1. **Problem Statement:** A manufacturer has 100 memory chips with 4% defective. A sample of 20 chips is selected. Let $X$ be the number of faulty chips in the sample. 2. **Binomial Distribution Formula:** The probability mass function (pmf) for a binomial random variable $X$ with parameters $n$ (number of trials) and $p$ (probability of success) is: $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $k = 0,1,2,...,n$. 3. **Given:** - $n = 20$ (sample size) - $p = 0.04$ (probability a chip is defective) --- **a) Probability mass function of $X$:** $$P(X = k) = \binom{20}{k} (0.04)^k (0.96)^{20-k}$$ for $k = 0,1,2,...,20$. --- **b) Probability exactly 15 computers are faulty:** Calculate $P(X=15)$: $$P(X=15) = \binom{20}{15} (0.04)^{15} (0.96)^5$$ This value is extremely small because $p=0.04$ is low and 15 faults is very unlikely. --- **c) Probability at least 3 but not more than 7 faulty:** Calculate $P(3 \leq X \leq 7)$: $$P(3 \leq X \leq 7) = \sum_{k=3}^7 \binom{20}{k} (0.04)^k (0.96)^{20-k}$$ This sums the probabilities for $k=3,4,5,6,7$. --- **d) Mean and standard deviation:** For binomial distribution: - Mean: $$\mu = np = 20 \times 0.04 = 0.8$$ - Variance: $$\sigma^2 = np(1-p) = 20 \times 0.04 \times 0.96 = 0.768$$ - Standard deviation: $$\sigma = \sqrt{0.768} \approx 0.876$$ --- **Summary:** - pmf: $$P(X=k) = \binom{20}{k} (0.04)^k (0.96)^{20-k}$$ - $P(X=15)$ is negligible. - $P(3 \leq X \leq 7)$ is sum of pmf from 3 to 7. - Mean $=0.8$, standard deviation $\approx 0.876$. This explains how to compute probabilities and statistics for the binomial distribution in this context.