A manufacturer produces a large number of toasters. From past experience, the manufacturer knows that approximately 2% are defective. In a quality control procedure, we randomly select 20 toasters for testing.
(a) Determine the probability that exactly one of the toasters is defective.
(b) Find the probability that at most two of the toasters are defective.
(c) Let X be the number of defective toasters in the sample of 20. Find the mean and standard deviation of X.
The correct answer and explanation is :
We are given a binomial distribution problem where:
- Each toaster has a probability $p = 0.02$ of being defective.
- The number of trials (toasters tested) is $n = 20$.
- Let $X$ be the number of defective toasters in the sample.
(a) Probability that exactly one toaster is defective
We use the binomial probability formula:
$$
P(X = k) = \binom{n}{k} p^k (1 – p)^{n – k}
$$
Substitute $n = 20$, $p = 0.02$, $k = 1$:
$$
P(X = 1) = \binom{20}{1} (0.02)^1 (0.98)^{19} \approx 20 \times 0.02 \times 0.698 \approx 0.279
$$
✅ Answer: $\boxed{P(X = 1) \approx 0.279}$
(b) Probability that at most two toasters are defective
This is the cumulative probability:
$$
P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)
$$
- $P(X = 0) = \binom{20}{0} (0.02)^0 (0.98)^{20} = (0.98)^{20} \approx 0.667$
- $P(X = 1) \approx 0.279$ (from part a)
- $P(X = 2) = \binom{20}{2} (0.02)^2 (0.98)^{18} \approx 190 \times 0.0004 \times 0.713 \approx 0.054$
$$
P(X \leq 2) \approx 0.667 + 0.279 + 0.054 = \boxed{1.000} \ (\text{rounded, more precisely } \approx 0.9999)
$$
✅ Answer: $\boxed{P(X \leq 2) \approx 0.999}$
(c) Mean and standard deviation of $X$
For a binomial distribution:
- Mean: $\mu = np = 20 \times 0.02 = \boxed{0.4}$
- Standard deviation: $\sigma = \sqrt{np(1 – p)} = \sqrt{20 \times 0.02 \times 0.98} \approx \sqrt{0.392} \approx \boxed{0.626}$
✍️ Explanation (300 words):
This is a classic example of a binomial probability problem, which models a fixed number of independent trials (here, testing 20 toasters), each with two outcomes: defective or not defective. The probability of a toaster being defective is constant at 2%.
In part (a), we calculate the probability of finding exactly one defective toaster using the binomial formula. The binomial coefficient $\binom{20}{1}$ gives the number of ways to choose 1 defective toaster out of 20. We then multiply this by the probability of that single event occurring (0.02) and the probability of the other 19 toasters being non-defective (0.98), raised to the 19th power. The resulting probability is approximately 0.279.
Part (b) asks for the cumulative probability of finding 0, 1, or 2 defective toasters. We compute each of these binomial probabilities and sum them. This is particularly useful in quality control, as it tells us how likely it is to encounter a small number of defective items. The total probability is close to 1, indicating that finding more than two defective toasters is very unlikely given the 2% defect rate.
In part (c), we determine the mean and standard deviation of the number of defective toasters. These give insight into the expected behavior and variability of defects in samples of 20 toasters. On average, we expect 0.4 defective toasters per batch. The standard deviation tells us that in most batches, the number of defects will vary by about 0.63 around the mean.
This statistical analysis helps manufacturers understand defect patterns and maintain quality control by setting expectations and thresholds for acceptable defect rates.