. Since 4p and 2p are the same variables, you can add them.

7(6p + 3r)

now you use the distributive property.

7 x 6p + 7 x 3r

42p + 21r

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8 0
3 years ago
Suppose that an airline quotes a flight time of 128 minutes between two cities. Furthermore, suppose that historical flight reco
ANTONII [103]

Answer:

(a) The probability density function of <em>X</em> is:

f_{X}(x)=\frac{1}{b-a};\ a

(b) The value of P (129 ≤ X ≤ 146) is 0.3462.

(c) The probability that a randomly selected flight between the two cities will be at least 3 minutes late is 0.4327.

Step-by-step explanation:

The random variable <em>X</em> is defined as the flight time between the two cities.

Since the random variable <em>X</em> denotes time interval, the random variable <em>X</em> is continuous.

(a)

The random variable <em>X</em> is Uniformly distributed with parameters <em>a</em> = 10 minutes and <em>b</em> = 154 minutes.

The probability density function of <em>X</em> is:

f_{X}(x)=\frac{1}{b-a};\ a

(b)

Compute the value of P (129 ≤ X ≤ 146) as follows:

Apply continuity correction:

P (129 ≤ X ≤ 146) = P (129 - 0.50 < X < 146 + 0.50)

                           = P (128.50 < X < 146.50)

                           =\int\limits^{146.50}_{128.50} {\frac{1}{154-102}} \, dx

                           =\frac{1}{52}\times \int\limits^{146.50}_{128.50} {1} \, dx

                           =\frac{1}{52}\times (146.50-128.50)

                           =0.3462

Thus, the value of P (129 ≤ X ≤ 146) is 0.3462.

(c)

It is provided that a randomly selected flight between the two cities will be at least 3 minutes late, i.e. <em>X</em> ≥ 128 + 3 = 131.

Compute the value of P (X ≥ 131) as follows:

Apply continuity correction:

P (X ≥ 131) = P (X > 131 + 0.50)

                 = P (X > 131.50)

                 =\int\limits^{154}_{131.50} {\frac{1}{154-102}} \, dx

                 =\frac{1}{52}\times \int\limits^{154}_{131.50} {1} \, dx

                 =\frac{1}{52}\times (154-131.50)

                 =0.4327

Thus, the probability that a randomly selected flight between the two cities will be at least 3 minutes late is 0.4327.

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Answer:

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Step-by-step explanation:

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