Download Vero Beach Statistics Individual Solutions 1. B. The difference

Vero Beach Statistics Individual Solutions
1. B. The difference between the two means are 80 – 69 = 11 and the difference
between the two standard deviations are 82  62 10 .
2. D. The population is broken down by grade level and a random sample is taken
for each sub group. This is the definition of a stratified sample.
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3. C. The formula for the standard deviation of a geometric distribution is
1
3
1

1 p
4  4  12  2 3 .

2
1
1
p
16
16
4. C. There are two z-scores for the boundaries. They are
83  72
63  72
 2.75 and
 2.25 . The percentiles for the two z-scores are
4
4
.9970201814 and .0122244334. The difference between them is .984795748. This
rounds to .9848.
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



5. A. The coefficient of determination is r 2 . The formula for slope of the line of best
sy  9  549
fit is r r 
r  .61 r 2  .3721.
sx  7  700

6. A. To change the standard deviation from 12 to 8, you must multiply the data by
2
2
2
128
. When you multiply the mean by , you get 64  
. To get up to a new
3
3
3
3
97
2
97
mean of 75, you must add
. So the transformation equation is y  x  .
3
3
3
Plugging in 76 produces the new test score of 83.
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7. D. Build a tree diagram with the information. There are two probabilities
 who do extracurricular activities and
 those who don’t. The
involving boys: boys
two probabilities are (.58)(.44) and (.42)(.38). Therefore, the solution is
.58(.44)
.2552 638
.


.58(.44)  (.42)(.38) .4148 1037
8. C. The mean of the data is 63, the median is 62.5 and the interquartile range is
34. The discriminant of the function is 62.52  4(63)(34) 12474.25 .
9. B. Find the mean of the data, which is 14. Subtract the mean from each value,
square the differences and add them up to get a total of 248. Divide by (n-1), or 9, to
248

get a variance of
. Take the square root and simplify completely to get the
9
solution.

10. D. The mean of the entire process is (8+5+6) = 19. The standard deviation of
18 19
 .08
the entire process is 32  4 2 122 13. Running a z score produces
13
to two decimal places. The probability to four decimal places is .4681.
11. B. Find
the mean by finding

1929
 X(P(X))  10(.19) 12(.15) 14(.12)  21(.23)  29(.22)  30(.09)  19.29  100 .

12. C. This is the definition of Simpson’s Paradox. We get one conclusion when the
data is in one combined set and we get the complete opposite conclusion when the
data is broken down into subgroups.
13. D. There are two z scores for this problem. They are 1) -1.27SD = 51 – mean
and 2) 1.53SD = 87 – mean. Solving by eliminating the mean produces a standard
90
deviation of
. When you plug this in and solve for the mean, you get a value of
7
67.32857143, which rounds to 67.33.
14. E. All of the statements are false. When all of the values in a sample data set are
the
same, the standard deviation is equal to zero. Standard deviation is equal to the
square root of the variance. Standard deviation is affected by outliers, therefore it is
not resistant.
sy 
8 
15. A. The line of best fit formula is y  y  r ( x  x )  y  76  (.73) x  5.
2 
sx 
This leads to y  76  2.92(x  5)  y  2.92x  61.4 .
16. B. The first two statements are true about both. Only in binomial situations are
 Only in geometric situations are we always looking
there a fixed number of trials.

for a successful trial. Only in binomial situations can it be approximately normal.


17. D. Because the two events are independent, P(A  B)  P(A)P(B) . Therefore,
P(A  B)  .76(.61)  .4636 . This creates an P(A  B)  .76  .61 .4636  .9064 .
Therefore, P(A'B')  .0936. Plugging into the expression gives
.9064  .4636  .0936  .3492 .
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18. B. First find the mean, which is 47.87. Then subtract the mean from each value,

square
the differences, multiply by their corresponding probabilities and add them
up. This gives a variance of 166.3331. The square root of this is the standard
deviation, which is 12.89701904, which rounds to 12.90.
19. A. The formula for binomial standard deviation is
np(1  p)  175(.63)(.37)  40.7925  6.386900657  6.39 .

20. C. The first 10 positive prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.
The mean of the set is 12.9 and the median of the set is 12. (12.9 – 12) = 0.9.
21. D. When you plug 74 into the equation, you get 87. This is the predicted score.
The residual is -6 = observed – predicted. Therefore, - 6 = observed – 87. This leads
to an answer of 81.
22. C. The test statistic at 90% is 1.28. Therefore, 1.28 
raw  66
raw  71.12 .
4
23. B. The total number of hot dogs sold at all sporting events is 1239. Therefore,
640
 .5165456013  51.65% .
the percentage of hot dogs sold at footballgames is
1239
24. D. The first success by the third attempt means that Rob could make it on his
first shot, second shot or third shot. Therefore, the solution is

.32  (.68)(.32)  (.68)2 (.32)  .685568 .
25. B. The new mean is 72(1.5) – 20 = 88. The new standard deviation is 5(1.5) =
7.5.
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26. A. The only way to show a cause and effect relationship is to do a controlled
experiment.
27. C. The prime numbers are 2, 3, 5 and 7. There are four suits, so there are 16
cards that are prime out of 52. The answer is the reduced solution.
28. C. His grades so far are 90(.1) + 81 (.2) + 78(.3) + 84(.15) = 61.2. To get to a
total grade of 85, Lirun must earn 23.8 points out of the 25 possible from the final
23.8
 .952 . Therefore, Lirun must earn at least a 95.2 on the final exam to
exam.
25
get an overall grade of 85.
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29. D. The range of the correlation coefficient is -1 to 1, inclusive.
30. A. Plug the numbers into a Venn diagram. There are three open parts: each of
the sections with two classes. Let a = math and science, b = math and history and
c = history and science. Therefore, a + b = 6, b + c = 9, and a + c = 7. Solving these
equations gives a = 2, b = 4 and c = 5. When you add these three numbers along
with the 6 students for all three classes produces a total of 17 students out of the 52
17
students in Mr. Smith’s class. Therefore, the answer is
.
52
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