Assume a binomial probability distribution has p = .60 and n = 200.
c. What is the probability of 100 to 110 successes (to 4 decimals)?
d. What is the probability of 130 or more successes (to 4 decimals)?
c. z = (100-120)/(200sqrt(120*.4)
z = (110-120)/200/sqrt(120*.4))
d. z = (130-120)/200/sqrt((120*.4))
Use Z table
To find the probabilities, we can use the binomial probability formula:
P(x) = nCx * p^x * (1-p)^(n-x)
Where:
- P(x) is the probability of getting x successes
- n is the total number of trials
- x is the number of successes
- p is the probability of success in a single trial
- nCx is the combination formula (n choose x), which calculates the number of ways to choose x successes out of n trials
c. To find the probability of 100 to 110 successes, we need to calculate the sum of probabilities from 100 to 110.
P(100 to 110) = P(100) + P(101) + P(102) + ... + P(110)
P(x) = nCx * p^x * (1-p)^(n-x)
Substituting the given values:
n = 200
p = 0.60
P(x) = 200Cx * 0.60^x * (1-0.60)^(200-x)
Now, we need to calculate P(x) for each individual value from 100 to 110 and sum them up.
d. To find the probability of 130 or more successes, we need to calculate the sum of probabilities from 130 to the maximum value of n (200 in this case).
P(130 or more) = P(130) + P(131) + P(132) + ... + P(n)
Following the same steps as above, calculate P(x) for each individual value from 130 to 200 and sum them up.