Answer:
The sum of the probabilities is greater than 100%; and the distribution is too uniform to be a normal distribution.
Step-by-step explanation:
The sum of the probabilities of a distribution should be 100%. When you add the probabilities of this distribution together, you have
22+24+21+26+28 = 46+21+26+28 = 67+26+28 = 93+28 = 121
This is more than 100%, which is a flaw with the results.
A normal distribution is a bell-shaped distribution. Graphing the probabilities for this distribution, we would have a bar up to 22; a bar to 24; a bar to 21; a bar to 26; and bar to 28.
The bars would not create a bell-shaped curve; thus this is not a normal distribution.
Answer:
a) 0.06 = 6% probability that a person has both type O blood and the Rh- factor.
b) 0.94 = 94% probability that a person does NOT have both type O blood and the Rh- factor.
Step-by-step explanation:
I am going to solve this question treating these events as Venn probabilities.
I am going to say that:
Event A: Person has type A blood.
Event B: Person has Rh- factor.
43% of people have type O blood
This means that 
15% of people have Rh- factor
This means that 
52% of people have type O or Rh- factor.
This means that 
a. Find the probability that a person has both type O blood and the Rh- factor.
This is

With what we have

0.06 = 6% probability that a person has both type O blood and the Rh- factor.
b. Find the probability that a person does NOT have both type O blood and the Rh- factor.
1 - 0.06 = 0.94
0.94 = 94% probability that a person does NOT have both type O blood and the Rh- factor.
Answer:
sorry
Step-by-step explanation:
I don't know what you want , do you want to know how to construct an inscribed angle or what
Answer:
Total cost of dinner = P + 0.15p = 23
Step-by-step explanation:
Consider the information provided.
The cost of dinner at a restaurant is $P.
The tip offered for the service was, $0.15p.
The total of the cost of dinner and the tip offered for the service is:
T = P + 0.15p = 23
The solution to this question is 20.6666666