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A teacher surveyed her class after they had taken a vocabulary test. Eighteen of the students claimed they had studied at least one hour for the test. The remaining twelve students admitted that they had not studied for the test at all. The test results (expressed as a percent) for the two groups are shown below. Studied: Did Not Study: Part 1: Calculate the mean of the group that studied, rounded to the nearest tenth. Part 2: Calculate the mean absolute deviation of the group that studied, rounded to the nearest tenth. Part 3: How many of the scores (from the group that studied) are within one mean absolute deviation of the mean.

Respuesta :

The evaluation of the figures needed for the considered survey are:

  • The mean of the considered data is 1667/18 ≈ 92.61
  • The mean of these 18 absolute deviations comes out as
  • [tex]91.8/18 \approx 5.1[/tex]
  • The number of scores which are within one mean absolute deviation of the mean is 10

How to find the mean absolute deviation of a discrete data?

Suppose the data specified is of n sized as [tex]x_1, x_2, \cdots , x_n[/tex]

Let this data's mean be [tex]\bar{x}[/tex]

Then, the mean absolute deviation of this data is the mean of the absolute differences of the observations from the mean of the data they belong to.

It is calculated as:

[tex]M.A.D = \dfrac{\sum_{i=1}^n|x_i-\bar{x}|}{n}[/tex]

The data is missing from the problem statement, and is as shown below:

  • Studied: 88, 100, 94, 79, 92, 100, 95, 83, 89, 99, 100, 91, 89, 95, 100, 93, 96, 84
  • Did Not Study: 82, 72, 45, 91, 58, 83, 65, 87, 90, 77, 73, 89

Calculating the needed figures part by part:
Part 1: Mean of the group that studied, rounded to the nearest tenth.

Mean is the sum of observations divided by number of observations.

The sum of 18 observations:   88, 100, 94, 79, 92, 100, 95, 83, 89, 99, 100, 91, 89, 95, 100, 93, 96, 84

is 1667. Thus, its mean is 1667/18 ≈ 92.61 ≈ [tex]\bar{x}[/tex]

Part 2: The mean absolute deviation of the group that studied:

The absolute deviations of 18 observations:  88, 100, 94, 79, 92, 100, 95, 83, 89, 99, 100, 91, 89, 95, 100, 93, 96, 84

from the mean  [tex]\bar{x}[/tex] ≈ 92.61 are approx:

[tex]|92.61 - 88| = 4.6\\|92.61 - 100| = 7.4\\|92.61 - 94| = 1.4\\|92.61 - 79| = 13.6\\|92.61 - 92| = 0.6\\|92.61 - 100| = 7.4\\|92.61 - 95| = 2.4\\|92.61 - 83| = 9.6\\|92.61 - 89| = 3.6\\|92.61 - 99| = 6.4\\|92.61 - 100| = 7.4\\|92.61 - 91| = 1.6\\|92.61 - 89| = 3.6\\|92.61 - 95| = 2.4\\|92.61 - 100| = 7.4\\|92.61 - 93| = 0.4\\|92.61 - 96| = 3.4\\|92.61 - 84| = 8.6\\[/tex]

Their sum is 91.8 approx.

Thus, the mean of these 18 absolute deviations comes out as

[tex]91.8/18 \approx 5.1[/tex]

Part 3: The number of scores which are within one mean absolute deviation of the mean

The range within one absolute deviation of the mean is [tex][92.61 - m.a.d, 92.61 + m.a.d] = [92.61 - 5.1, 92.61 + 5.1] = [87.51, 97.71][/tex]

The values from the studied group which is in this range are:

88, 94, 92, 95, 89, 91, 89, 95, 93, 96

They are total 10 in counts.

Thus, the number of scores which are within one mean absolute deviation of the mean is 10

Thus, the evaluation of the figures needed for the considered survey are:

  • The mean of the considered data is 1667/18 ≈ 92.61
  • The mean of these 18 absolute deviations comes out as
  • [tex]91.8/18 \approx 5.1[/tex]
  • The number of scores which are within one mean absolute deviation of the mean is 10

Learn more about mean absolute deviation here:

https://brainly.com/question/26462759

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