To solve this problem, we’ll use the normal approximation to the binomial distribution since the problem asks for it and also implies that it may be appropriate given certain conditions.

Step 1: Verify the conditions for normal approximation

For the normal approximation to apply, both $np$ and $n(1 - p)$ should be greater than 5, where:

• $n = 62$ (total number of products),
• $p = 0.78$ (probability of failure).

Calculating these values:

• $np = 62 \times 0.78 = 48.36$,
• $n(1 - p) = 62 \times 0.22 = 13.64$.

Since both values are greater than 5, the normal approximation is appropriate.

Step 2: Set up the normal approximation parameters

The binomial distribution $X \sim \text{Binomial}(n = 62, p = 0.78)$ can be approximated by a normal distribution with:

• Mean $\mu = np = 48.36$,
• Standard deviation $\sigma = \sqrt{np(1 - p)} = \sqrt{48.36 \times 0.22} \approx 3.2586$.

Step 3: Solve each part using the normal distribution

(a) Probability that 47 or more fail

We need $P(X \geq 47)$. For the continuity correction, this becomes $P(X \geq 46.5)$.

Using the normal approximation: $Z = \frac{46.5 - 48.36}{3.2586} \approx -0.5738$ Looking up this Z-score in the standard normal table, we get $P(Z \geq -0.5738) \approx 0.716$.

(b) Probability that 58 or fewer fail

We need $P(X \leq 58)$. For the continuity correction, this becomes $P(X \leq 58.5)$.

Using the normal approximation: $Z = \frac{58.5 - 48.36}{3.2586} \approx 3.116$ Looking up this Z-score, $P(Z \leq 3.116) \approx 0.9991$.

(c) Probability that 15 or more succeed

Since the probability of failure is $0.78$, the probability of success is $1 - 0.78 = 0.22$.

Define $Y$ as the number of successes, where $Y = 62 - X$. Thus, $Y \sim \text{Binomial}(n = 62, p = 0.22)$, with:

• Mean $\mu_Y = 62 \times 0.22 = 13.64$,
• Standard deviation $\sigma_Y = \sqrt{13.64 \times 0.78} \approx 3.2586$ (same as before).

We need $P(Y \geq 15)$, or equivalently $P(Y \geq 14.5)$ with the continuity correction.

Calculating the Z-score: $Z = \frac{14.5 - 13.64}{3.2586} \approx 0.2645$ Looking up this Z-score, $P(Z \geq 0.2645) \approx 0.395$.

(d) Probability that fewer than 10 succeed

This is $P(Y < 10)$, or equivalently $P(Y \leq 9.5)$ with the continuity correction.

Calculating the Z-score: $Z = \frac{9.5 - 13.64}{3.2586} \approx -1.271$ Looking up this Z-score, $P(Z \leq -1.271) \approx 0.1021$.

Summary of Answers

1. (a) $P(X \geq 47) \approx 0.716$
2. (b) $P(X \leq 58) \approx 0.9991$
3. (c) $P(Y \geq 15) \approx 0.395$
4. (d) $P(Y < 10) \approx 0.1021$

Would you like more details on any of these steps, or do you have further questions?

Follow-up Questions

1. How would the probabilities change if the failure rate were 70% instead of 78%?
2. Can we calculate the exact binomial probabilities without the normal approximation for these cases?
3. Why is the continuity correction necessary when using the normal approximation?
4. What is the impact of sample size $n$ on the accuracy of the normal approximation?
5. How would we find these probabilities if we used a statistical software instead of manual calculations?

Tip

For binomial probabilities involving large $n$ and extreme $p$, the normal approximation becomes more accurate if $np$ and $n(1-p)$ are both significantly greater than 5.