When a truckload of apples arrives at a packing plant, a random sample of 250 is selected and examined for bruises, and other defects. The whole truckload will be rejected if more than 5% of the sample is unsatisfactory. Although it is not known, in fact 8% of the apples on the truck do not meet the desired standard. What is the probability the shipment will be accepted anyways.
To find the probability that the shipment will be accepted, we need to determine the probability that the sample contains 5% or fewer unsatisfactory apples.
Given that 8% of the apples on the truck do not meet the desired standard, we can assume that the apples are independently and identically distributed (i.i.d). Additionally, we know that the sample size is 250.
To solve this problem, we can use the binomial distribution. The binomial distribution is used for discrete, independent trials with two possible outcomes (success or failure) and a fixed number of trials.
In this case, a success would be an apple that meets the desired standard, and a failure would be an apple that does not meet the desired standard. The probability of success is 1 - 0.08 (since 8% of the apples do not meet the standard), and the probability of failure is 0.08.
Let's denote X as the number of unsatisfactory apples in the sample. We want to calculate P(X ≤ 0.05*250), which is the probability of having 5% or fewer unsatisfactory apples.
Using the binomial distribution, the probability mass function (PMF) for X is given by:
P(X=k) = C(n, k) * p^k * (1-p)^(n-k)
where n is the sample size, p is the probability of success, and C(n, k) represents the binomial coefficient.
In our case, we have n = 250, k = 0.05*250 = 12.5. Since we cannot have a fractional number of unsatisfactory apples, we need to consider the probability of having 12 or fewer unsatisfactory apples:
P(X ≤ 12) = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 12)
Calculating this sum would be an exhaustive process, but luckily, we can use a normal approximation to estimate the probability.
For a large sample size (n) and a low probability of success (p), the binomial distribution can be approximated by a normal distribution with mean μ = n*p and standard deviation σ = √(n*p*(1-p)).
In our case, μ = 250*0.08 = 20 and σ = √(250*0.08*0.92) ≈ 4.08.
Now, we can use the normal distribution to estimate the probability:
P(X ≤ 12) ≈ P(Z ≤ (12 - 20)/4.08)
Using a standard normal table or calculator, we find that P(Z ≤ -1.96) ≈ 0.025.
Therefore, the probability that the shipment will be accepted is approximately 0.025 or 2.5%.