1.
This can also be done on the TI-83/84 by putting the X values in L1 and the probabilities in L2 . Then
1-Var Stats L1,L2 will give the above values for the mean and standard deviation.
(a) P (A and B) = P (A) · P (B | A) = (0.6)(0.5) = 0.30.
(b) P (A or B) = P (A) + P (B) – P (A and B) = 0.6 + 0.3 – 0.3 = 0.6
(Note that the 0.3 that is subtracted came from part (a).)
(c) P (B) = 0.3, P (B|A) = 0.5. Since P (B) ≠ P (B|A), events A and B are not independent.
- (a) Let X represent the score a student earns. We know that X has approximately N (500, 90). What
we are looking for is shown in the following graph.
P (X > 600) = 1 – 0.8667 = 0.1333.
The calculator answer is normalcdf(600,10,000,500.900) =0.1333 (remember that the upper
bound must be “big”—in this exercise, 10,000 was used to get a sufficient number of standard
deviations above 600).
(b) We already know, from part (a), that the area to the left of 600 is 0.8667. Similarly we determine
the area to the left of 450 as follows:ThenP (450 < X < 600) = 0.8667 – 0.2877 = 0.5790.There are 0.5790 (9000) ≈ 5211 scores.
[This could be done on the calculator as follows: normalcdf (450,600, 500,90) = 0.5775.]
(c) This situation could be pictured as follows.