Assume that the duration in minutes of a telephone conversation follows the exponential distribution fx-15e-x-5- x- 0- The probability that the conversation will exceed five minutes is

If f(x) is p.d.f of x, then nbsp;nbsp; P(a≤ x≤ b)=∫_a^bf(x)dx So, nbsp; nbsp; nbsp;nbsp; P(5 lt;x lt;∞ )=∫_5^∞f(x)dx=∫_5^∞15e^ x/5dx=1e ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Other

Engineering Mathematics

Discrete and Continuous RV

Assume that t...

Question

Assume that the duration in minutes of a telephone conversation follows the exponential distribution $f (x) = \frac{1}{5} e^{- x / 5}, x \geq 0.$ The probability that the conversation will exceed five minutes is

\frac{1}{e}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1 - \frac{1}{e}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{e^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 - \frac{1}{e^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{1}{e}$
if f(x) is p.d.f of x, then
$P (a \leq x \leq b) = \int_{a}^{b} f (x) d x$

So, $P (5 < x < \infty) = \int_{5}^{\infty} f (x) d x = \int_{5}^{\infty} \frac{1}{5} e^{- x / 5} d x = \frac{1}{e}$

Suggest Corrections

Similar questions

It was observed that $150$ vehicles crossed a particular location of a highway in a duration of $30$ minutes. Assuming that vehicle arrival follows a negative exponential distribution, find out the number of times headway is greater than $5$ seconds in the above observation ?

Q. Arrivals at a telephone booth are considered to be Poisson, with an average time of 10 minutes between successive arrivals. The length of a phone call is distributed exponentially with mean 3 minutes. The probability that an arrival does not have to wait before service is

Q. A telephone switch board receiving number of phone calls follows Poisson distribution with parameter

3

per hour. The probability that it receives

5

calls in

2

hours duration is

Q. At a telephone enquiry system the number of phone calls received regarding relevant inquiry follows probability distribution with an average of 5 phone calls during 10 minute time interval. The probability that there is at the most one phone call during a 10 minute period is___

Q. At a telephone enquiry system the number of phone calls regarding relevant enquiry follow Poisson distribution with a average of 5 phone calls during IO-minute time intervals. The probability that there is at the most one phone call during a 10-minute time period is

Join BYJU'S Learning Program

Assume that the duration in minutes of a telephone conversation follows the exponential distribution f(x)=15e−x/5, x≥0. The probability that the conversation will exceed five minutes is

The correct option is A 1eif f(x) is p.d.f of x, then P(a≤x≤b)=∫baf(x)dx So, P(5<x<∞)=∫∞5f(x)dx=∫∞515e−x/5dx=1e

Assume that the duration in minutes of a telephone conversation follows the exponential distribution $f (x) = \frac{1}{5} e^{- x / 5}, x \geq 0.$ The probability that the conversation will exceed five minutes is

The correct option is A $\frac{1}{e}$
if f(x) is p.d.f of x, then
$P (a \leq x \leq b) = \int_{a}^{b} f (x) d x$

So, $P (5 < x < \infty) = \int_{5}^{\infty} f (x) d x = \int_{5}^{\infty} \frac{1}{5} e^{- x / 5} d x = \frac{1}{e}$