Assume that about 45% of all U.S. adults try to pad their insurance claims. Suppose that you are the director of an insurance adjustment office. Your office has just received 110 insurance claims to be processed in the next few days. What is the probability that fewer than 45 of the claims have been padded?
Your values are the following:
p = .45, q = 1 - p = .55, x = 45, and n = 110
Use the normal approximation to the binomial distribution.
Find mean and standard deviation.
mean = np = (100)(.45) = ?
sd = √npq = √(100)(.45)(.55) = ?
Once you determine the mean and standard deviation, use z-scores and z-table to find probability:
z = (x - mean)/sd
Substitute the values into the formula. Once you have the z-score, determine your probability using a z-table.
I hope this will help get you started.
Correction:
Use 110 for n.
To solve this problem, we can use the binomial probability formula. The binomial probability formula calculates the probability of getting a specific number of successes (padded claims) in a fixed number of trials (total claims).
Let's break down the problem using the given information:
Probability of a padded claim: 45%
Total claims received: 110
The probability that fewer than 45 claims have been padded can be calculated by summing the probabilities of having 0, 1, 2, ..., 44 padded claims.
The formula for the binomial probability mass function is:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
where:
P(X = k) is the probability of having exactly k successes,
C(n, k) is the binomial coefficient, representing the number of ways to choose k successes from n trials (given by C(n, k) = n! / (k! * (n-k)!)),
p is the probability of success (padding an insurance claim),
k is the number of successes (padded claims), and
n is the number of trials (total claims received).
Using this formula, we can calculate the probability as follows:
To find the probability that fewer than 45 of the claims have been padded, we can use the binomial probability formula.
The binomial probability formula is given by:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
Where:
- P(X = k) represents the probability of getting exactly k successes
- n is the total number of trials or in this case, the number of insurance claims
- k is the number of successes (number of padded claims)
- p is the probability of success (probability of a claim being padded)
- C(n, k) is the number of possible combinations of n items taken k at a time
In this case:
- n = 110 (total number of insurance claims)
- k represents the number of padded claims
- p = 0.45 (probability of a claim being padded)
We need to calculate the probability for k = 0, 1, 2, ..., 44 and then add them up to find the probability that fewer than 45 claims have been padded.
Now, let's calculate the probability.