1. The relative frequency of scores between 60 and 100 is .15+.25 +.35+.10=.85. So the relative
frequency of a score from 0 to 59 is 1-.85=.15 (Incorrect answer a). The frequency of scores in
this range is 140*.15=21 (Answer d)
2. 𝑧 =
𝑥− 𝑥̅
𝑠
=
9−25
√16
= −4 (Answer d). Note: If you forget to take the square root of 16 you will
get answer choice b. If you subtract in the wrong order in the numerator, then you could get
answer c or, if you also forgot the square root, answer a.
3. 𝑠 2 =
∑ 𝑥𝑖2 − (∑ 𝑥𝑖 )2 /𝑛
𝑛−1
=
16,057−4552 /25
24
=
7,776
24
= 324 → 𝑠 = √324 = 18 (Answer b). Note: If
you forget to take the square root you get incorrect answer e. Other common algebra mistakes
may get you one of the other incorrect answer choices.
4. There are 6*6=36 possible outcomes that are equally likely (“balanced die”). Six of these
outcomes correspond to the event described ( (1,1), (2,2), … , (6,6) ). For an equally likely model
the probability of an event is the ratio of the number of sample points in the event to the total
number of sample points: probability =
6
36
=
1
6
(Answer c).
5. For the threatening stimulus, the upper edge of the box (i.e. the third quartile or 75th percentile)
is 1.8. So 25% of the observations exceed 1.8 since 75% are at or below 1.8 (Answer b). Note:
You will get incorrect answer d if you fail to recognize that the event in question is
complementary to the one that determines the 75th percentile (ie . 75 = 𝑃(𝑇 ≤ 1.8) ).
6. The incorrect answer choice is Answer e. Note: The data actually goes from a low of 1.3 to the
high value 2.5 where you find the outlier. Answer a: The inter-quartile range is the difference
between the 75th percentile (upper edge of box at 1.8) and the 25th percentile (lower edge of
box at 1.5). Answer b: The 25th for non-threatening is 1.9. That exceeds the 75th for threatening
which is 1.8. Answer c: The respective medians are 2.0 and 1.7. So the difference is .3. Answer d:
The lone asterisk indicates only one outlier.
7. There are 80 observations (an even number). So the median is the average of the two “middle”
numbers in the ordered data (ie the 40th and the 41st observations). By inspecting the
cumulative frequency column at the far left you can see that the 38th observation was (end of
row) 36.9. So the two observations we need to compute the median are the second (40th = 37.0)
and third (41st = 37.1) in the next row. The average of these two numbers is 37.05 (Answer c).
8. The mean is determined by dividing the sum of all observations (given as 2,968.3) by the total
number of observations (given as 80): 𝑥̅ =
2,968.3
80
= 37.103 (Answer c). Note: you would get
incorrect answer choice a if you thought there were 100 observations (text example).
Example Final Exam Solutions
1
9. Using the bottom-up portion of the cumulative frequency column, you can see that 16
16
observations were 39.0 or higher. So the percentage is: 80 ∗ 100% = 20% (Answer d).
10!
10∗9
10
10. There are ( ) = 2! 8! = 1∗2 = 45 equally likely (ie phrase “selected at random”) outcomes to
2
the selection process. Your wager is on one possible outcome: 𝐸 = {(1, 3)}. So the probability
is: 𝑃(𝐸) =
number of outcomes in E
number of sample points
=
1
45
(Answer d). Note: This was a “combination” problem. If
the order of selection mattered, then there would be 10 ∗ 9 = 90 distinct outcomes. In this
case you might incorrectly choose answer c.
11. The event A does not occur if the sum is 2, 3, …, 6: 𝑃(𝐴) =
12. The question asks for 𝑃(𝐴|𝐵) =
mistakenly compute 𝑃(𝐵|𝐴 ) =
1+2+3+4+5
36
=
15
36
=
5
12
(Answer a).
𝑃(𝐴 ∩ 𝐵)
𝑃(7,8)
11/36
11
= 𝑃(6,7,8) = 16/36 = 16 (Answer a). Note: If you
𝑃(𝐵)
𝑃(𝐴 ∩ 𝐵)
11/36
11
= 21/36 = 21 , then you get incorrect answer b.
𝑃(𝐴)
Answer choice c is the unconditional probability of event 𝐴 ∩ 𝐵. Answer d is the unconditional
probability of event B. Answer e is the unconditional probability of event A.
