February Regional – Statistics Team – Answers and Solutions 1. 0.6925
2. -0.24
3.
!
!
4. 0.984
!
5. 𝑜𝑟 2.5
!
6. 1
7. 0.9982
8. 24,916
9. 0.0024
10. 1.83
11. 2,646
12. 0
13. 1
14. 0.83
15. 2.885
1. Answer: 0.6925
Solution: Enter each data set into an empty list in the calculator and run the 1-Variable Statistics function.
The means for each data set are A = 4.10, B = 4.30, C = 3.40, and D = 5.40 and the sample standard
deviations for each data set are SA = 3.11, SB = 2.31, SC = 3.13, and SD = 2.99, when rounded to the nearest
hundredth. Dividing each standard deviation by the corresponding mean gives the following four coefficients
of variation: CVA = 3.11/4.1 = .76, CVB = 2.31/4.3 = .54, CVC = 3.13/3.4 = .92, and CVD = 2.99/5.4 = .55,
when rounded to the nearest hundredth. The sum of these results divided by 4 is 2.77 / 4 = 0.6925.
2. Answer: -0.24
Solution: Clear List 1 in the calculator and enter the data. Run the 1-Variable Statistics function on List 1 to
get all the necessary statistics for the given formulas: 𝑥 = 5, 𝑠! = 3.3333, 𝑄! = 2, 𝑄! = 5.5, 𝑎𝑛𝑑 𝑄! = 8.
For A, clear List 2 and use it to get the sum of the cubed deviations from the mean by entering (L1 – 5)^3 into
the heading of List 2. Note that the main part of the formula is now just the mean of List 2, so run the
1-Variable Statistics function on List 2 to get the mean of -6. Finally, A = g1 = -6/3.33333 = -0.1620 when
rounded. Notice that B is just the same as A except multiplied by the factor of
result from A times this factor yields B = 𝐺! = −0.1620
!! !!!(!.!)
!
!" !
!
!(!!!)
!!!
.
Thus, taking the
= −0.1708 when rounded. For part C:
!(!!!.!)
𝐺𝑆 = = − = −0.1667 when rounded. Finally, part D is: 𝑆𝑘! = = −0.4500 when
!!!
!
!.!!!!
rounded. The mean of these four results is (-.1620 – .1708 – .1667 – .4500) / 4 = -0.24 when rounded.
3. Answer:
!
!
Solution:
!
A) . 5 = !
B) 1 − . 5! = !
C) 1 − . 5 = D) 1 − . 5! = !
!
!
!
!"
!"
Summary: The square root of the mean of the above 4 results is:
! ! ! !"
! ! ! ! ! ! !"
!
=
!"
!"
!
= !
February Regional – Statistics Team – Answers and Solutions 4. Answer: 0.984
Solution: Each of these linear functions is of the form y = kx and forms a right triangle between the graph of
the function and the x-axis on the domain [0, b], where b is an element of the set {A, B, C, D}. Therefore, by
!
using the formula for the area of a right triangle, each solution is found by solving: 𝑘𝑏 ! = 1. This yields the
!
following solutions:
A)
B)
C)
D)
!!
= 1, 𝑠𝑜 𝐴 = 2 !
!! !
!
!! !
!
!! !
!
= 1, 𝑠𝑜 𝐵 = 1 = 1, 𝑠𝑜 𝐶 = = 1, 𝑠𝑜 𝐷 = !
!
!
!
Summary: The mean of the above values is
!!!!
!
!
!
!
!
!
= 0.984 when rounded to the nearest thousandth.
!
5. Answer: 𝑜𝑟 2.5
!
Solution: Define events as follows: P = Pre-Calculus, A = AB Calculus, B = BC Calculus, M = Multivariate
Calculus, and S = AP Statistics. A close inspection of the given information yields the following additional
information: P(P only) = 8/80, P(A only) = 23/80, P(B only) = 7/80, P(M only) = 2/80, and
P(S only) = 20/80 = ¼ or 0.25 = A. The sum of all five of these values gives us 60/80 = ¾ or 0.75 = B. By
design, a student who is taking two of the classes MUST be in AP Statistics, therefore, C = 1. By contrast,
exactly half of those taking AP Statistics are also taking another math class, making D = ½ or 0.5. A Venn
diagram with 5 circles arranged such that the center circle (representing set S) is the only one that intersects
the other 4 circles would help visualize the solution as well. Thus, the summary answer is A + B + C + D =
0.25 + 0.75 + 1 + 0.5 = 2.5 or 5/2.
6. Answer: 1
Solution: A = -1 This statement is always false since the neither the mean nor the variance of X are 0.
B = 1 This statement is always true for two iid’s unless E(X) = 0, which is not the case.
C = 1 This statement is always true since E(X1) = E(X2) = E(X).
D = 1 This statement is always true for any random variables, independent or not.
E = 0 This statement is only true when X and Y are independent, which is not specified.
F = -1 This statement would be true if the 2 was in a square root or SD(Y) = 0, which is not the case.
G = 1 This statement is always true for any linear combination of random variables.
H = -1 This statement is always false since a standard deviation is never negative.
Summary: The sum is: -1 + 1 + 1 + 1 +0 – 1 + 1 – 1 = 1.
