Suppose that a sample of n = 1,600 tires of the same type are obtained at random from an ongoing production process in which 8% of all such tires produced are defective. What is the probability that in such a sample 150 or fewer tires will be defective

Well, the probability of a tire being defective is 8%, so the probability of a tire not being defective is 92%.

To find the probability that in a sample of 1,600 tires, 150 or fewer will be defective, we need to add up the probabilities of having 0, 1, 2, 3, ..., 150 defective tires.

However, calculating all those probabilities individually can be a bit tedious, so let me tell you a joke to lift your spirits!

Why did the tire go to the doctor?

Because it was feeling a little flat!

Now, let's get back to the problem. To calculate the probability, we can use a binomial distribution. The formula is:

P(X ≤ k) = Σ (n C k) * p^k * q^(n-k)

Where:
- P(X ≤ k) is the probability of getting k or fewer defective tires in a sample of n tires.
- n is the number of trials (in this case, the number of tires we sample, which is 1,600).
- k is the number of successful events (defective tires) we're interested in (up to 150 in this case).
- (n C k) is the binomial coefficient, which represents the number of ways to choose k out of n items.
- p is the probability of success (a tire being defective), 8% in this case.
- q is the probability of failure (a tire not being defective), which is 92%.

So, using this formula, you can calculate the probability that in a sample of 1,600 tires, 150 or fewer will be defective. Just remember to include the binomial coefficient for each value of k. Good luck with your calculations!

To find the required probability, we can use the binomial distribution formula.

The binomial distribution formula is given by:

P(X = k) = (n C k) * p^k * (1 - p)^(n - k)

Where:
- P(X = k) is the probability of getting exactly k successes (defective tires in this case) in a sample of size n.
- (n C k) is the binomial coefficient, which represents the number of ways to choose k items from a set of n.
- p is the probability of success (probability of getting a defective tire in this case).
- (1 - p) is the probability of failure (probability of getting a non-defective tire in this case).

First, let's calculate the probability of getting a defective tire (p) and the probability of getting a non-defective tire (1 - p):

p = 8% = 0.08
q = 1 - p = 1 - 0.08 = 0.92

Now, we need to calculate the probability of having 150 or fewer defective tires in a sample of size 1600. This can be done by summing up the probabilities of having 0, 1, 2, ..., 150 defective tires.

P(X ≤ 150) = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 150)

We can use a calculator or software to calculate this sum, or we can approximate it using a normal distribution if n is large.

If you provide the level of approximation acceptable, I can proceed with the calculation accordingly.

To find the probability that in a sample of 1,600 tires, 150 or fewer will be defective, we can use the binomial distribution formula.

The binomial distribution formula is given by:

P(X = k) = (nCk) * p^k * q^(n-k)

Where:
- P(X = k) is the probability of getting exactly k successes (defective tires) in a sample of n tires.
- nCk represents the combination formula, also known as "n choose k," which calculates the number of ways to choose k items from a set of n items.
- p is the probability of success (probability of getting a defective tire).
- q is the probability of failure (1 - p), where p is the probability of success.

Here, n = 1,600 (sample size) and p = 0.08 (probability of a defective tire). We want to find the probability that 150 or fewer tires will be defective, so we need to calculate the probabilities for k = 0, 1, 2, ..., 150 and sum them up.

P(X <= 150) = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 150)

Let's calculate this probability step by step using the formula:

1. Calculate p (probability of success) and q (probability of failure):
p = 0.08
q = 1 - p = 1 - 0.08 = 0.92

2. Calculate the probability for each value of k from 0 to 150:
For k = 0:
P(X = 0) = (1600C0) * (0.08^0) * (0.92^1600)

For k = 1:
P(X = 1) = (1600C1) * (0.08^1) * (0.92^1599)

Continue this calculation for k = 2, 3, ..., 150.

3. Sum up the probabilities:
P(X <= 150) = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 150)

This will give you the probability that 150 or fewer tires will be defective in the sample of 1,600 tires.