a). The percentile of a score of \(172\) is \(\boxed{79{\text{th}}\,{\text{percentile}}}\).
b). The Z score for a test of 200 is \(\boxed{1.5}\).
c). The percentage of score between \(100\) to \(160\) is \(\boxed{53.3\% }\).
d). The proportion of respondents should score above \(190\) is \(\boxed{0.10565}\).
e). The corresponding score for \(67{\text{th}}\) percentile is \(\boxed{142.7}\).
Further Explanation:
The Z score of the standard normal distribution can be obtained as,
\(\boxed{{\text{Z}} = \dfrac{{X - \mu }}{\sigma }}\)
Given:
The mean of test is \(\boxed{140}\).
The standard deviation of the score is \(\boxed{40}\).
Explanation:
Part (a).
The Z score of \(172\) can be obtained as,
\(\begin{aligned}{\text{Z}}&= \frac{{172 - 140}}{{40}} \\ &= \frac{{32}}{{40}} \\ &= 0.8 \\ \end{aligned}\)
The percentile of a score of \(172\) can be obtained as,
\({\text{Percentile}} = {\text{P}}\left( {{\text{Z}<0.8} \right)\)
From Z-table the percentile is \(78.81\%\).
Approximately \(\boxed{79{\text{th}}\,{\text{percentile}}}\).
Part (b).
The Z score for a test score \(200\) can be obtained as,
\(\begin{aligned} {\text{Z}}&= \frac{{200 - 140}}{{40}} \\ &= \frac{{60}}{{40}} \\ &= 1.5 \\\end{aligned}\)
The Z-score is \(\boxed {1.5}\).
Part (c).
The percentage of scores fall between \(100\) and \(160\) can be obtained as,
\(\begin{aligned} {\text{Percentage}} &= {\text{P}}\left( {\frac{{100 - 140}}{{40}} < {\text{Z}} < \frac{{160 - 140}}{{40}}} \right) \\ &= {\text{P}}\left( {\frac{{ - 40}}{{40}} < {\text{Z}} < \frac{{20}}{{40}}} \right) \\ &= {\text{P}}\left( { - 1 < {\text{Z}} < 0.5} \right) \\ &= {\text{P}}\left( {{\text{Z} < 0}}{\text{.5}}} \right) - {\text{P}}\left( {{\text{Z} > }} - {\text{1}}} \right) \\ \end{gathered}\)
Further solve above equation.
\(\begin{aligned}{\text{Percentage}} &= {\text{P}}\left( {{\text{Z} < 0.5}}\right) -\left[ 1 - P({Z<1})}\right] \\ &= {\text{P}}\left( {{\text{Z}<0.5}\right){\text{ + P}}\left( {{\text{Z} < 1}}} \right) - 1 \\ &= 0.69146 + 0.84134 - 1 \\ &= 0.5328 \\\end{gathered}\)
Approximately the percentage is \(\boxed{53.3\% }\).
Part (d).
The proportion of respondent score above \(190\) can be obtained as,
\(\begin{aligned}{\text{Proportion}} &= {\text{P}}\left( {{\text{Z}} > \frac{{190 - 140}}{{40}}} \right) \\ &= {\text{P}}\left( {{\text{Z}} > \frac{{50}}{{40}}} \right) \\ &= {\text{P}}\left( {{\text{Z}} > 1.25} \right) \\ &= 1 - {\text{P}}\left( {\text{Z}< 1.25} \right) \\ &= 1 - 0.89435 \\ &= 0.10565 \\\end{gathered}\)
The proportion of respondents should score above \(190\) is \(\boxed{0.10565}\).
Part (e).
The Z-value of \(67{\text{th}}\) is \(0.0675\).
The corresponding score for \(67{\text{th}}\) percentile can be obtained as,
\(\begin{aligned}0.0675 &= \frac{{X - 140}}{{40}} \\ 40\left( {0.0675} \right) &= X - 140 \\ 2.7 + 140 &= X \\ 142.7 &= X \\ \end{aligned}\)
The corresponding score for \(67{\text{th}}\) percentile is \(\boxed{142.7}\).
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.