Questions: Find the probability that the claim will be rejected, assuming that the manufacturer's claim is true. Use the normal distribution to approximate the binomial distribution if possible. The probability is [Round to four decimal places as needed.]

Solution Steps

For a binomial distribution with parameters \( n = 100 \) and \( p = 0.02 \), we calculate the mean \( \mu \) and standard deviation \( \sigma \) as follows:

\[ \mu = n \cdot p = 100 \cdot 0.02 = 2.0 \]

The variance \( \sigma^2 \) is given by:

\[ \sigma^2 = n \cdot p \cdot q = 100 \cdot 0.02 \cdot (1 - 0.02) = 100 \cdot 0.02 \cdot 0.98 = 1.96 \]

Thus, the standard deviation \( \sigma \) is:

\[ \sigma = \sqrt{1.96} \approx 1.4 \]

To find the probability that the claim will be rejected, we need to determine the probability that the number of successes \( X \) is greater than or equal to 5. This can be approximated using the normal distribution:

\[ P(X \geq 5) = P\left(Z \geq \frac{5 - \mu}{\sigma}\right) \]

Calculating the Z-score:

\[ Z_{start} = \frac{5 - 2.0}{1.4} \approx 2.142857 \]

Since we are looking for \( P(X \geq 5) \), we need to find:

\[ P(Z \geq 2.142857) = 1 - P(Z < 2.142857) \]

Using the properties of the standard normal distribution, we find:

\[ P(Z < 2.142857) \approx 0.9836 \]

Thus,

\[ P(Z \geq 2.142857) = 1 - 0.9836 = 0.0164 \]

However, since we are interested in the probability of rejection, we also consider the upper tail:

\[ P(Z \geq 21.4286) \approx 0.0 \]

Final Answer