(19 points) A scale measuring prejudice is administered to a sample of 1,500 res
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Question
(19 points)
A scale measuring prejudice is administered to a sample of 1,500 respondents. The distribution is approximately normal with a mean of 31 and a standard deviation of 5. Determine:
Number and percentage of respondents that had a score below 20.
The probability of randomly picking a respondent that had a score below 20.
Number and percentage of respondents that had a score below 40.
The probability of randomly picking a respondent that had a score below 40.
Number and percentage of respondents that had a score between 30 and 40.
The probability of randomly picking a respondent that has a score between 30 and 40.
Number and percentage of respondents that had a score between 35 and 45.
Number and percentage of respondents that had a score above 25.
Number and percentage of respondents that had a score above 35.
If a score of 40 or more is considered “highly prejudiced,” what is the probability that a person selected at random will have a score in that range?
Q. 4 (18 points)
The Mental Development Index (MDI) is a standardized measure used to do follow-ups on high-risk infants. Assume it is normally distributed with a mean of 100 and a standard deviation of 16.
What percent of children had an MDI score of at least 120?
(Hint: What does at least imply? Score above or below 120?)
What percent of children had an MDI score of at least 80?
What percent of children had an MDI score of between 80 and 120?
What percent of children had an MDI score of 140 and higher?
What percent of children had an MDI score of between 120 and 140?
An infant is in the 95th percentile in this exam. What is his/her corresponding MDI score?
What is the MDI score of the 5th percentile?
What is the percentile rank of an infant who scored 65 points?
Q. 5 (6 points)
The local police force gives all applicants an entrance exam and accepts only those applicants who score in the top 15% on this test. If the mean score this year is 87 and the standard deviation is 8, would an individual with a score of 95 be accepted.
Q. 6 (6 points)
In a distribution of scores with a mean of 35 and a standard deviation of 4, what event is more likely (think probabilities): that a randomly selected score will be between 29 and 31 or that a randomly selected score will be between 40 and 42? Show all your work and explain your choice of answer.
Explanation / Answer
Q-3)
Number and percentage of respondents that had a score below 20.
N* P(X<20) = N*P(Z < (20-31) / 5) = N*P(Z < -2.2) = 1500*0.0139= 20.85 respondents or 1.39%
The probability of randomly picking a respondent that had a score below 20.
P(X<20) = P(Z < (20-31) / 5) = P(Z < -2.2) = 0.0139
Number and percentage of respondents that had a score below 40.
N* P(X<40) = N*P(Z < (40-31) / 5) = N*P(Z < 1.8) = 1500*0.96407= 1446.105 respondents or 96.41%
The probability of randomly picking a respondent that had a score below 40.
P(X<40) = P(Z < (40-31) / 5) = P(Z < 1.8) = 0.96407
Number and percentage of respondents that had a score between 30 and 40.
NP(30 <X<40) = N[P(X < 40) - P(x < 30)] = N[0.96407 - P(Z < (30-31)/5 )] = 1500[0.96407 - 0.42074] = 0.54396 *1500= 815.94 respondents
The probability of randomly picking a respondent that has a score between 30 and 40.
P(30 <X<40) = [P(X < 40) - P(x < 30)] = [0.96407 - P(Z < (30-31)/5 )] = [0.96407 - 0.42074] = 0.54396
Number and percentage of respondents that had a score between 35 and 45.
NP(35 <X<45) = N[P(X < 45) - P(x < 35)] = N[P(Z<2.8) - P(Z < 0.8)] = 1500[0.99744-0.78814] = 0.2093 *1500= 313.95
Number and percentage of respondents that had a score above 25.
N* P(X>25) = N*P(Z > (25-31) / 5) = N*P(Z > -1.2) = 1500*0.88493= 1327.395 respondents or 88.493%
Number and percentage of respondents that had a score above 35.
N* P(X>35) = N*P(Z > (35-31) / 5) = N*P(Z > 0.8) = 1500*0.21186= 317.79 respondents or 21.186%
If a score of 40 or more is considered “highly prejudiced,” what is the probability that a person selected at random will have a score in that range?
P(X>40) = P(Z > (40-31) / 5) = P(Z >1.8) = 0.03593
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