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The probability that it will take no more than 2 seconds to scan one customer's bags is approximately 32.97%.
Solución
To solve this problem, we need to use the properties of the exponential distribution. The exponential distribution is defined by its probability density function (PDF):
\[
f(t; \lambda) = \lambda e^{-\lambda t}
\]
where \( \lambda \) is the rate parameter, which is the reciprocal of the mean. Given that the mean time to scan one customer's bags is 5 seconds, we can calculate \( \lambda \) as follows:
\[
\lambda = \frac{1}{\text{mean}} = \frac{1}{5} \text{ seconds}^{-1}
\]
Next, we want to find the probability that the time \( T \) to scan a customer's bags is less than or equal to 2 seconds. This is given by the cumulative distribution function (CDF) of the exponential distribution:
\[
P(T \leq t) = 1 - e^{-\lambda t}
\]
Substituting \( \lambda \) and \( t = 2 \) seconds into the CDF:
\[
P(T \leq 2) = 1 - e^{-\frac{1}{5} \cdot 2}
\]
Now, let's calculate this probability.
1. Calculate \( -\frac{1}{5} \cdot 2 \).
2. Calculate \( e^{-\frac{2}{5}} \).
3. Finally, compute \( 1 - e^{-\frac{2}{5}} \).
Let's perform these calculations step by step.
First, we calculate \( -\frac{2}{5} \):
\[
-\frac{2}{5} = -0.4
\]
Next, we calculate \( e^{-0.4} \):
Now, I will calculate \( e^{-0.4} \) and then find \( 1 - e^{-0.4} \).
Simplify the expression by following steps:
- step0: Simplify:
\(e^{-0.4}\)
- step1: Convert the expressions:
\(e^{-\frac{2}{5}}\)
- step2: Express with a positive exponent:
\(\frac{1}{e^{\frac{2}{5}}}\)
- step3: Transform the expression:
\(\frac{1}{\sqrt[5]{e^{2}}}\)
- step4: Multiply by the Conjugate:
\(\frac{\sqrt[5]{e^{3}}}{\sqrt[5]{e^{2}}\times \sqrt[5]{e^{3}}}\)
- step5: Multiply the numbers:
\(\frac{\sqrt[5]{e^{3}}}{e}\)
Calculate or simplify the expression \( 1 - e^{-0.4} \).
Calculate the value by following steps:
- step0: Calculate:
\(1-e^{-0.4}\)
- step1: Convert the expressions:
\(1-e^{-\frac{2}{5}}\)
- step2: Evaluate the power:
\(1-\frac{1}{e^{\frac{2}{5}}}\)
- step3: Calculate:
\(1-\frac{e^{\frac{3}{5}}}{e}\)
- step4: Reduce fractions to a common denominator:
\(\frac{e}{e}-\frac{e^{\frac{3}{5}}}{e}\)
- step5: Transform the expression:
\(\frac{e-e^{\frac{3}{5}}}{e}\)
- step6: Simplify:
\(\frac{e+\sqrt[5]{e^{3}}}{e}\)
The calculation of \( e^{-0.4} \) gives us approximately \( 0.67032 \).
Now, we can find the probability that it will take no more than 2 seconds:
\[
P(T \leq 2) = 1 - e^{-0.4} \approx 1 - 0.67032 \approx 0.32968
\]
Thus, the likelihood that it will take no more than 2 seconds to scan one customer's bags is approximately \( 0.32968 \) or \( 32.97\% \).
Respondido por UpStudy AI y revisado por un tutor profesional
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