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Statistics Test Individual Solutions
February Regional
1. D: The function factors to (𝑥 − 1)(𝑥 − 2)2 and the sum of the distinct roots is 3,
therefore A=.3 and B=.35. The mean of the distribution using a variety of methods
is 3.5.
2. B: A and B were found in the above solution. The standard deviation of the
distribution is
√235
10
3. A: 𝑉𝑎𝑟(𝑥) = 𝐸(𝑥 2 ) − [𝐸(𝑥)]2 therefore 15 = 25 − [𝐸(𝑥)]2 and [𝐸(𝑥)]2 = 𝟏𝟎.
4. D:
(10−1)!
2!∗3!∗3!
= 𝟓𝟎𝟒𝟎
5. A: Since the mode is greater than 2, the distribution has a tail on the left side. This
makes it left-skewed.
6. A: Definition of systematic sampling.
7. C: The situation present is a geometric distribution and the standard deviation of
1−𝑝
1−.75
𝟐
a geometric distribution is √ 𝑝2 . Thus √ .752 = 𝟑.
𝑝(1−𝑃)
8. C: 𝑀𝐸 = 𝑧 ∗ √
𝑛
. Since there is no previous information on proportions a p-
value of .5 will be used. The 𝑧 ∗ for a 95% confidence interval is 1.96 and since the
required width is .1 the margin in .05. Solving for n and rounding up results in 385.
9. C: 𝐸(𝑋 + 𝑌) = 5 + 6 = 11. 𝑉(𝑋 − 𝑌) = 3 + 2 − (2 ∗ .8 ∗ √3 ∗ √2) = 1.08. 11 +
1.08 = 𝟏𝟐. 𝟎𝟖.
10. B: There are 57 combinations with at least two girls and there are 31
𝟑𝟏
combinations with a girl being first, thus the probability is 𝟓𝟕.
11. E: Since the sum of the residuals is 0 (𝐴𝐵)0 = 𝟏
1
1
12. A: The probabilities follow the following geometric sequence 𝑝𝑛 {1 + 10 + 100 … }
9
which has to sum up to 1. This means that 𝑝𝑛 = 10 and the probability he makes
9
9
9
more than 2 errors is 1 − [10 + 100 + 1000] =. 𝟎𝟎𝟏
13. D: Calculate the raw score by solving: 1.645 =
𝑟𝑎𝑤−670
25
√154
, 𝑟𝑎𝑤 = 673.3. Next
calculate the z-score using the raw score and the alternative hypothesis;
673.3−676
−1.34. the proportion of the normal curve greater than this score is 0.909.
25
√154
=
Statistics Test Individual Solutions
February Regional
14-16. Consider a Venn diagram with variables x, y, z, and c. Let x represent the
region of only A, y represent the overlap between A and B, z represent the region of
only B, and c representing the region out side of both A and B. We know that 𝑥 + 𝑦 +
𝑦
𝑧 + 𝑐 = 1. From the first probability given we know that 𝑦+𝑧 = .25, from the second
𝑥
we know 𝑐 = .1, and from the third probability we know that 𝑥+𝑦 = .3. We now have
four simultaneous equations that can be solved to determine x, y, z, and c.
𝑥+𝑦+𝑧+𝑐 =1
. 75𝑦 − .25𝑧 = 0
𝑐 = .1
{ . 7𝑥 − .3𝑦 = 0
27
63
189
1
Solving results in 𝑥 = 310 ; 𝑦 = 310 ; 𝑧 = 310 ; 𝑐 = 10
27
63
90
9
14. A: 310 + 310 = 310 = 31
63
15. B: 310
16. D:
63
310
27 63
+
310 310
63
7
= 90 = 10
17. D: The first six triangular numbers are {1,3,6,10,15,21}, there are four odd
numbers and the set of odd Pythagorean numbers are {3,15,21}, therefore the
𝟑
probability is 𝟒.
18. E: Probability that you reject a false null hypothesis = 1 − 𝑇𝑦𝑝𝑒 𝐼 𝐸𝑟𝑟𝑜𝑟 = 1 −
.05 = . 𝟗𝟓
19. B: The sample size from the information given is 4. If the sum of the mean and
standard deviation is 60, then the standard deviation of the distribution must be 5.
𝜎
𝜎
The standard deviation of a distribution is 𝑛 → 2 = 5 → 𝜎 = 𝟏𝟎
√
20. C: 𝑉𝑎𝑟(𝑥) = 𝐸(𝑥 2 ) − [𝐸(𝑥)]2 and the distribution is binomial. Therefore
1
1
1 2
𝐸(𝑋 2 ) = 𝑛𝑝(1 − 𝑝) + (𝑛𝑝)2 = 32 ∗ 2 ∗ 2 + (32 ∗ 2) = 𝟐𝟔𝟒.
21. D: I: The correlation could be positive or negative, so the statement is false. II: A
geometric distribution is never approximately normal, so the statement is false. III:
To increase power the alternative and null should be farther away from each other,
so the statement is false.
1
22. B: 𝐴 = 3 because it is simply the probability of guessing the door correctly
initially. To calculate B, consider that there is a 2/3 chance that the car is behind one
of the two doors you did not initially choose. Once the host opens one of the doors
and shows that it is a camel that door now has a probability of 0 to contain the car.
Statistics Test Individual Solutions
February Regional
Now the entire 2/3 chance transfers to the remaining door. This means that you
2
𝟏
have a 2/3 chance of winning a car if you switch doors. 𝐵 = 3. 𝑨 − 𝑩 = − 𝟑.
23. E: The only value that always has to be positive is sample size; all other values
can be negative or 0.
24. C: The P(X=7) is 0 and the 𝑃(𝑋 < 8|𝑋 > 6)= 1/2. 1/2 + 0= 2/5.
25. A: The probability of a full house is
13 4 12 4
( )( )( )( )
1 3 1 2
52
( )
5
=. 001
26. A: I is a true statement. II is false the mean is simply k, the degrees of freedom. III
is false the probability is 0 or 1 since the value of e is determined and fixed.
27. B: The probability of the scenario occurring using conditional probability
formulas is
𝑥
5
∗
𝑥+6 10
6
4
𝑥
5
∗ +
∗
𝑥+6 10 𝑥+6 10
5
= 13, solving for x results in a value of 3.
28. D: The probability Megan eventually wins is the following geometric sequence:
1
2 3 1
2 3 2 3 1
2
+ (3 ∗ 4 ∗ 3) + (3 ∗ 4 ∗ 3 ∗ 4 ∗ 3) + ⋯ = 3. The probability that he wins before the
3
4th overall throw is equal to the probability he wins on his first or third throw which
1
2
3
1
1
1 2
equals 3 + (3 ∗ 4 ∗ 3) = 2.
2
3
𝟑
=𝟒
29. C: The information given results in two simultaneous equations:
36−𝜇
𝜎
24−𝜇
𝜎
= .5 and
= 2 solving the equations results in 𝜇 = 20 𝑎𝑛𝑑 𝜎 = 8, 𝑖 20+8 = 1
30. B: There is a 364/365 probability that a person doesn’t have a birthday on
August 26. Using complementary probability the equation that needs to be solved is
364 𝑛
364 𝑛
1 − (365) > .5 → .5 > (365) →
ln(.5)
ln(
364
)
365
> 𝑛 → 𝑛 = 𝟐𝟓𝟑