Question

bob owns a shoe shop and he is planning to buy 500 pairs of shoes to sell next summer. the manufacturer claims that it averages only 6 pairs of inferior - quality shoes for every 10,000 pairs of shoes it produces. what is the probability that his order will contain one or more pairs of inferior - quality shoes?____% (round your answer to the nearest hundredth)

Explanation:

Step1: Calculate the probability of a single pair being inferior

The probability \(p\) of a pair of shoes being inferior is \(p=\frac{6}{10000}=0.0006\). The number of trials \(n = 500\).

Step2: Use the Poisson approximation

Since \(n\) is large (\(n = 500\)) and \(p\) is small (\(p=0.0006\)), we can use the Poisson approximation \(\lambda=np\). So \(\lambda=500\times0.0006 = 0.3\).
The probability mass function of a Poisson distribution is \(P(X = k)=\frac{e^{-\lambda}\lambda^{k}}{k!}\), where \(X\) is the random - variable representing the number of inferior - quality shoes.
The probability that the order contains zero inferior - quality shoes is \(P(X = 0)=\frac{e^{- 0.3}(0.3)^{0}}{0!}\). Since \(0!=1\) and \((0.3)^{0}=1\), then \(P(X = 0)=e^{-0.3}\approx0.7408\).

Step3: Calculate the probability of at least one inferior - quality shoe

The probability that \(X\geq1\) is \(P(X\geq1)=1 - P(X = 0)\).
Substitute \(P(X = 0)=e^{-0.3}\) into the formula: \(P(X\geq1)=1 - e^{-0.3}\).
Using a calculator, \(e^{-0.3}\approx0.7408\), so \(P(X\geq1)=1 - 0.7408=0.2592\)

Answer:

\(25.92\)