13. The event described is the union of events A and B. So there are 2 possible ways to proceed.
Method 1: List the sample points in the union and then add up their probabilities. Here we have:
𝐴 ∪ 𝐵 = {6, 7, 8, … , 12} → 𝑃(𝐴 ∪ 𝐵) =
5+6+5+4+3+2+1
36
=
26
36
(Answer d). Method 2: Use the
additive probability law along with results determined in the preceding solutions. Here we have:
𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵) =
21
36
+
16
11
− 36
36
=
26
36
. Note: If you confused the verbal
description of union with intersection, then you would get answer a. If you forgot to subtract
the probability of the intersection in the additive law, then you would get the absurd
(probability cannot be greater than 1) answer e.
14. Since the three probabilities must add up to 1, the correct answer is: 1 − (. 24 + .33) = .43
(Answer c). You might choose d if you thought in terms of the percent chance.
15. The first statement tells you that 𝑃(𝐸 ∪ 𝐹) = .60. You are also given: 𝑃(𝐸) = .48, 𝑃(𝐹) = .57.
You are asked to calculate 𝑃(𝐸 ∩ 𝐹) (ie both occur). According to the additive law, we have:
𝑃(𝐸 ∪ 𝐹) = 𝑃(𝐸)
+ 𝑃(𝐹)
𝑃(𝐸 ∩ 𝐹) → 𝑃(𝐸 ∩ 𝐹) = .48 + .57 − .60 = .45 (Answer d).
⏟ −⏟
⏟
⏟
.60
.48
.57
?
Note: Answer e is absurd (ie >1). It is the sum of 𝑃(𝐸) and 𝑃(𝐹).
16. The mean is computed as: 𝜇 = ∑ 𝑥 ∗ 𝑝(𝑥) = 1 ∗ .6 + 2 ∗ .3 + 3 ∗ .1 = 1.5 (Answer b). Note: If
you confuse population mean with sample mean then you would get incorrect answer c. If you
added up the probabilities without multiplying by 𝑥, then you get answer a. If you calculated
∑ 𝑥 2 ∗ 𝑝(𝑥), then you will get answer d.
Example Final Exam Solutions
2
17. The variance is calculated first: 𝜎𝑋2 = ∑ 𝑥 2 ∗ 𝑝(𝑥) − 𝜇2 = 2.7 − 1.52 = .45 (incorrect answer
a). Next you calculate the square root: 𝜎𝑋 = √𝜎𝑋2 = . 451/2 = .6708 (Answer b). If you
mistakenly computed the sample variance 𝑠 2 assuming a sample of 1, 2, 3, then you will get
𝑠 2 = 1, 𝑠 = 1 (answer c). Answer e would result from the mistake: 𝜎𝑋2 = ∑ 𝑥 2 ∗ 𝑝(𝑥) = 2.7.
And answer d would result from taking the square root of the preceding mistake.
18. The number of heads in 5 flips follows a binomial distribution with 𝑛 = 5 trials and 𝑝 = .5 (ie a
5
“fair” coin). The event at least 4 heads is 𝑋 ≥ 4. The probability is calculated as: ( ) ∗. 54 ∗. 51 +
4
5+1
6
5
5
0
( ) ∗. 5 ∗. 5 = 32 = 32 (Answer c). Answer e is the probability that 𝑋 is at most 4. Answer
5
d is the probability that 𝑋 is less than 4. Answer a is the probability that 𝑋 exceeds 4.
19. The sample mean is normally distributed with mean 188 and variance: 𝜎𝑋2̅ =
the variable
𝑋̅−188
√36
2
𝜎𝑋
𝑛
=
122
4
= 36. So
194−188
is standard normal: 𝑃(𝑋̅ > 194) = 𝑃 (𝑍 >
) = 𝑃(𝑍 > 1) = .5 −
36
√
.3413 = .1587 (Answer a). Note: Answer d was a preliminary result in the correct solution. If
you reasoned that the sample is too small to apply the Central Limit Theorem, then you might
have mistakenly selected answer e. The point is: Since the population is normally distributed the
sample mean is exactly normally distributed regardless of the sample size. Also, if a question
intends for you to apply the CLT it will instruct you to “approximate” a probability.
20. The mean of a binomial distribution with 𝑛 = 20 (rolls) and 𝑝 = 1/6 (ie “balanced” die) is 𝜇 =
1
𝑛𝑝 = 20 ∗ (6) = 10/3 (Answer b). It is incorrect to round this answer to 3 (answer a). The
mean of a binomial is not necessarily a possible value for the number of successes.
