It has been found that 2% of the tools produced by a certain machine are defective. What is the probability that in a shipment of 400 such tools, 3% or more will be defective? (Probability that the normal variate lies between 0 and 1.43 is 0.4236.)
0.0764
We use the normal approximation to the binomial. With n = 400 and p = 0.02 (defective rate):
Mean: \(\mu = np = 400 \times 0.02 = 8\)
Standard deviation: \(\sigma = \sqrt{npq} = \sqrt{400 \times 0.02 \times 0.98} = \sqrt{7.84} = 2.8\)
3% of 400 = 12 defective tools. The z-score is:
\(z = \dfrac{X - \mu}{\sigma} = \dfrac{12 - 8}{2.8} = 1.43\)
We need \(P(Z \geq 1.43)\). Using the given value \(P(0 \leq Z \leq 1.43) = 0.4236\):
\(P(Z \geq 1.43) = 0.5 - 0.4236 = \mathbf{0.0764}\)
Hence the required probability is 0.0764.
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