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Tallahassee Statewide
Statistics Individual Test Solutions
1/14/2017
1. D. This is an example of a stratified sample. A population is broken up by a
common factor, then an SRS is done for each subset of the common factor.
æs ö
2. C. The formula for the line of best fit is y - y = r ç y ÷ ( x - x ) . Plugging the
è sx ø
values in produces
æ 5000 ö
y - 43000 = (-.62)ç
÷ ( x - 8) ® y - 43000 = -1550(x - 8) ® y = -1550x + 55400 .
è 2 ø
3. B. Since P(A'Ç B') = .28 , P(A È B) =.72 . Using the formula
P(AÈ B) = P(A)+ P(B)- P(AÇ B) produces
.72 =.59 +.54 - P(AÇ B) ® P(AÇ B) =.41.
4. C. Mean, median and correlation coefficient can be negative. Range,
interquartile range and standard deviation can never be negative. The minimum
value for each of these three statistics is zero.
5. B. Since you weren’t told that the two variables were independent, the
formula for s x+y = s x2 + s y2 + 2rs xs y . Plugging in produces
s x+y = 6 2 + 4 2 + 2 (.72) ( 6) ( 4) =
2164 2 541
.
=
25
5
6. D. Create two z-scores from the given information. Those z-scores are
62 - mean
87 - mean
= -1.16 and
=1.53. When you solve for the standard
sd
sd
2500
deviation by eliminating the mean, you get sd =
. When you plug that into
269
19578
either equation, the mean =
» 72.78 » 72.8 .
269
7. A. The five number summary for the given set of data is (1, 20, 35,5, 93, 99).
So the IQR = 73. 73(1.5) = 109.5. Subtracting that value from the first quartile
and adding that value to the third quartile gives (-89.5, 202.5).
8. D. Find the mean of a spin of the wheel. This results in
æ 10 ö æ 5 ö æ 3 ö
æ1ö
æ1ö
1ç ÷ + 2 ç ÷ + 5ç ÷ +10 ç ÷ + 20 ç ÷ = .5 +.5+.75+.5 +1 = 3.25. Since it costs $5
è 20 ø è 20 ø è 20 ø
è 20 ø
è 20 ø
to spin the wheel, Mr. Smith is expected to profit $1.75.
9. A. Find the probability of getting one question correct and the other two
æ 1 öæ 2 öæ 1 ö æ 3 öæ 1 öæ 1 ö æ 3 öæ 2 öæ 1 ö 2 + 3+ 6 11
wrong. This leads to ç ÷ç ÷ç ÷ + ç ÷ç ÷ç ÷ + ç ÷ç ÷ç ÷ =
= .
è 4 øè 3 øè 2 ø è 4 øè 3 øè 2 ø è 4 øè 3 øè 2 ø
24
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Tallahassee Statewide
Statistics Individual Test Solutions
1/14/2017
10. E. The sum of the probabilities of the distribution is .99. The sum must be
equal to one in order for the standard deviation to be calculated.
zs
. The z-score needed for this
n
problem is at the 97.5th percentile. This z-score is 1.9599, which rounds to 1.96.
1.96(12)
® n = 22.12 » 23 .
Plugging in produces 5 =
n
11. D. The formula for margin of error is m =
12. A. The mean of the data is 13. When you subtract the mean from each
value, square the differences and add them up, you get a total of 288. Divide by
(n-1), or 9 in this case, to get a variance of 32. The square root of that is the
standard deviation.
13. C. The z-score needed for the 83.4th percentile is .97. Plugging into the z74 - 62
score formula produces .97 =
® sd =12.3711 » 12.37 .
sd
14. B. This is a cumulative binomial problem. We want to find the probability
that he completed between 17 and 21 passes. Therefore, we use the cumulative
binomial to produce binomcdf(28, .59, 21) – binomcdf(28, .59, 16) = .483014,
which rounds to .4830.
15. A. Since p(x=2, y=2) = .1581, .31(y=2) = .1581, so (y=2) = .51. Therefore,
(y=3) = .27. So, P(x =4, y=3) = .0675 = P(x=4)(.27). So (x = 4) = .25. Therefore,
(x = 1) = .26. Finally, P(x=1, y=2) = (.26)(.51) = .1326.
