Probability and Random Variable Free Response Practice Problems for AP Statistics (page 2)

Solutions

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.

2.  
   1. P(A and B) = P(A)·P(B | A) = (0.6)(0.5) = 0.30.
   2. 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).)

   3. P(B) = 0.3, P(B|A) = 0.5. Since P(B) ≠ P(B|A), events A and B are not independent.
3.  
   1. 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.
      

   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).

   2. 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:
      

   Then

   P(450 < X < 600) = 0.8667 – 0.2877 = 0.5790.

   There are 0.5790 (9000) ≈ 5197 scores.

   [This could be done on the calculator as follows: normalcdf(450,600, 500,90) = 0.5775.]

   3. This situation could be pictured as follows.
      

   The z-score corresponding to an area of 0.99 is 2.33 (invNorm(0.99) on the calculator). So, zx = 2.33. But, also,

   

   Thus,

   

   Solving algebraically for x, we get x = 709.7. Rachel needs a score of 710 or higher.

   Remember that this type of problem is usually solved by expressing z in two ways (using the definition and finding the area) and solving the equation formed by equating them. On the TI-83/84, the answer could be found as follows: invNorm(0.99,500, 90)= 709.37.

4.  
   1. μ_3+6X = 3 + 6μ_X = 3 + 6(3) = 21.
   2. Because infogana

   Thus,

   
5. 

Because we are given that the target was hit, we only need to look at those outcomes. P(the person who hit the target was Laura |the target was hit)

6. μ_{X – Y} = μ_X – _Y = 3 – 5 = – 2.

Since X and Y are independent, we have Note that the variances add even though we are subtracting one random variable from another.

7.  
   1. is not true. This is an approximation based on the empirical rule. The actual value proportion within two standard deviations of the mean, to 4 decimal places is 0.9545.
   2. is true. This is a property of the normal curve.
   3. is true. This is because the bell shape of the normal curve means that there is more area under the curve for a given interval length for intervals closer to the center.
8.  
   1. No, the events are not independent. The probability of B changes depending on what happens with A. Because there are 12 face cards, if the first card drawn is a face card, then P(B) = 11/51. If the first card is not a face card, then P(B) = 12/51. Because the probability of B is affected by the outcome of A, A and B are not independent.
   2. P(A) = 12/52 = 3/13. P(A|C) = 3/13 (3 of the 13 diamonds are face cards). Because these are the same, the events "draw a face card on the first draw" and "the first card drawn is a diamond" are independent.
9. The area to the right of x is 0.6, so the area to the left is 0.4. From the table of Standard Normal Probabilities, . Also

So,

60% of the area is to the right of 687.5. The calculator answer is given by invNorm(0.4,700,50)=687.33.

10. Let D = "a home has a desktop computer"; L = "a home has a laptop computer." We are given that P(D) = 0.8 and P(D and L) = 0.3. Thus,
    
11. The situation can be pictured as shown below.

The shaded area is a trapezoid whose area is

12. The fact that Harv's pumpkin is at the 90th percentile means that it is larger than 90% of the pumpkins in the contest. From the table of Standard Normal Probabilities, the area to the left of a term with a z-score of 1.28 is about 0.90. Thus,
    

So, Harv's pumpkin weighed about 148 pounds (for your information, the winning pumpkin at the Half Moon Bay Pumpkin Festival in 2008 weighed over 1528 pounds!). Seven of the 62 pumpkins weighed more than 1000 pounds!

13. Since ∑P(X) = 1, we have P(X = 4) = 1 – P(X = 2) – P(X = 3) = 1 – 0.3 – 0.5 = 0.2. Thus, filling in the table for X, we have
    

Since X and Y are independent, P(X = 4 and Y = 3) = P(X = 4) · P(X = 3). We are given that P(X = 4 and Y = 3) = 0.03. Thus, P(X = 4) · P(Y = 3) = 0.03. Since we now know that P(X = 4) = 0.2, we have (0.2). P(Y = 3) = 0.03, which gives us P(Y = 3) = = 0.15. Now, since ∑P(Y) = 1, we have P(Y = 5) = 1 – P(Y = 3) – P(Y = 4) – P(Y = 6) = 1 – 0.15 – 0.1 – 0.4 = 0.35.

14. Let A = "the number is divisible by 24" = {24, 48, 72, 96}. Let B = "the number is divisible by 36" = {36, 72}.

Note that P(A and B) = (72 is the only number divisible by both 24 and 36). P(win a pizza) = P(A or B) = P(A) + P(B) – P(A and B) = .

15. Because the numbers of men and women in the school are about equal (that is, P(women) = 0.5), let an even number represent a female and an odd number represent a male. Begin on the first line of the table and consider groups of 12 digits.

Count the even numbers among the 12. This will be the number of females among the group. Repeat five times. The relevant part of the table is shown below, with even numbers underlined and groups of 12 separated by two slanted bars (\\):

In the five groups of 12 people, there were 3, 6, 3, 8, and 6 women. (Note: The result will, of course, vary if a different assignment of digits is made. For example, if you let the digits 0 – 4 represent a female and 5 – 9 represent a male, there would be 4, 7, 7, 5, and 7 women in the five groups.) So, in 40% of the trials there were 4 or fewer women in the class even though we would expect the average to be 6 (the average of these 5 trials is 5.2). Hence, it seems that getting only 4 women in a class when we expect 6 really isn't too unusual because it occurs 40% of the time. (It is shown in the next chapter that the theoretical probability of getting 4 or fewer women in a group of 12 people, assuming that men and women are equally likely, is about 0.19.)

16. Because P(girl) = 0.6, let the random digits 1, 2, 3, 4, 5, 6 represent the birth of a girl and 0, 7, 8, 9 represent the birth of a boy. Start on the second row of the random digit table and move across the line until you find the third digit that represents a girl. Note the number of digits needed to get three successes. Repeat 5 times and compute the average. The simulation is shown below (each success, i.e., girl, is underlined and separate trials are delineated by \\).
    

It took 4, 7, 3, 6, and 5 children before they got their three girls. The average wait was 5. (The theoretical average is exactly 5—we got lucky this time!). As with Exercise 15, the result will vary with different assignment of random digits.

17.  
    1. P(x ≤ 22) = P(x = 20) + P(x = 21) + P(x = 22) = 0.2 + 0.3 + 0.2 = 0.7.
    2. P(x > 21) = P(x = 22) + P(x = 23) + P(x = 24) = 0.2 + 0.1 + 0.2 = 0.5.
    3. P(21 ≤ x < 24) = P(x = 21) + P(x = 22) + P(x = 23) = 0.3 + 0.2 + 0.1 = 0.6.
    4. P(x ≤ 21 or x > 23) = P(x = 20) + P(x = 21) + P(x = 24) = 0.2 + 0.3 + 0.2 = 0.7.
18.  
    1. 18 of the 38 slots are winners, so P(win if bet on red)
    2. The probability distribution for this game is
       

    The player will lose 5.2, on average, for each dollar bet.

19. 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).)
20. 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,