The time to pass through a security screening at an airport follows an exponential distribution. The mean time to pass through the security screening is 15 minutes. To catch the flight, a passenger must clear the security screening within 15 minutes. The probability that the passenger will miss the flight is _______. [round off to 3 decimal places]
Concept:
The probability density function of the exponential distribution is,
f(t) = λe-λt
Where, mean = \(\frac{1}{\lambda }\), t = time,
Calculation:
Given:
Maen = 15 minutes,
\(\lambda =\frac{1}{15}\)
The probability that passenger will miss the flight = P( t >15)
P(t > 15) = 1 - P(t ≤ 15)
= 1 - \(\int_{0}^{15}\lambda e^{-\lambda t}dt\)
= 1 - \(\lambda \frac{{{e^{ - \lambda t}}}}{{ - \lambda \;}}\left| {\begin{array}{*{20}{c}} {15}\\ 0 \end{array}} \right.\)
= \(e^{-(\frac{1}{15})15}\)
= \(\frac{1}{e}\)
P( t > 15) = 0.368
Important Questions from Continuous Distributions
Probability density function of a random variable X is given below
\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {0.25}&{if\;1 \le x \le 5}\\ 0&{otherwise} \end{array}} \right.\)
P (X ≤ 4) is
The variable x takes a value between 0 and 10 with uniform probability distribution. The variable y takes a value between 0 and 20 with uniform probability distribution. The probability of the sum of variables (x + y) being greater than 20 is _________
A nationalized bank has found that the daily balance available in its savings accounts follows a normal distribution with a mean of Rs. 500 and a standard deviation of Rs. 50. The percentage of savings account holders, who maintain an average daily balance more than Rs 500 is _______
The number of parameters in the univariate exponential and Gaussian distributions, respectively are
Find the value of λ such that the function f (x) is a valid probability density function. _______
\(f\left( x \right)\begin{array}{*{20}{c}} { = \lambda \left( {x - 1} \right)\left( {2 - x} \right)}&{for1 \le x \le 2}\\ { = 0}&{otherwise} \end{array}\)
