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 model
p(x = x)=\frac{e^{-16.8}(16.8)^x}{x!}quad x = 0,1,2,cdots
part (a) what does 16.8 represent in the probability model? select the most appropriate explanation below.
a. 16.8 is the standard deviation of the distribution of people entering the security check - in lineup every 15 - minutes.
b. 16.8 represents the average number of people who enter the a security check - in lineup every 15 - minutes.
c. 16.8 represents a weighted - average of the number of people who enter the a security check - in lineup every 15 - minutes.
d. 16.8 represents how much skewed the distribution of values is.
e. 16.8 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)=0.1222
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.
0.5595
part (d) in the past 15 - minutes, you have been told that somewhere between 15 and 19 people, inclusive, have entered the secunty lineup. compute the probability that this uncertain number is 17.
0.2432 (use four decimals in your answer)

Explanation:

Step1: Identify the distribution

The probability model $P(X = x)=\frac{e^{-16.8}(16.8)^x}{x!}$ for $x = 0,1,2,\cdots$ is a Poisson - distribution formula, where the parameter $\lambda$ represents the average number of events in the given interval. Here, $\lambda = 16.8$, so 16.8 is the average number of people who enter the security check - in lineup every 15 minutes.

Step2: Calculate $P(X = 17)$ for part (b)

Substitute $x = 17$ and $\lambda=16.8$ into the Poisson - distribution formula $P(X = x)=\frac{e^{-\lambda}\lambda^x}{x!}$.

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Step3: Adjust $\lambda$ for 5 - minute interval in part (c)

Since the original $\lambda = 16.8$ is for a 15 - minute interval, for a 5 - minute interval, $\lambda_{new}=\frac{16.8}{3}=5.6$.
The probability that at least 5 people enter is $P(X\geq5)=1 - P(X\lt5)=1-(P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4))$.

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$P(X\lt5)=0.0037 + 0.0207+0.0579 + 0.1078+0.1509 = 0.341$
$P(X\geq5)=1 - 0.341 = 0.659$

Step4: Calculate conditional probability in part (d)

The probability that $15\leq X\leq19$ is $P(15\leq X\leq19)=P(X = 15)+P(X = 16)+P(X = 17)+P(X = 18)+P(X = 19)$.

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$P(15\leq X\leq19)=0.1104 + 0.1159+0.1122 + 0.1043+0.0912=0.534$
The conditional probability $P(X = 17|15\leq X\leq19)=\frac{P(X = 17)}{P(15\leq X\leq19)}=\frac{0.1122}{0.534}\approx0.2101$

Answer:

Part (a): B. 16.8 represents the average number of people who enter the a security check - in lineup every 15 - minutes.
Part (b): 0.1122
Part (c): 0.6590
Part (d): 0.2101