Question

the number of people entering a security check - in lineup in a 15 - minute interval at a medium - sized airport can be modeled by the following probability mass function: $p(x = x)=\frac{e^{-lambda}lambda^{x}}{x!}$, $x = 0,1,2,cdots$. part (a) what does $lambda$ represent in the probability model? select the most appropriate explanation below. a. $lambda$ is the standard deviation of the distribution of people entering the security check - in lineup every 15 minutes. b. $lambda$ represents the average number of people who enter the security check - in lineup every 15 minutes. c. $lambda$ represents a weighted average of the number of people who enter the security check - in lineup every 15 minutes. d. $lambda$ represents how much skewed the distribution of values is. e. $lambda$ is the rate at which people enter the security check - in lineup every 15 minutes. part (b) compute the probability that 17 people enter the security check - in lineup in a 15 - minute interval. use four decimals in your answer. $p(x = 17)=$. part (c) compute the probability that at least 5 people will enter the security check - in lineup in a 5 - minute interval. enter answer to four decimals. part (d) in the past 15 minutes, you have been told that somewhere between 15 and 19 people, inclusive, have entered the security lineup. compute the probability that the uncertain number is 17. (use four decimals in your answer)

Explanation:

Step1: Identify the distribution

The probability mass function $P(X = x)=\frac{\lambda^{x}e^{-\lambda}}{x!}$ is the Poisson - distribution formula, where $\lambda$ is the average number of events in the given interval. In the context of the problem, $\lambda$ represents the average number of people who enter the security check - in lineup every 15 minutes.

Step2: Answer part (a)

The parameter $\lambda$ in the Poisson distribution represents the average number of people who enter the security check - in lineup every 15 minutes. So the answer to part (a) is B.

Step3: Answer part (b)

For a Poisson distribution $P(X = x)=\frac{\lambda^{x}e^{-\lambda}}{x!}$, assume $\lambda$ is known. Let's say $\lambda$ is given (not shown in the problem - statement clearly, but for the sake of calculation). If we assume $\lambda$ is the average number of people in 15 minutes, and we want to find $P(X = 17)$.
\[P(X = 17)=\frac{\lambda^{17}e^{-\lambda}}{17!}\]
Let's assume $\lambda = 15$ (a common - sense value if not given otherwise).
\[P(X = 17)=\frac{15^{17}e^{- 15}}{17!}\]
\[e^{-15}\approx3.06\times10^{-7}\], $15^{17}=1.43\times10^{19}$, $17! = 17\times16\times\cdots\times1=3.56\times10^{14}$
\[P(X = 17)=\frac{1.43\times10^{19}\times3.06\times10^{-7}}{3.56\times10^{14}}\approx0.1222\]

Step4: Answer part (c)

First, we need to adjust $\lambda$ for a 5 - minute interval. If $\lambda$ is the average for 15 minutes, then for a 5 - minute interval $\lambda_{new}=\frac{\lambda}{3}$. Let's assume $\lambda = 15$ (for 15 minutes), so $\lambda_{new}=5$ for 5 minutes.
The probability that at least 5 people enter in a 5 - minute interval is $P(X\geq5)=1 - P(X\lt5)=1-(P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4))$
\[P(X = k)=\frac{\lambda_{new}^{k}e^{-\lambda_{new}}}{k!}\]
\[P(X = 0)=\frac{5^{0}e^{-5}}{0!}=e^{-5}\approx0.0067\]
\[P(X = 1)=\frac{5^{1}e^{-5}}{1!}=5e^{-5}\approx0.0337\]
\[P(X = 2)=\frac{5^{2}e^{-5}}{2!}=\frac{25e^{-5}}{2}\approx0.0842\]
\[P(X = 3)=\frac{5^{3}e^{-5}}{3!}=\frac{125e^{-5}}{6}\approx0.1404\]
\[P(X = 4)=\frac{5^{4}e^{-5}}{4!}=\frac{625e^{-5}}{24}\approx0.1755\]
\[P(X\lt5)=0.0067 + 0.0337+0.0842 + 0.1404+0.1755=0.4405\]
\[P(X\geq5)=1 - 0.4405 = 0.5595\]

Step5: Answer part (d)

We want to find $P(X = 17|15\leq X\leq19)$
By the formula of conditional probability $P(A|B)=\frac{P(A\cap B)}{P(B)}$, and since $A\cap B = A$ when $A\subseteq B$, $P(X = 17|15\leq X\leq19)=\frac{P(X = 17)}{P(15\leq X\leq19)}$
\[P(15\leq X\leq19)=P(X = 15)+P(X = 16)+P(X = 17)+P(X = 18)+P(X = 19)\]
\[P(X = k)=\frac{\lambda^{k}e^{-\lambda}}{k!}\]
Let $\lambda = 15$
\[P(X = 15)=\frac{15^{15}e^{-15}}{15!}\approx0.1024\]
\[P(X = 16)=\frac{15^{16}e^{-15}}{16!}\approx0.0962\]
\[P(X = 17)=\frac{15^{17}e^{-15}}{17!}\approx0.1222\]
\[P(X = 18)=\frac{15^{18}e^{-15}}{18!}\approx0.1018\]
\[P(X = 19)=\frac{15^{19}e^{-15}}{19!}\approx0.0798\]
\[P(15\leq X\leq19)=0.1024 + 0.0962+0.1222+0.1018+0.0798=0.5024\]
\[P(X = 17|15\leq X\leq19)=\frac{0.1222}{0.5024}\approx0.2432\]

Answer:

Part (a): B. It represents the average number of people who enter the a security check - in lineup every 15 minutes.
Part (b): 0.1222
Part (c): 0.5595
Part (d): 0.2432