An unnoticed mechanical failure has caused one-fourth of a machine shop’s production of 10000 pistol firing pins to be defective. A random sample of 25 firing pins was drawn from the population.
a.) Explain why this random variable has a binomial distribution? (2 points)
b.) What are the mean and the standard deviation of the random variable? (2 points)
c.) Find P(X > 4). Use the binomcdf function on your calculator. (2 points)
d.) Is it appropriate to use the normal approximation to the binomial for this problem? Support your answer with numbers and a test. (3 points)
e.) Using the normal approximation with the continuity correction, calculate the probability that you’ll observe more than 4 defective firing pins in the random sample of 25. Indicate the X-values and the z-scores involved. (3 points)
f.) How does your answer compare with the one you have in part c.)? (2 points)
g.) How do you account for the difference in parts c.) and e.)? (2 points )
h.) Sketch the probability distribution and the area you found using the normal approximation with the continuity correction. (2 points)
a.) This random variable has a binomial distribution because it satisfies the four conditions for a binomial experiment:
1) There are a fixed number of trials: drawing 25 firing pins.
2) Each trial has two possible outcomes: defective or non-defective.
3) The probability of success (defective firing pin) is constant for each trial: one-fourth of the production is defective.
4) The trials are independent: the outcome of one trial does not affect the outcome of the next trial.
b.) The mean 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 = 25 * (1/4) = 6.25.
The standard deviation of a binomial distribution is given by σ = sqrt(n * p * (1 - p)). In this case, σ = sqrt(25 * (1/4) * (3/4)) ≈ 2.5.
c.) To find P(X > 4), we can use the binomcdf function on the calculator, which calculates the cumulative probability up to a given value. In this case, we want to find the probability that the number of defective firing pins is greater than 4.
d.) To determine if it is appropriate to use the normal approximation to the binomial, we can check if the conditions for the normal approximation are met. The conditions are:
1) n * p ≥ 10: 25 * (1/4) = 6.25 ≥ 10 (Not met, since it is less than 10)
2) n * (1 - p) ≥ 10: 25 * (3/4) = 18.75 ≥ 10 (Met)
Since the first condition is not met, it is not appropriate to use the normal approximation for this problem.
e.) Since we are not using the normal approximation, we cannot calculate the probability using z-scores. Instead, we can calculate the probability using the binomial distribution directly.
f.) The answer obtained in part c.) was based on using the binomial distribution and the binomcdf function, while the answer obtained in part e.) was based on using the normal approximation with the continuity correction.
g.) The difference in parts c.) and e.) can be accounted for by the approximation used. The normal approximation assumes a continuous distribution and uses the approximation of the binomial distribution to a normal distribution. The normal approximation is only an approximation and may not be as accurate as using the binomial distribution directly.
h.) The probability distribution for this problem would be a binomial distribution with the number of defective firing pins on the x-axis and the probability on the y-axis. The area found using the normal approximation with the continuity correction would be represented by shaded area under the normal curve.
a.) This random variable has a binomial distribution because it satisfies the following conditions:
1. The number of trials is fixed at 25 (random sample size).
2. Each trial is independent of the others (each firing pin is inspected separately).
3. There are only two possible outcomes for each trial (defective or not defective).
4. The probability of success (firing pin being defective) is constant for each trial at 1/4 (one-fourth of the production is defective).
b.) The mean 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, n = 25 and p = 1/4. So, the mean is μ = 25 * 1/4 = 6.25.
The standard deviation of a binomial distribution is given by σ = √(n * p * (1-p)). Plugging in the values, we get σ = √(25 * 1/4 * 3/4) ≈ 2.50.
c.) To find P(X > 4), we can use the binomcdf function on a calculator. binomcdf(n, p, k) gives us the cumulative probability of obtaining k or fewer successes in n trials.
For this problem, n = 25, p = 1/4, and we want to find P(X > 4). Using the binomcdf function, we can calculate P(X ≤ 4) as 1 - binomcdf(25, 1/4, 4). Subtracting this value from 1 will give us P(X > 4).
d.) To determine if it is appropriate to use the normal approximation to the binomial, we need to check if the conditions are met. The conditions for using the normal approximation to the binomial are:
1. n * p >= 5: n = 25, p = 1/4 -> 25 * 1/4 = 6.25. This condition is met.
2. n * (1-p) >= 5: n = 25, p = 1/4 -> 25 * 3/4 = 18.75. This condition is met.
Additionally, for the normal approximation, np and n(1-p), the mean and variance of the binomial distribution respectively, should both be sufficiently large (usually > 10). np = 25 * 1/4 = 6.25, n(1-p) = 25 * 3/4 = 18.75. Both are greater than 10, so the conditions for the normal approximation are met.
e.) Using the normal approximation with the continuity correction, we can calculate the probability that we observe more than 4 defective firing pins. To do this, we calculate the z-score corresponding to 4 and find the probability of obtaining a value greater than this z-score.
The formula for z-score is: z = (x - μ) / σ, where x is the number of successes, μ is the mean, and σ is the standard deviation. In this case, x = 4, μ = 6.25, and σ = 2.50. Calculating the z-score: z = (4 - 6.25) / 2.50 = -0.90.
Using a standard normal table or a calculator, we can find the cumulative probability for a z-score of -0.90. Let's call this probability P(Z < -0.90). To find the probability of observing more than 4 defective firing pins, we subtract this cumulative probability from 1 because we want P(Z > -0.90).
f.) In part c, we used the binomcdf function to find P(X > 4). In part e, using the normal approximation with continuity correction, we found P(Z > -0.90), where Z is the standard normal random variable.
g.) The difference between parts c and e is due to the different approaches used to calculate the probabilities. In part c, we used the binomial distribution directly, while in part e, we used the normal approximation with the continuity correction. The binomial distribution gives an exact probability, while the normal approximation provides an estimate.
h.) Unfortunately, as a text-based AI, I am unable to sketch the probability distribution. However, I can describe it for you. The probability distribution for this binomial random variable would be a discrete distribution with possible values from 0 to 25 on the x-axis (representing the number of defective firing pins in the sample) and the corresponding probabilities on the y-axis. The area of interest calculated using the normal approximation with the continuity correction would be to the right of the x-value 4 on the distribution.