Answer to Question #219944 in Statistics and Probability for reitu

Question #219944

QUESTION 4

The time between customers arriving at a book store is exponentially distributed with a mean of 5

minutes.

(a) What is the probability that no customers arrive during a 30-minute period? (5)

(b) If the store opens at 9:00, what is the probability that the first customer of the day arrives

between 9:10 and 9:20? (6)

(c) If the store opens at 9:00, what is the probability that the first customer arrives at exactly

9:15? (4)



1
Expert's answer
2021-07-25T17:12:35-0400

Suppose a random variable X denotes the time(in minutes) between customer's arrivals.

Therefore:

XExp(λ=15=0.2)X\approx Exp\left(\lambda =\frac{1}{5}=0.2\right)

f(x)=λeλxf\left(x\right)=\lambda e^{-\lambda x}\:

f(x)=0.2e0.2xforx0f\left(x\right)\:=0.2e^{-0.2x}\:for\:x\ge 0


(a). No customers arrive during a 30 - minute period if the the time between customer's arrivals is greater than 30 minutes.

Required probability is given by;


P(X>30)=30P\left(X>30\right)=\int _{30}^{\infty } 0.2e0.2xdx0.2e^{-0.2x}dx\:


=[e0.2x]30=\left[-e^{-0.2x}\:\right]^{^{\infty }}_{30}


=[e0.2x]30=\left[e^{-0.2x}\:\right]^{^{30\:}}_{\infty }


=e0.2×30e=e^{-0.2\times 30}-e^{-\infty }


=e60=e^{-6}-0\:


=0.0025=0.0025

Hence, 0.00250.0025 is the probability that no customers arrive during a 30 - minute period.


(b). Considering 9:00 as the origin of time(i.e. time zero), we obtain 9:10 as 10 minutes and 9:20 as 20 minutes . Required probability is given by;


P(10<X<20)=10200.2e0.2xdxP\left(10<X<20\right)=\int _{10}^{20}0.2e^{-0.2x}\:dx


=[e0.2x]1020=\left[-e^{-0.2x}\right]^{^{20}}_{10}



=[e0.2x]2010=\left[e^{-0.2x}\right]^{^{10}}_{20}


=e0.2×10e0.2×20=e^{-0.2\times 10}-e^{-0.2\times 20}


=e2e4=e^{-2}-e^{-4}\:


=0.13533530.0183156=0.1353353-0.0183156


=0.1170=0.1170

Hence, 0.11700.1170 is the probability that the first customer arrives between 9:10 and 9:20.


(c). Considering 9:00 as the origin of time (i.e. time zero), we obtain 9:15 as 15 minutes.

Required probability is given by;

P(X=15)=15150.2e0.2xdxP\left(X=15\right)=\int _{15}^{15}0.2e^{-0.2x}dx


=[e0.2x]1515=\left[e^{-0.2x}\right]^{^{15}}_{15}


=e0.2×15e0.2×25=e^{-0.2\times 15}-e^{-0.2\times 25}


=e3e3=e^{-3}-e^{-3}


=0.049787070.04978707=0.04978707-0.04978707


=0=0

Hence, is the probability that the first customer arrives at exactly 9:15

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