Consider a binomial experiment with 20 trials and probability 0.45 on a single trial. Use the normal distribution to find the probability of exactly 10 successes. Round your answer to the thousandths place.
Try the binomial probability function:
P(x) = (nCx)(p^x)[q^(n-x)]
x = 10
n = 20
p = 0.45
q = 1 - p = 1 - 0.45 = 0.55
Substitute the values and calculate.
An easier way would be to use a binomial probability table with the values stated above.
I hope this will help.
To find the probability of exactly 10 successes in a binomial experiment with 20 trials and a probability of success of 0.45, we can use the normal approximation to the binomial distribution.
The mean (μ) of a binomial distribution is given by:
μ = n * p
where n is the number of trials and p is the probability of success.
In this case, μ = 20 * 0.45 = 9.
The standard deviation (σ) of a binomial distribution is given by:
σ = sqrt(n * p * (1 - p))
In this case, σ = sqrt(20 * 0.45 * 0.55) ≈ 2.19089.
To use the normal distribution, we need to standardize the value of 10, which is the number of successes we want to find the probability for.
Z = (X - μ) / σ
where Z is the standard score, X is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.
In this case, Z = (10 - 9) / 2.19089 ≈ 0.456.
Now, we can use a standard normal distribution table or a calculator with the cumulative distribution function (CDF) to find the probability associated with Z = 0.456.
The probability of exactly 10 successes in this binomial experiment, using the normal approximation, is the same as the probability associated with Z ≤ 0.456.
Using a standard normal distribution table or a calculator, the probability associated with Z ≤ 0.456 is approximately 0.675.
Therefore, the probability of exactly 10 successes in this binomial experiment, using the normal distribution approximation, is approximately 0.675 (rounded to the thousandths place).
To solve this problem using the normal distribution, we need to check if the conditions for using the normal approximation to the binomial distribution are met. The conditions are:
1. The number of trials, denoted by n, is large. In this case, n = 20, which is considered large.
2. The probability of success on a single trial, denoted by p, is not too close to 0 or 1. In this case, p = 0.45, which is not too close to 0 or 1.
Since the conditions are met, we can approximate the binomial distribution with a normal distribution.
To find the probability of exactly 10 successes, we can use the formula for the mean and standard deviation of a binomial distribution:
Mean (μ) = n * p = 20 * 0.45 = 9
Standard Deviation (σ) = √(n * p * (1 - p)) = √(20 * 0.45 * (1 - 0.45)) = √(4.95) ≈ 2.22
Next, we convert the problem to a standard normal distribution by standardizing the value X = 10:
Z = (X - μ) / σ
Z = (10 - 9) / 2.22 ≈ 0.45
Now, we need to find the probability of Z being less than or equal to 0.45. We can use a standard normal distribution table or a calculator to find this probability.
Using a standard normal distribution table, we find that the probability corresponding to 0.45 is approximately 0.673.
Rounding this answer to the thousandths place, the probability of exactly 10 successes in a binomial experiment with 20 trials and a probability of success of 0.45 is approximately 0.673.