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2016 FAMAT State Convention Hustle Statistics Solutions
1. 62. The sum of the nine numbers is 558. When you divide by the number of
numbers (9), you get the solution.
10
. The mean of the data is 2. When you subtract the mean from each value,
2
square the differences and add them up, you get a total of 10. When you divide by
5
(n-1), or 4 in this case, you get a variance of . When you take the square root, you
2
get the standard deviation.
2.

3. y = 1.2x + 6 or y =
6
x + 6 . The formula
for the line of best fit is

5
sy 
y  y  r x  x . When you plug in the given values, you get
sx 
3 8 
6
6
6

y  36   x  25  y  36  (x  25)  y  36  x  30  y  x  6 .
4 5 
5
5
5





10
12
. The probability of getting a face card is
. The probability of getting a
13
52
16
24
prime number is
. The probability of neither event occurring is
. The
52
52
12  16 
24  120 128  288 40
10
8 12 

 .
expected value of the game is 10  
52  52 
52 
52
52
13
4. -


5. 2 3 . The formula for the standard deviation of a geometric distribution is
1
3
1

1 p
4  4  12  2 3 .
. Plugging in the value produces
1
1 2
p2
 
16
4 
41
. The expected value for each value of the die is 10 times. Plugging the
5

(obs  exp) 2
values into the chi square formula, 
, produces
exp
6. 8.2 or

10 10  8 10  6 10  7 10  12 10  17 10
, which leads to
10
0  4 16  9  4  49 82 41
 
 .
10
10 5
2


2
2
2
21
. Given that the events are independent,
25
P(A  B)  P(A)P(B)  .8(.6)  .48 . So,
7. .84 or


2
2
P(A  B)  P(A)  P(B)  P(A  B)  .8  .6  .48  .92. Then,
P(A'B') 1 P(A  B) 1 .92  .08. Therefore, the final answer is .92 - .08 = .84.
8. 80. Given the information, there are two equations:
92  mean
69  mean
 2.4 and
 2.2 . When you solve the equations by
sd
sd
eliminating the mean, you get 23 = 4.6sd. This leads to a standard deviation of 5.
When you plug that into one of the equations, you get a mean of 80.



9. 39.7. To find the mean, you find
 X(P(X)) . Plugging the values in produces
23(.1)  36(.3)  38(.2)  47(.3)  49(.1)  39.7 .
10. 12 2 . The standard deviation of (3x – 4y) is


(3(4)) 2  4(3)  288  12 2 .
11. 11.65. The formula for a two sample t test is t 

2
x1  x 2
. Plugging into the
s12 s22


n1 n 2
48.75  25.45
23.3
23.3
formula produces t 


 11.65 .
2
2
2
144  256
12 16

100 100
100
12. -30. A residual = observed – predicted. When you plug 70 into the linear
equation, you get a predicted weight of 215. The observed weight is 185. Plugging

into the formula
gives the solution.
79
. When you put the numbers in order, you get 18, 41, 46, 54, 68, 75,
2
41 46
 43.5 . The
83, 83, 90. The median of the data is 68. The first quartile is
2
third quartile is 83. When you subtract the two values, you get the solution.

3 14
14.
. This is a binomial situation. The standard
 deviation of a binomial
2
distribution is np(1  p) . Plugging the values in produces
13. 39.5 or




 3  7 
3
3 7 3 14
.
150  

37 
10 10 
2
2
2

4
15. . First, find the probability that a student plays sports. The probability that a
7
male plays sports is (.4)(.6)= .24. The probability that a female plays sports is
(.6)(.3) = .18. So, 42% of students play sports. Therefore, the final solution is
.24 4
 .
.42 7
p(1  p)
. There wasn’t a p value
n
given, so the conservative estimate of p = .5 will be used. Plugging the values in
.5(.5)
produces .07 1.96
.07 n  .98  n 14 n 196 .
n

16. 196. The formula for margin of error is m  z
17. 14. If 16 students like neither sport, then 64 students are in the union between
the two sports. If the two sports were mutually exclusive, then there would be 78

students.
The difference, 78 – 64, is the number of students who like both sports.
18. 95. Since each part of the grade is weighted equally, she needs a total of 340
points to earn an overall grade of 85. With her three grades of 81, 86 and 78
already, her final exam grade must be at least 340 – (81+86+78) = 95.
1
1 P(A  B) P(A  B)
. Since P(A | B)  

P(A  B)  .2 . Therefore,
4
3
P(B)
.6
P(A  B)  P(A)  P(B)  P(A  B)  .5  .6  .2  .9. So, P(A'B')  .1. Therefore,
P(A'B') .1 1
P(A'| B') 
  .
P(B')
.4 4



1
20. y = x + 53 . To go from a standard deviation of 12 down to 4, you must divide
3
the scores by 3. Once you do that, the mean goes from 66 to 22. You must add 53 to
get up to the new mean of 75. Hence, the equation is produced.
19. .25 or




sy 
4
. The formula for the slope of the line of best fit is r . Plugging the
25
sx 


4
4
6 2
 r r 
 . Coefficient of determination is r 2 . When
values in produces
6 
15
15 5
you
 square the correlation coefficient, you get the solution.

1

22. .5 or . The probability that Jay gets at least three heads is
2
1 31 2
1 4 1 
1 5 10  5 1 1
 .
5 C3      5 C4     5 C5   
2  2 
2  2 
2 
32
2

23. 2.65. Using 68-95-99.7 as the empirical percentages, less than two standard
deviations below the mean is half of 5%, or 2.5%. Greater than 3 standard
deviations above the mean is half of .3%, or .15%. The sum of the two percentages
is the solution.
21. .16 or
1364
. The expected value for a chi square test of independence is
5
620(440) 1364
(row total)(column total)

. Plugging those values in produces
.
1000
5
total in the table
24. 272.8 or



25. 68. When you put the numbers in order, you get 18, 41, 46, 54, 68, 75, 83, 83,
90. The median of the data is the solution.