4. ( 3 points) If the time required to scan one customer's bags is exponentially distributed with mean of 5 seconds, then what is the likelihood that it will take no more than 2 seconds?

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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 $$ 32.97\% $$.
The probability that it will take no more than 2 seconds to scan one customer's bags is approximately 32.97%.

4. ( 3 points) If the time required to scan one customer's bags is exponentially distributed with mean of 5 seconds, then what is the likelihood that it will take no more than 2 seconds?

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