AP Statistics
Section 6.1 Worksheet
Name: _____________________________________
1. Suppose you roll a pair of fair, six-sided dice. Let T = the sum of the spots showing on the up-faces.
a) Find the probability distribution of T.
T
P(T)
2
1
36
3
2
36
4
3
36
5
4
36
6
5
36
7
6
36
8
5
36
9
4
36
10
3
36
11
2
36
12
1
36
b) Find P(T ≥ 5) and interpret the result.
6 30
 1 2 3
P  T  5   1  P T  4   1       1  
 0.833
36
36
36
36
36


This means that about five-sixths of the time, when you roll a pair of 6-sided dice, you will have a sum of 5 or more.
2. In 2010, there were 1319 games played in the National Hockey League’s regular season (Don’t forget that the Blackhawks won it
all though!!!!). Imagine selecting one of these games at random and then randomly selecting one of the two teams that played in the
game. Define the random variable X = number of goals scored by a randomly selected team in a randomly selected game. The table
below gives the probability distribution of X:
Goals:
Probability:
0
0.061
1
0.154
2
0.228
3
0.229
4
0.173
5
0.094
6
0.041
7
0.015
8
0.004
9
0.001
a) What is the probability that the number of goals scored by a randomly selected team in a randomly selected game is at least 6?
P(X ≥ 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) = 0.041 + 0.015 + 0.004 + 0.001 = 0.061.
b) Compute the mean of the random variable X and interpret this value in context.
 X  E ( X )  (0)(0.061)  (1)(0.154) 
 (9)(0.001) = 2.851. The mean number of goals for a randomly selected team in a randomly
selected game is 2.851. If you were to repeat the random sampling process over and over again, the mean number of goals scored
would be about 2.851 in the long run.
c) Compute the standard deviation of the random variable X and interpret this value in context.
X 
 0  2.851  0.061  ...   9  2.851  0.001  1.632. The standard deviation of X is 1.632. The number of goals scored by a
2
2
randomly selected team typically varies by about 1.6 goals from the mean, 2.851.
3. A life insurance company sells a term insurance policy to a 21-year-old male that pays $100,000 if the insured dies within the next 5
years. The probability that a randomly chosen male will die each year can be found in mortality tables. The company collects a
premium of $250 each year as payment for the insurance. The amount Y that the company earns on this policy is $250 per year, less
the $100,000 that it must pay if the insured dies. Here is a partially completed table that shows information about risk of mortality and
the values of Y = profit earned by the company:
a) Fill in the missing values of Y.
b) Fill in the missing probability.
Death age
Profit
Probability
21
–$99, 750
0.00183
22
–$99,500
0.00186
23
–$99,250
0.00189
24
–$99,000
0.00191
25
–$98,750
0.00193
26/more
$1,250
0.99058
c) Calculate the mean Y . Interpret this value in context.
Y  99750  0.00183  ...  1250  0.99058  $303.35. If the company insures many, many people, they will make about $303.35 per
life insurance policy, on average.
d) Calculate the mean  Y . Interpret this value in context.
Y 
 99750  303.35  0.00183  ...  1250  303.35  0.99058   $9707.57.
2
The amount of money the insurance company will
make on a randomly selected insurance policy will typically differ from the mean (303.34) by about $9707.57.
4. A study of 12,000 able-bodied male students at the University of Illinois found that their times for the mile run were approximately
Normal with mean 7.11 minutes and standard deviation 0.74 minute. Choose a student at random from this group and call his time for
the mile Y. Find P(Y < 6) and interpret the result.
The time Y of a randomly chosen student has the N(7.11, 0.74) distribution. We want to find
6  7.11 

P Y  6   P  z 
  P  z  1.50   0.0668.
0.74 

Using technology: normalcdf (lower: −1000, upper:6, μ:7.11, σ:0.74) = 0.0668.
There is about a 7% chance that this student will run the mile in under 6 minutes.
PICTURE!!!!