7. Answer: 0.9982
Solution: Part A is normal distribution probability contained within a binomial. First we need to get the
probability that one randomly selected cable will break with a force over 12,400 N by using:
normalcdf(12400, ∞, 12432, 25) = 0.8997 which rounds to 0.90. Now it becomes a binomial with n = 400,
p = 0.90, and x ≥ 349 which is 1 – binomcdf(400, .9, 348) = 0.9689 which rounds to 0.97. Using the normal
approximation to the binomial with mean 0.9(400) = 360 and standard deviation of SQRT(400*.9*.1) = 6 and
normalcdf(349, ∞, 360, 6) = 0.9666, which rounds to 0.97 as well. Part B is a sampling distribution situation
with n = 400, mean 12,432, and standard deviation 25/SQRT(400) = 1.25. Since the population is normal and
the sample is large, B = normalcdf(12430, 12435, 12432, 1.25) = 0.9370 which rounds to 0.94. Summary:
P(A or B) = P(A) + P(B) – P(A)*P(B) = 0.97 + 0.94 – (0.97)(0.94) = 0.9982 since A and B are independent.
8. Answer: 24,916
Solution: Using the T-Interval function in the calculator with 𝑥: 12440, 𝑠! : 16, 𝑛: 36, and C-Level: 95 we get
(A, B) = (12435, 12445). Inserting 𝑠! = 16 𝑎𝑛𝑑 𝑛 = 36 into the given confidence interval formula and with
February Regional – Statistics Team – Answers and Solutions taking square roots to get standard deviations gives:
(!")!"!
!".!"#
<𝜎<
(!")!"!
!".!"#
and 12.98 < 𝜎 < 20.87. Thus.
(C, D) = (13, 21). Since neither confidence interval contains the original parameters of 12,432 for the mean
and 25 for the standard deviation, they both provide evidence that the improvements worked as intended and
so E = 2. The summary sum of A + B + C + D + E is 12,435 + 12,445 + 13 + 21 + 2 = 24,916
9. Answer: .0024
Solution: A = normalcdf (-∞, 60, 60, 7) = .5000, B = normalcdf (-∞, 60, 55, 5) = .8413, C = normalcdf (-∞,
60, 45, 10) = .9332, and D = normalcdf (-∞, 60, 65, 2) = .0062.
Summary: Since they are all independent, we can take the product of A*B*C*D = 0.0024, when rounded.
10. Answer: 1.83
Solution: 0.9505 = P(Z ≤ A) so A = invNorm(.9505, 0, 1) = 1.65
0.0044 = P(Z ≤ B) so B = invNorm(.0044, 0, 1) = -2.62
0.1922 = P(Z ≤ C) so C = invNorm(.1922, 0, 1) = -0.87
0.6700 = P(Z ≤ D) so D = invNorm(.6700, 0, 1) = 0.44
Summary: Place the four results into an empty list in the calculator and run the 1-Variable Statistics function
to get the sample standard deviation of 1.83, when rounded to hundredths place.
11. Answer: 2,646
Solution: A = .5*.5(1.96 / .03)2 = 1067.111111 which rounds up to 1068. For B, it is necessary to get the
population standard deviation for the die first. Entering the numbers 1, 2, 3, 4, 5, and 6 into a list in the
calculator and running the 1-Variable Statistics function gives 𝜎 = 1.7078 when rounded to 4 decimal places.
Thus, B = (1.96*1.7078 / .3)2 = 124.49 which rounds up to 125. A standard normal distribution has a
standard deviation of 1, so C = (1.96*1 / .1)2 = 384.16 which rounds up to 385. D is essentially the same as
A, so D = 1068. Their sum is 1068 + 125 + 385 + 1068 = 2,646.
12. Answer: 0
Solution: Although there is insufficient information to compute any of these probabilities exactly, parts A
and B are exactly the same since each can be thought of as
!
!
where n is the total number of songs stored on
! !
the i-pod. Along a similar line of thinking, parts C and D are both
as time t approached infinity.
!
Therefore, according to the Law of Large Numbers, parts C and D both approach 0 as time approaches
!
!
infinity. Thus making the summary answer A – B + C + D = − + 0 + 0 = 0.
!
!
13. Answer: 1
Solution: Since X is continuous, A, B, and C are all essentially 0 and D is 1. Therefore, their sum is 1.
However, if you make the mistake of treating X as a discrete random variable, you will get the same answer
as follows: P(A) = 4/11, P(B) = 5/11, P(C) = 1/11, and P(D) = 1/11 and their sum is 11/11 = 1.
14. Answer: 0.83
Solution: Sum each individual data set and divide the sum by 10 or enter each data set into an empty list in
the calculator and run the 1-Variable Statistics function. Either way, the resulting means for each data set are:
A = 4.1, B = 4.3, C = 3.4, and D = 5.4. Take these results and enter them into an empty list and again run the
1-Variable Statistics function to obtain the sample standard deviation of 0.83, when rounded to the nearest
hundredth.
15. Answer: 2.885
Solution: Enter each data set into an empty list in the calculator and run the 1-Variable Statistics function.
The resulting sample standard deviations for each data set are: A = 3.11, B = 2.31, C = 3.13, and D = 2.99,
when rounded to the nearest hundredth. The sum of these results and divided by 4 is 11.54 / 4 = 2.885.
Note: 2.89 is not an acceptable answer since the rounding instructions are clearly given.