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AleksandrR [38]
2 years ago
8

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 :
87
,

100
,

94
,

79
,

92
,

100
,

95
,

83
,

89
,

99
,

100
,

91
,

89
,

95
,

100
,

93
,

96
,

83


Did Not Study :
82
,

72
,

45
,

91
,

58
,

83
,

65
,

87
,

90
,

77
,

73
,

89

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.
Mathematics
1 answer:
andrezito [222]2 years ago
3 0

Step-by-step explanation:

part1

the mean value is

(87+100+94+79+92+100+95+83+89+99+100+91+89+95+100+93+96+83) / 18

= 1665 / 18 = 92.5

part2

this is the sum of the absolute value of the differences bergen the individual sites and the mean value divided by the number of data points :

(5.5+7.5+1.5+13.5+0.5+7.5+2.5+9.5+3.5+6.5+7.5+1.5+3.5+2.5+7.5+0.5+3.5+9.5)/18

= 94/18 = 5.222222222... ≈ 5.2

part3

how many scores are less than 5.2 away (plus or minus) from the mean value ? in other words, how many differences in the part2 calculation were smaller than or equal to 5.2 ?

there are 9 scores meeting that criteria :

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

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Step-by-step explanation:

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Regression Equation = ŷ = bX + a

b = SP/SSX = 436.71/53.71 = 8.13032

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Part A:

Given that lie <span>detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector correctly determined that a selected person is saying the truth has a probability of 0.85
Thus p = 0.85

Thus, the probability that </span>the lie detector will conclude that all 15 are telling the truth if <span>all 15 applicants tell the truth is given by:

</span>P(X)={ ^nC_xp^xq^{n-x}} \\  \\ \Rightarrow P(15)={ ^{15}C_{15}(0.85)^{15}(0.15)^0} \\  \\ =1\times0.0874\times1=0.0874
<span>

</span>Part B:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.25
Thus p = 0.15

Thus, the probability that the lie detector will conclude that at least 1 is lying if all 15 applicants tell the truth is given by:

P(X)={ ^nC_xp^xq^{n-x}} \\ \\ \Rightarrow P(X\geq1)=1-P(0) \\  \\ =1-{ ^{15}C_0(0.15)^0(0.85)^{15}} \\ \\ =1-1\times1\times0.0874=1-0.0874 \\  \\ =0.9126


Part C:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15

The mean is given by:

\mu=npq \\  \\ =15\times0.15\times0.85 \\  \\ =1.9125


Part D:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15

The <span>probability that the number of truthful applicants classified as liars is greater than the mean is given by:

</span>P(X\ \textgreater \ \mu)=P(X\ \textgreater \ 1.9125) \\  \\ 1-[P(0)+P(1)]
<span>
</span>P(1)={ ^{15}C_1(0.15)^1(0.85)^{14}} \\  \\ =15\times0.15\times0.1028=0.2312<span>
</span>
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3 years ago
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