21. The event “at least 4 sixes” is 4 ≤ 𝑋 ≤ 20 where 𝑋 is binomial with 𝑛 = 20 trials and 𝑝 = .20.
The associated probability is: 𝑃(4 ≤ 𝑋 ≤ 20) = 1 − 𝑃(0 ≤ 𝑋 ≤ 3) = 1 − .411 = .589
(Answer c). Note: Answer b would result if you forgot to subtract from 1. Answer a would result
if you mistakenly calculated: 1 − 𝑃(𝑋 ≤ 4) = 1 − .630.
22. The z-score of 75 is: 𝑧 =
75−84
√9
= −3. The probability is: 𝑃(𝑍 < −3) = .0013 (Answer b).
Note: If you failed to take the square root of 9, then you get a z-score of -1 and answer c results.
Answer d is 𝑃(0 ≤ 𝑍 ≤ 1).
23. There is a general relationship for converting a z-score to an observation: 𝑥 = 𝑥̅ + 𝑧𝑠 = 25 −
.5 ∗ √16 = 23 (Answer d). Note: Answer a would result if you forgot to take the square root of
16. Answer e would result if you used a z-score of +.5 instead of -.5.
Example Final Exam Solutions
3
24. Let 𝑋 be the number in the sample of size 30 that are over their desirable weight. This variable is
binomial with 𝑛 = 30, 𝑝 = .20 (“success” means more than 40% above ideal weight). The mean
of this distribution is 𝜇 = 𝑛𝑝 = 30 ∗ .2 = 6 (Answer c). Note: Normal cannot be correct. It is a
continuous random variable whereas a count is discrete. If you were confused and used 𝑝 = .4,
then answer d would result.
25. You are given: 𝜇𝑋 = 192, 𝜎𝑋 = 16. Since the sample size is 𝑛 = 256 = 142 , we have:
𝜇𝑋̅ = 𝜇𝑋 = 192, 𝜎𝑋̅ =
̅ − 𝟏𝟗𝟐
𝑿
𝟏
𝜎𝑋
√𝑛
16
= 16 = 1. According to the Central Limit Theorem, the distribution of
is approximately standard normal (Answer c). Note: The other answers are based on
screwing up the square root factors in the denominator.
26. 𝑧 =
𝑥̅ −𝜇𝑋
𝜎𝑋 /√𝑛
=
73−72
√36/√81
= 1.5 (Answer e).
27. The incorrect statement is Answer b. It should say “divided by the square root of the sample
𝑥̅ − 𝜇
√𝑛
size.” Note: The t-distribution is appropriate for the ratio 𝑠/
computed for a sample from a
normal population. It is not appropriate for 𝑥̅ itself.
28. The standard error of the mean is: 𝜎𝑥̅ =
have: . 01 =
100
𝑛
→ 𝑛=
100
.01
𝜎𝑋
√𝑛
=
10
√𝑛
. If this ratio is equal to .1, then squaring, we
= 10,000 (Answer d). Algebra mistakes could give one of the
other answers.
29. The incorrect statement is answer b. The sampling error is one-half of the width of the interval.
Note: For answer a note that: 𝑧𝛼⁄2
𝜎𝑋
√4𝑛
= 𝑧𝛼⁄2
𝜎𝑋
2∗√𝑛
=
1
2
∗ 1.96
𝜎𝑋
√𝑛
.
30. Since the test is one-sided (𝐻𝑎 : 𝜇 > 30), the rejection region for an 𝛼 level test is of the form:
𝑡 > 𝑡𝛼 . For 15 degrees of freedom we have: 𝑡.025 = 2.131 > 2.125
⏟
> 𝑡.050 = 1.753. So we
statistic
would reject at a 5% level, but we would fail to reject at a 2.5% level (Answer c). Note: Other
answer choices could result if you put the rejection region in both tails. Or, if you incorrectly
used 16 degrees of freedom (𝑡.025 = 2.120 < 2.125
⏟
: reject at a 2.5% level, but fail to reject at a
statistic
1% level), then answer d would result.
31. The sampling error for a small sample t-interval with confidence level 98% is: 𝑡𝛼⁄2
8
√16
𝑠
√𝑛
= 2.602 ×
= 5.204 (Answer e). Note: Answer a would result if you forgot to take the square root of
16. Answer choices b or d might result if you used the t-percentile for 16 degrees of freedom.
Answer choice c results if you forget that the sampling error involves the standard error of the
mean.
