A scale measuring prejudice has been administered to a large sample of respondents. The distribution of scores is approximately normal with a mean of 31 and a standard deviation of 5. What percentage...

Mean ,μ= 31
Standard deviation , σ = 5.
Probability that sample had scores above 20 = P(x>20)
=P((x-μ)/σ>(20-31)/5)
=P(z>-2.2)
=1-0.0139 = 0.9861
Hence, percentage of sample that had scores above 20 = 98.61%
Probability that sample had scores below 20 = P(x<20)
=P((x-μ)/σ<(20-31)/5)
=P(z<-2.2)
=0.0139
Hence, percentage of sample that had scores below 20 = 1.39%
Probability that sample had scores above 40 =...