Respuesta :
Answer:
a) Probability of getting exactly 7 hits = 0.001075
b) Probability of getting fewer than seven = 0.999
c) Probability of getting at least seven hits = 0.001
Step-by-step explanation:
The process is a Poisson process.
Rate, [tex]\lambda = 0.4[/tex]
The probability is to be calculated between 8:27 PM and 8:31 PM
t = 4
[tex]\lambda t = 0.4 * 4 = 1.6[/tex]
If the number of hits is represented by X
[tex]P(X = x) = \frac{e^{- \lambda t} (\lambda t)^x}{x!}[/tex]
a) Probability of getting exactly 7 hits
[tex]P(X = x) = \frac{e^{- 1.6} (1.6)^7}{7!}\\P(X= 7) = 0.001075[/tex]
b) Probability of getting fewer than seven
[tex]P(X < 7) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6)\\P(X < 7) = e^{- 1.6} [ \frac{1.6^0}{0!} + \frac{1.6^1}{1!} + \frac{1.6^2}{2!} + \frac{1.6^3}{3!} + \frac{1.6^4}{4!} + \frac{1.6^5}{5!} + \frac{1.6^6}{6!} ][/tex]
P(X < 7) = 0.999
c) Probability of getting at least seven hits
[tex]P(X \geq 7) = 1 - P(x < 7)\\P(X \geq 7) = 1 - 0.999\\P(X \geq 7) = 0.001[/tex]
Using the Poisson distribution, it is found that:
a) 0.0011 = 0.11% probability of exactly 7 hits in a 4 minute interval, which means that over many 4 minute intervals between 7 and 10 PM, 0.11% of them will have exactly 7 hits.
b) 0.9985 = 99.85% probability of fewer than seven hits in a 4 minute interval, which means that over many 4 minute intervals between 7 and 10 PM, 99.95% of them will have fewer than 7 hits.
c) 0.0015 = 0.15% probability of at least seven hits in a 4 minute interval, which means that over many 4 minute intervals between 7 and 10 PM, 0.15% of them will have at least 7 hits.
In a Poisson distribution, the probability is:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
The parameters are:
- x is the number of successes
- e = 2.71828 is the Euler number
- [tex]\mu[/tex] is the mean in the given interval.
In this problem, we have a mean of 0.4 hits per minute, interval of 4 minutes, thus:
[tex]\mu = 0.4(4) = 1.6[/tex]
The probabilities we are going to use are:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-1.6}1.6^{0}}{(0)!} = 0.2019[/tex]
[tex]P(X = 1) = \frac{e^{-1.6}1.6^{1}}{(1)!} = 0.3230[/tex]
[tex]P(X = 2) = \frac{e^{-1.6}1.6^{2}}{(2)!} = 0.2584[/tex]
[tex]P(X = 3) = \frac{e^{-1.6}1.6^{3}}{(3)!} = 0.1378[/tex]
[tex]P(X = 4) = \frac{e^{-1.6}1.6^{4}}{(4)!} = 0.0551[/tex]
[tex]P(X = 5) = \frac{e^{-1.6}1.6^{5}}{(5)!} = 0.0176[/tex]
[tex]P(X = 6) = \frac{e^{-1.6}1.6^{6}}{(6)!} = 0.0047[/tex]
[tex]P(X = 7) = \frac{e^{-1.6}1.6^{7}}{(7)!} = 0.0011[/tex]
Item a:
P(X = 7) = 0.0011, thus:
0.0011 = 0.11% probability of exactly 7 hits in a 4 minute interval, which means that over many 4 minute intervals between 7 and 10 PM, 0.11% of them will have exactly 7 hits.
Item b:
This probability is:
[tex]P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)[/tex]
Then, with the values we found:
[tex]P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.2019 + 0.3230 + 0.2584 + 0.1378 + 0.0551 + 0.0176 + 0.0047 = 0.9985[/tex]
0.9985 = 99.85% probability of fewer than seven hits in a 4 minute interval, which means that over many 4 minute intervals between 7 and 10 PM, 99.95% of them will have fewer than 7 hits.
Item c:
This probability is:
[tex]P(X \geq 7) = 1 - P(X < 7) = 1 - 0.9985 = 0.0015[/tex]
0.0015 = 0.15% probability of at least seven hits in a 4 minute interval, which means that over many 4 minute intervals between 7 and 10 PM, 0.15% of them will have at least 7 hits.
A similar problem is given at https://brainly.com/question/24098004
