Respuesta :
a). The percentile of a score of [tex]172[/tex] is [tex]\boxed{79{\text{th}}\,{\text{percentile}}}[/tex].
b). The Z score for a test of 200 is [tex]\boxed{1.5}[/tex].
c). The percentage of score between [tex]100[/tex] to [tex]160[/tex] is [tex]\boxed{53.3\% }[/tex].
d). The proportion of respondents should score above [tex]190[/tex] is [tex]\boxed{0.10565}[/tex].
e). The corresponding score for [tex]67{\text{th}}[/tex] percentile is [tex]\boxed{142.7}[/tex].
Further Explanation:
The Z score of the standard normal distribution can be obtained as,
[tex]\boxed{{\text{Z}} = \dfrac{{X - \mu }}{\sigma }}[/tex]
Given:
The mean of test is [tex]\boxed{140}[/tex].
The standard deviation of the score is [tex]\boxed{40}[/tex].
Explanation:
Part (a).
The Z score of [tex]172[/tex] can be obtained as,
[tex]\begin{aligned}{\text{Z}}&= \frac{{172 - 140}}{{40}} \\ &= \frac{{32}}{{40}} \\ &= 0.8 \\ \end{aligned}[/tex]
The percentile of a score of [tex]172[/tex] can be obtained as,
[tex]{\text{Percentile}} = {\text{P}}\left( {{\text{Z}<0.8} \right)[/tex]
From Z-table the percentile is [tex]78.81\%[/tex].
Approximately [tex]\boxed{79{\text{th}}\,{\text{percentile}}}[/tex].
Part (b).
The Z score for a test score [tex]200[/tex] can be obtained as,
[tex]\begin{aligned} {\text{Z}}&= \frac{{200 - 140}}{{40}} \\ &= \frac{{60}}{{40}} \\ &= 1.5 \\\end{aligned}[/tex]
The Z-score is [tex]\boxed {1.5}[/tex].
Part (c).
The percentage of scores fall between [tex]100[/tex] and [tex]160[/tex] can be obtained as,
[tex]\begin{aligned} {\text{Percentage}} &= {\text{P}}\left( {\frac{{100 - 140}}{{40}} < {\text{Z}} < \frac{{160 - 140}}{{40}}} \right) \\ &= {\text{P}}\left( {\frac{{ - 40}}{{40}} < {\text{Z}} < \frac{{20}}{{40}}} \right) \\ &= {\text{P}}\left( { - 1 < {\text{Z}} < 0.5} \right) \\ &= {\text{P}}\left( {{\text{Z} < 0}}{\text{.5}}} \right) - {\text{P}}\left( {{\text{Z} > }} - {\text{1}}} \right) \\ \end{gathered}[/tex]
Further solve above equation.
[tex]\begin{aligned}{\text{Percentage}} &= {\text{P}}\left( {{\text{Z} < 0.5}}\right) -\left[ 1 - P({Z<1})}\right] \\ &= {\text{P}}\left( {{\text{Z}<0.5}\right){\text{ + P}}\left( {{\text{Z} < 1}}} \right) - 1 \\ &= 0.69146 + 0.84134 - 1 \\ &= 0.5328 \\\end{gathered}[/tex]
Approximately the percentage is [tex]\boxed{53.3\% }[/tex].
Part (d).
The proportion of respondent score above [tex]190[/tex] can be obtained as,
[tex]\begin{aligned}{\text{Proportion}} &= {\text{P}}\left( {{\text{Z}} > \frac{{190 - 140}}{{40}}} \right) \\ &= {\text{P}}\left( {{\text{Z}} > \frac{{50}}{{40}}} \right) \\ &= {\text{P}}\left( {{\text{Z}} > 1.25} \right) \\ &= 1 - {\text{P}}\left( {\text{Z}< 1.25} \right) \\ &= 1 - 0.89435 \\ &= 0.10565 \\\end{gathered}[/tex]
The proportion of respondents should score above [tex]190[/tex] is [tex]\boxed{0.10565}[/tex].
Part (e).
The Z-value of [tex]67{\text{th}}[/tex] is [tex]0.0675[/tex].
The corresponding score for [tex]67{\text{th}}[/tex] percentile can be obtained as,
[tex]\begin{aligned}0.0675 &= \frac{{X - 140}}{{40}} \\ 40\left( {0.0675} \right) &= X - 140 \\ 2.7 + 140 &= X \\ 142.7 &= X \\ \end{aligned}[/tex]
The corresponding score for [tex]67{\text{th}}[/tex] percentile is [tex]\boxed{142.7}[/tex].
Learn more:
1. Learn more about normal distribution https://brainly.com/question/12698949
2. Learn more about standard normal distribution https://brainly.com/question/13006989
Answer details:
Grade: College
Subject: Statistics
Chapter: Normal distribution
Keywords: Z-score, standard normal distribution, standard deviation, criminologist, test, measure, probability, low score, mean, repeating, indicated, normal distribution, percentile, percentage, undesirable behavior, proportion.
Using the normal distribution, it is found that:
a) The percentile rank of a score of 172 is 79.
b) The z-score for a test score of 200 is Z = 1.5.
c) 53.28% of scores fall between 100 and 160.
d) 0.1056 = 10.56% of respondents should score above 190.
e) His or her corresponding recidivism score is of 157.6.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem:
- Mean of 140, thus [tex]\mu = 140[/tex].
- Standard deviation of 40, thus, [tex]\sigma = 40[/tex].
Item a:
This percentile is the p-value of Z when X = 172, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{172 - 140}{40}[/tex]
[tex]Z = 0.8[/tex]
[tex]Z = 0.8[/tex] has a p-value of 0.79.
The percentile rank of a score of 172 is 79.
Item b:
The z-score is Z when X = 200, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{200 - 140}{40}[/tex]
[tex]Z = 1.5[/tex]
The z-score for a test score of 200 is Z = 1.5.
Item c:
The proportion is the p-value of Z when X = 160 subtracted by the p-value of Z when X = 100, thus:
X = 160:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{160 - 140}{40}[/tex]
[tex]Z = 0.5[/tex]
[tex]Z = 0.5[/tex] has a p-value of 0.6915.
X = 100:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{100 - 140}{40}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a p-value of 0.1587.
0.6915 - 0.1587 = 0.5328.
0.5328 x 100% = 53.28%.
53.28% of scores fall between 100 and 160.
Item d:
This proportion is 1 subtracted by the p-value of Z when X = 190, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{190 - 140}{40}[/tex]
[tex]Z = 1.25[/tex]
[tex]Z = 1.25[/tex] has a p-value of 0.8944.
1 - 0.8944 = 0.1056.
0.1056 = 10.56% of respondents should score above 190.
Item e:
The z-score for the 67th percentile is Z = 0.44, thus, this score is X when Z = 0.44.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.44 = \frac{X - 140}{40}[/tex]
[tex]X - 140 = 0.44(40)[/tex]
[tex]X = 157.6[/tex]
His or her corresponding recidivism score is of 157.6.
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