The player will lose 5.2¢, on average, for each dollar bet.
From the tables, we see that P (z < – 2.5) = P (z > 2.5) = 0.0062. So the probability that we are
more than 2.5 standard deviations from the mean is 2(0.0062) = 0.0124. (This can be found on the
calculator as follows: 2 normalcdf (2.5,1000).)
- The situation can be pictured as follows:
If 90% of the area is to the right of 50, then 10% of the area is to the left. So,
Solutions to Cumulative Review Problems
II is true: the mean is most likely greater than the median. This is because the mean, being
nonresistant, is pulled in the direction of outliers or skewness. Because the given histogram is clearly
skewed to the right, the mean is likely to be to the right of (that is, greater than) the median.
- The kind of sample you want is a stratified random sample . The sample should have 25 Asian
students, 8 African American students, 12 Latino students, and 55 Caucasian students. You could get a
list of all Asian students from the data clerk and randomly select 25 students from the list. Repeat this
process for percentages of African American, Latino, and Caucasian students. Now the proportion of
each ethnic group in your sample is the same as its proportion in the population. - The five-number summary is [20, 32, 42, 46, 50]. The box plot looks like this:
- (a) The LSRL line is: = – 41.430 + 0.12486(#boats).
(b) For each additional registered powerboat, the number of manatee deaths is predicted to increase
by 0.12. You could also say that the number increases, on average, by 0.12.