Plz help me i need to pass I've been trying this lesson all day and I can't figure this out
2 answers:
9514 1404 393
Answer:
see the attachment
Step-by-step explanation:
The problem statement tells you where you're starting (6·0 = 0) and where you want to end up (6·-10 = -60). These equations are the first and last steps in the right-hand column.
You should be able to recognize the step that was performed to get from one line to the next. The description of that is what goes in the left-hand column.
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Answer:
The correct option is 1 - [(0.8)¹⁰+10*0.2*(0.8)⁹]= 0.6242
Step-by-step explanation:
Hello!
Given the distribution of probabilities for blood types for African-Americans:
O: 0.4
A: 0.2
B: 0.32
AB: 0.08
A random sample of 10 African-American is chosen, what is the probability that 2 or more of them have Type A blood?
Let X represent "Number of African-Americans with Type A blood in a sample of 10.
Then you have two possible outcomes,
"Success" the person selected has Type A blood, with an associated probability p= 0.2
"Failure" the selected person doesn't have Type A blood, with an associated probability q= 0.8
(You can calculate it as "1-p" or adding all associated probabilities of the remaining blood types: 0.4+0.32+0.08)
Considering, that there is a fixed number of trials n=10, with only two possible outcomes: success and failure. Each experimental unit is independent of the rest and the probability of success remains constant p=0.2, you can say that this variable has a Binomial distribution:
X~Bi(n;p)
You can symbolize the asked probability as:
P(X≥2)
This expression includes the probabilities: X=2, X=3, X=4, X=5, X=6, X=7, X=8, X=9, X=10
And it's equal to
1 - P(X<2)
Where only the probabilities of X=0 and X=1 are included.
There are two ways of calculating this probability:
1) Using the formula:
With this formula, you can calculate the point probability for each value of X=x₀ ∀ x₀=1, 2, 3, 4, 5, 6, 7, 8, 9, 10
So to reach the asked probability you can:
a) Calculate all probabilities included in the expression and add them:
P(X≥2)= P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) + P(X=8) + P(X=9) + X=10
b) Use the complement rule and calculate only two probabilities:
1 - P(X<2)= 1 - [P(X=0)+P(X=1)]
2) Using the tables of the binomial distribution.
These tables have the cumulative probabilities listed for n: P(X≤x₀)
Using the number of trials, the probability of success, and the expected value of X you can directly attain the corresponding cumulative probability without making any calculations.
>Since you are allowed to use the complement rule I'll show you how to calculate the probability using the formula:
P(X≥2) = 1 - P(X<2)= 1 - [P(X=0)+P(X=1)] ⇒
⇒ 1 - (0.1074+0.2684)= 0.6242
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Using the table:
P(X≥2) = 1 - P(X<2)= 1 - P(X≤1)
You look in the corresponding table of n=10 p=0.2 for P(X≤1)= 0.3758
1 - P(X≤1)= 1 - 0.3758= 0.6242
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Full text in attachment.
I hope it helps!
