Suppose that the amount of time a hospital patient must wait for a nurse's help is described by a continuous random variable with density function f(t) = e-t/3 where t≥ 0 is measured in minutes. (a) What is the probability that a patient must wait for more than 4 minutes? (b) A patient spends a week in the hospital and requests nurse assistance once each day. What is the probability that the nurse will take longer than 5 minutes to respond on (exactly) two occasions? (c) What is the probability that on at least one call out of seven, the nurse will take longer than 7 minutes to respond?

MATLAB: An Introduction with Applications
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Chapter1: Starting With Matlab
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The given problem illustrates a situation where the time a hospital patient must wait for a nurse's assistance is modeled by a continuous random variable with a specific density function. The function is given by:

\[ f(t) = \frac{1}{3} e^{-t/3} \]

for \( t \geq 0 \) (time is measured in minutes).

### Questions:

**(a)** What is the probability that a patient must wait for more than 4 minutes?

To find this, you would calculate the probability \( P(T > 4) \).

**(b)** A patient spends a week in the hospital and requests nurse assistance once each day. What is the probability that the nurse will take longer than 5 minutes to respond on (exactly) two occasions?

Here, we look at a binomial distribution scenario with 7 trials (one for each day), where the probability of taking longer than 5 minutes is calculated using the given density function. We aim to find the probability of exactly two such incidents.

**(c)** What is the probability that on at least one call out of seven, the nurse will take longer than 7 minutes to respond?

This involves calculating the probability of the complement (nurse taking less than or equal to 7 minutes) for all seven calls and subtracting from 1 to find the probability of at least one over 7 minutes.

### Explanation of the Density Function:

The function \( f(t) = \frac{1}{3} e^{-t/3} \) is an exponential density function. The exponential distribution is often used to model waiting times or the time until an event occurs. Here, the rate parameter \( \lambda = \frac{1}{3} \) indicates the average rate of the event (i.e., nurse response) occurrence.
Transcribed Image Text:The given problem illustrates a situation where the time a hospital patient must wait for a nurse's assistance is modeled by a continuous random variable with a specific density function. The function is given by: \[ f(t) = \frac{1}{3} e^{-t/3} \] for \( t \geq 0 \) (time is measured in minutes). ### Questions: **(a)** What is the probability that a patient must wait for more than 4 minutes? To find this, you would calculate the probability \( P(T > 4) \). **(b)** A patient spends a week in the hospital and requests nurse assistance once each day. What is the probability that the nurse will take longer than 5 minutes to respond on (exactly) two occasions? Here, we look at a binomial distribution scenario with 7 trials (one for each day), where the probability of taking longer than 5 minutes is calculated using the given density function. We aim to find the probability of exactly two such incidents. **(c)** What is the probability that on at least one call out of seven, the nurse will take longer than 7 minutes to respond? This involves calculating the probability of the complement (nurse taking less than or equal to 7 minutes) for all seven calls and subtracting from 1 to find the probability of at least one over 7 minutes. ### Explanation of the Density Function: The function \( f(t) = \frac{1}{3} e^{-t/3} \) is an exponential density function. The exponential distribution is often used to model waiting times or the time until an event occurs. Here, the rate parameter \( \lambda = \frac{1}{3} \) indicates the average rate of the event (i.e., nurse response) occurrence.
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