Advertisements
Advertisements
प्रश्न
The length of time (in minutes) that a certain person speaks on the telephone is found to be random phenomenon, with a probability function specified by the probability density function f(x) as
f(x) = `{{:("Ae"^((-x)/5)",", "for" x ≥ 0),(0",", "otherwise"):}`
What is the probability that the number of minutes that person will talk over the phone is (i) more than 10 minutes, (ii) less than 5 minutes and (iii) between 5 and 10 minutes
Advertisements
उत्तर
(i) more than 10 minutes
`int_10^00 "f"(x) "d"x`
= `1/5 int_10^oo "e"^(x/5) "d"x`
= `1/5 ("e"^((-x)/5)/(((-1)/5)))^oo`
= `- ["e"^((-x)/5)]_10^oo`
= `- ["e"^-oo - "e"^((-10)/5)]`
= `- [0 - "e"^-2]`
= `"e"^-2`
= `1/"e"^2`
(ii) less than 5 minutes
`int_0^5 f(x) "d"x = int_0^5 "Ae"^((x)/5)`
= `1/5 int_0^5 "e"^((-x)/5) "d"x`
= `1/5 ["e"^((-x)/5)/((-1)/5)]_0^5`
= `- ["e"^((-x)/5)]_0^5`
= `- ["e"^((-5)/5) - "e"^0]`
= `- ("e"^-1 - 1)`
= `1 - "e"^-1`
= `1 - 1/"e"`
= `("e" - 1)/"e"`
(iii) between 5 and 10 minutes
`int__5^10 "f"(x) "d"x = int_5^10 "Ae"^((-x)/5) "d"x`
= `int_5^10 1/5 "e"^((-x)/5) "d"x`
= `1/5 ["e"^((-x)/5)/((-1)/5)]_5^10`
= `- ["e"^((-x)/5)]_5^10`
= `- ["e"^((-10)/5) - "e"^((-5)/5)]`
= `[-"e"^-2 - "e"^-1]`
= `"e"^-1 - "e"^-2`
= `1/"e"- 1/"e"^2`
= `("e" - 1)/"e"^2`
APPEARS IN
संबंधित प्रश्न
A six sided die is marked ‘2’ on one face, ‘3’ on two of its faces, and ‘4’ on remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find the values of the random variable and number of points in its inverse images
Let X be a discrete random variable with the following p.m.f
`"P"(x) = {{:(0.3, "for" x = 3),(0.2, "for" x = 5),(0.3, "for" x = 8),(0.2, "for" x = 10),(0, "otherwise"):}`
Find and plot the c.d.f. of X.
The discrete random variable X has the probability function.
| Value of X = x |
0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(x) | 0 | k | 2k | 2k | 3k | k2 | 2k2 | 7k2 + k |
Evaluate p(x < 6), p(x ≥ 6) and p(0 < x < 5)
The discrete random variable X has the probability function.
| Value of X = x |
0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(x) | 0 | k | 2k | 2k | 3k | k2 | 2k2 | 7k2 + k |
If P(X ≤ x) > `1/2`, then find the minimum value of x.
What do you understand by continuous random variable?
Distinguish between discrete and continuous random variables.
Explain the terms probability density function
Choose the correct alternative:
The probability function of a random variable is defined as
| X = x | – 1 | – 2 | 0 | 1 | 2 |
| P(x) | k | 2k | 3k | 4k | 5k |
Then k is equal to
Choose the correct alternative:
The probability density function p(x) cannot exceed
Consider a random variable X with p.d.f.
f(x) = `{(3x^2",", "if" 0 < x < 1),(0",", "otherwise"):}`
Find E(X) and V(3X – 2)
