Question 1 of 20
0.0/ 1.0 Points
If random variable X has a binomial distribution with n =10 and P(success) =p =0.2, find the
probability that X is less than 6. (That is, find P(X<6) Answer: (round to 4 decimal places)

Question 2 of 20
0.0/ 1.0 Points
Approximately 8% of all people have blue eyes. Out of a random sample of 20 people, what is

Question 3 of 20
1.0/ 1.0 Points
It is known that 40% of adult workers have a high school diploma. If a random sample of 10
adult workers is selected, what is the expected number of adult workers with a high school

Question 4 of 20
1.0/ 1.0 Points
Approximately 10% of all people are left-handed. If 200 people are randomly selected, what is
the expected number of left-handed people? Round to the whole number. Do not use decimals.

Answer Key:20
Feedback:
200*.10
Question 5 of 20
0.0/ 1.0 Points
If random variable X has a binomial distribution with n=15 and P(success) =p= 0.6, find the

Part 2 of 6 - Contingency Table Knowledge Check Practice
1.0/ 1.0 Points
Question 6 of 20
1.0/ 1.0 Points
The table of data obtained from shows hit information
for four well known baseball players. Suppose that one hit from the table is randomly selected.
NAME
Single
Double
Triple
Home Run
TOTAL HITS
Babe Ruth
1,517
506
136
714
2,873
Jackie Robinson
1,054
273
54
137
1,518
Ty Cobb
3,603
174
295
114
4,189
Hank Aaron
2,294
624
98
755
3,771
TOTALS
8,468
1,577
583
1,720
12,351
Find
P
(hit was a Home Run|The hit was made by Ty Cobb ).

0.027
C.
0.009
D.
0.973
Answer Key:B
Feedback:114/4189
Part 3 of 6 - Counting Principle Knowledge Check Practice
0.0/ 3.0 Points
Question 7 of 20
0.0/ 1.0 Points
The number of M&M’s for each color found in a case were recorded in the table below.
Blue = 481, Brown = 371, Green = 483, Orange = 544, Red = 372, Yellow = 369
Total = 2620

Question 8 of 20
0.0/ 1.0 Points
Find the probability of rolling a sum of two dice that is a 7 or a 12. Round answer to 4 decimal
places. Answer:

Feedback:
Dice outcomes
12
3
4
5
6
1
(1, 1)
(1,
2)
(1, 3)(1, 4)(1, 5)(1, 6)
2
(2, 1)
(2,
2)
(2, 3)(2, 4)(2, 5)(2, 6)
3
(3, 1)
(3,
2)
(3, 3)(3, 4)(3, 5)(3, 6)
4
(4, 1)
(4,
2)
(4, 3)(4, 4)(4, 5)(4, 6)
5
(5, 1)
(5,
2)
(5, 3)(5, 4)(5, 5)(5, 6)
6
(6, 1)
(6,
2)
(6, 3)(6, 4)(6, 5)(6, 6)
Only 7 ways to sum to exactly 7 or exactly 12
7/36