16. C. The standard deviation of the set (5x – 3y) is
( 5·6) + (3·10)
2
2
= 1800 = 30 2 .
17. B. When you create a Venn diagram with the information listed, there are
three openings for the students who take two classes exactly. Let a = biology
and physics, b = biology and chemistry and c = chemistry and physics. The
equations are a + b = 19, b + c = 14 and a + c = 13. When you solve those
equations, a = 9, b = 10 and c = 4. Adding up the individual totals produce
(20 + 27 + 32 + 6 + 9 + 10 + 4) = 108. Therefore, 22 students take none of the
three science classes.
18. E. The resistant measures in Statistics are median and interquartile range.
19. A. This is a geometric distribution situation. You find the probability that
Dakota is successful on the first, second or third try. This produces .36 +
2
(.64)(.36) + (.64) (.36) = .737856 .
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Tallahassee Statewide
Statistics Individual Test Solutions
20. C. The equations from the information is np =139.5 and
1/14/2017
np(1- p) =
3 589
.
10
When you substitute for n in the second equation, you get
æ 139.5 ö
3 589
5301
®139.5(1- p) =
® (1- p) = .38 ® p = .62 . Plugging p
ç
÷ p(1- p) =
10
100
è p ø
into the first equation produces n(.62) =139.5 ® n = 225.
21. B. In order for the standard deviation to go from 12 to 6, you must divide the
data by 2. When you divide the mean by 2, you get 29. So you must add 46 to
1
get to the new mean of 75. Therefore, the transformation equation is y = x + 46 ,
2
where x is the original score and y is the curved score. When you plug 66 in for
x, you get the solution of 79.
22. D. Given that the events are independent,
P ( AÇ B) = P(A)P(B) = (.68)(.47) =.3196 . Therefore,
P(AÈ B) =.68+.47-.3196 =.8304 and P(A'Ç B) =.1504 and P(AÇ B') =.3604 .
P(B Ç A') P(A Ç B') .1504 .3604
P(A' | B) + P(B' | A) =
+
=
+
= .85 .
P(B)
P(A)
.47
.68
23. A. The empirical percentages for a normal distribution are 68-95-99.7. The
probability of being one standard deviation below the mean is .16. The
probability of being three standard deviations above the mean is .0015. The sum
of the two probabilities is the solution.
24. D. There are six treatments for the experiment. There are three types of
food and two levels for the vitamin C (yes or no). 3X2 = 6.
3
25. C. When you plug 80 into the equation, you get y = (80) + 52 = 82 . A
8
residual = observed – predicted. 73 – 82 = -9.
26. B. The probability that a student who does not drive to school is (.44)(.92) +
(.56)(.85) = .8808. The probability that a student who does not drive is a girl is
.476 595
(.56)(.85) = .476. The final solution is
.
=
.8808 1101
27. C. Since the mean is to the right of the median, the mean is larger due to
the skew of scores to the right, raising the value of the mean.
28. C. The probability that a student scores less than 1300 comes from
1300 -1030
z=
= 2.16 . This leads to a p value of .9846137204. The probability
125
1100 -1030
that a student scored greater than 1100 comes from z =
= .56 . This
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Tallahassee Statewide
Statistics Individual Test Solutions
1/14/2017
leads to a p value of .287739682. The p value from 1100 to 1300 is .287739682
– (1 - .9846137204) = .2723534024. The final solution is
.2723534024
= .2766093918 » .2766 .
.9846137204
29. A. This is a geometric situation. The formula for the standard deviation is
1- p
15 1
. The probability of choosing a red M+M is
= . Plugging the value in
2
p
45 3
1
3=
2
æ1ö
ç ÷
è 3ø
1produces
2
3 = 6.
1
9
30. B. The integers in the data set X are the perfect squares. Every perfect
square has an odd number of integral factors. There are 14 perfect squares less
than 200. The mean is produced by
1+ 4 + 9 +16 + 25+ 36 + 49 + 64 + 81+100 +121+144 +169 +196
= 72.5 .
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