Question

a survey finds customers are overcharged by price scanning systems, on average, on 1.75% of items. suppose a customer purchases 17 items. find the following probability.
the customer is overcharged on at least 3 items.
the probability that the customer is overcharged on at least 3 items is
(round to four decimal places as needed.)

Explanation:

Step1: Identify the binomial distribution parameters

Let \(n = 17\) (number of trials, i.e., number of items), \(p=0.0175\) (probability of over - charging on an item), and \(q = 1 - p=1 - 0.0175 = 0.9825\). The probability mass function of a binomial distribution is \(P(X = k)=C(n,k)\times p^{k}\times q^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\).

The probability that the customer is overcharged on at least 3 items is \(P(X\geq3)=1 - P(X = 0)-P(X = 1)-P(X = 2)\)

Step2: Calculate \(P(X = 0)\)

$$
LATEXBLOCK0
$$

Using a calculator, \(0.9825^{17}\approx0.7397\)

Step3: Calculate \(P(X = 1)\)

$$
LATEXBLOCK1
$$

\(17\times0.0175 = 0.2975\), and \(0.9825^{16}\approx0.7529\), so \(P(X = 1)\approx0.2975\times0.7529\approx0.2240\)

Step4: Calculate \(P(X = 2)\)

$$
LATEXBLOCK2
$$

\(136\times0.00030625 = 0.04165\), and \(0.9825^{15}\approx0.7665\), so \(P(X = 2)\approx0.04165\times0.7665\approx0.0319\)

Step5: Calculate \(P(X\geq3)\)

$$
LATEXBLOCK3
$$

Answer:

\(0.0044\)