smith is a weld inspector at a ship yard. he knows from keeping track of good and substandard welds that for the afternoon shift 5% of all welds done will be substandard. If smith checks 300 of the 7500 welds completed that shift, what is the probability that he will find less than 20 substandard welds?

To solve this problem, we can use the binomial distribution. The binomial distribution is used to calculate the probability of getting a certain number of successes in a fixed number of independent Bernoulli trials.

In this case, the Bernoulli trial is each weld being either substandard or good, and the success is a substandard weld. The probability of success in one trial is given as 5%, which is 0.05.

First, we need to calculate the expected number of substandard welds. Since there are 7500 welds completed, and 5% of them are substandard, the expected number of substandard welds is:

Expected substandard welds = 7500 * 0.05 = 375

Next, we can use the binomial distribution formula to calculate the probability of finding less than 20 substandard welds when checking 300 welds:

P(X < 20) = Σ (nCr) * p^r * (1-p)^(n-r)

Where:
- n is the number of trials (300 in this case)
- r is the number of successes (0 to 19 in this case)
- nCr is the number of combinations of n items taken r at a time, calculated as n! / (r! * (n-r)!)
- p is the probability of success (0.05)
- (1-p) is the probability of failure (1 - 0.05 = 0.95)

We need to calculate this probability for each value of r from 0 to 19, and then sum them up.

P(X < 20) = Σ (300Cr) * 0.05^r * 0.95^(300-r) for r = 0 to 19

To calculate this probability, you can use statistical software, a calculator with binomial distribution function, or refer to pre-calculated tables for binomial probabilities.

Note: Since the upper limit is 19 (less than 20), you don't need to sum the probabilities for r > 19.

I hope this explanation helps you understand how to solve the problem.

To find the probability that Smith will find less than 20 substandard welds, we can use the binomial distribution formula.

The formula for the probability of finding exactly X substandard welds out of N welds, given a probability p of finding a substandard weld, is:

P(X) = (N choose X) * p^X * (1-p)^(N-X)

In this case, N is the number of welds Smith checked (300), X is the number of substandard welds (less than 20), and p is the probability of finding a substandard weld (5% or 0.05).

We need to calculate the probability for each possible number of substandard welds that is less than 20 and then sum them up.

P(X < 20) = P(X=0) + P(X=1) + P(X=2) + ... + P(X=19)

Let's calculate each term and then add them up:

P(X=0) = (300 choose 0) * 0.05^0 * (1-0.05)^(300-0)
= 1 * 1 * 0.95^300

P(X=1) = (300 choose 1) * 0.05^1 * (1-0.05)^(300-1)

Continuing this pattern, we calculate P(X) for X = 2, 3, ..., 19.

Finally, we add up all the individual probabilities:

P(X < 20) = P(X=0) + P(X=1) + P(X=2) + ... + P(X=19)

Note: We can use software or a calculator to calculate the individual probabilities and add them up.

We construct a binomial distribution.

The probability of "success" (a substandard rod) is .05. We have 300 trials.

B(.05, 300)
P(X < 20) = .881