Example Final Exam Solutions
4
32. Since the test is 2-sided, p-value = 2𝑃(𝑍 ≥ 1.54) = .1236 (Answer c). Note: Answer a results if
the test were one-sided with: 𝐻𝑎 : 𝜇 > 𝜇0 . Answer d is equal to: 𝑃(0 ≤ 𝑍 ≤ 1.54). Answer e is
equal to: 𝑃(−1.54 ≤ 𝑍 ≤ 1.54).
33. Since the test is one-sided (left tail), the p-value is: 𝑃(𝑍 < −2.26) = .0119 (Answer b). Note:
The negative answers a and d are absurd! Answer e is equal to: 𝑃(0 ≤ 𝑍 ≤ 2.26). Answer c is
equal to: 𝑃(𝑍 < −2.25).
34. Answer choice a is a false statement. It is incorrect to ascribe a probability to a numerical
confidence interval since there is no randomness. The other statements are all correct.
35. Statement I is the true definition of 𝛼, the level of significance. Statement II confuses the level of
significance with the probability of a type II error. Statement III is true. So Answer e is the
correct choice.
36. Applying Chebychev’s Rule with 𝑘 = 1.5, the probability is at least 1 −
1
𝑘2
=1−
4
9
5
9
= (Answer
e). Note: Answers a and b are incorrect uses of the normal distribution.
37. The sample proportion 𝑝̂ is the midpoint of this interval: 𝑝̂ =
.48402+.54598
2
= .515 =
515
1,000
(Answer b).
38. The true proportion is likely to be in the interval. However, some numbers in the interval are
less than .5 and some are greater than .5. So Answer d is the correct statement. Note: Nothing
is ever “proven” by a statistical procedure.
39. Since the level is 5% and the test is one-sided (left tail), the rejection region is : 𝑡 < 𝑡.05 =
−1.703 (Answer e). Note: Answer a would be the rejection region for a 5% level, 2-sided test
since 𝑡.025 = 2.052. Answer d would be the rejection region for a 10% level 2-sided test.
𝑋̅−26
√28
40. The given value 𝑠/
= −2.473 is the value of the t-statistic used in this test. Since the test is
one-sided (left tail), the p-value is: 𝑃(𝑡 < −2.473 = −𝑡.01 ) = .01 (see the table) (Answer
b).Note: Due to the limited number of t-percentiles in the table, the only way the problem could
be answered is if the given value of the statistic is ± one of these standard percentiles.
41. The general principles are: Reject the null hypothesis when the p-value is below alpha (be sure
you’re not comparing a decimal with a percent), and, for a 2-sided test the p-value is double
that for a 1-sided test. So Answer d is the only false statement.
42. The false statement is Answer c. To be correct, it should say that the midpoint of the intyerval is
always equal to the sample mean.
Example Final Exam Solutions
5
43. The false statement is Answer c. These formulas are derived from properties of the sampling
distribution. Note on answer e: The interval straddles 0, so it is possible that 𝜇1 − 𝜇2 = 0 (ie it is
possible that the means are equal).
44. The correct answer is Answer e. Note: The other answer choices are missing the phrase “at
least” (part of Chebychev’s Rule) or the qualifier “approximately” (part of the Empirical Rule).
45. The false statement is Answer d. The small sample test assumes that the population sampled is
normally distributed.
46. The proportion of students in the sample on probation is:
15+20
500
= .07 (Answer c).
47. The proportion of female students on probation in the sample is:
Note: If you thought the proportion was
20
260
20
20+260
= .071 (Answer c).
then you mistakenly get answer d.
48. There are 35=15+20 students in the sample on probation. The proportion of females in this
20
group is 35 = .571 (Answer d). Note: This calculation assumes an “equally likely” model (ie
“selected at random”).
49. The total number of males in the sample of size 500 is 15+205=220. The proportion of males in
the sample is thus:
220
500
= .44 (Answer d).
50. The standard deviation is: 𝜎 = √14.11 − 3.492 = √1.9299 = 1.39 (Answer b).
51. The interval 𝜇 ± 𝜎 is equal to: (2.1 , 4.9). The probability sought is: 𝑃(2.1 < 𝑋 < 4.9) =
𝑝(3) + 𝑝(4) = .58 (Answer c). Note: Answer d is an approximate answer for mounded and
symmetric distributions (the Empirical Rule). This distribution is mounded, but it is skewed to
the right.
52. 𝑃(win) = 𝑃(𝑅1 ∩ 𝐵2 ) = 𝑃(𝑅1 ) 𝑃(𝐵2 |𝑅1 ) = .44 × .25 = .11. So the probability that you will
not win is: 𝑃(win𝑐 ) = 1 − 𝑃(win) = .89 (Answer e).
Example Final Exam Solutions
6