Respuesta :
Answer:
The probability that all 70 messages are transmitted in less than 12 minutes is 0.117.
Step-by-step explanation:
Let X = the transmission time of each message.
The random variable X follows an Exponential distribution with parameter λ = 5 minutes.
The expected value of X is:
[tex]E(X)=\frac{1}{\lambda}=\frac{1}{5}=0.20[/tex]
The variance of X is:
[tex]V(X)=\frac{1}{\lambda^{2}}=\frac{1}{5^{2}}=0.04[/tex]
Now define a random variable T as:
T = X₁ + X₂ + ... + X₇₀
According to a Central limit theorem if a large sample (n > 30) is selected from a population with mean μ and variance σ² then the sum of random variables X follows a Normal distribution with mean, [tex]\mu_{s} = n\mu[/tex] and variance, [tex]\sigma^{2}_{s}=n\sigma^{2}[/tex].
Compute the probability of T < 12 as follows:
[tex]P(T<12)=P(\frac{T-\mu_{T}}{\sqrt{\sigma_{T}^{2}}}<\frac{12-(70\times0.20)}{\sqrt{70\times 0.04}})\\=P(Z<-1.19)\\\=1-P(Z<1.19)\\=1-0.883\\=0.117[/tex]
*Use a z-table for the probability.
Thus, the probability that all 70 messages are transmitted in less than 12 minutes is 0.117.
Following are the solution to the given question:
Let X signify the letter's transmission time [tex]i^{th}[/tex] , and [tex]i =1,2,3,...,70.[/tex] this is assumed that the [tex]X_i \sim Exp (\lambda =5)[/tex].
[tex]\to Mean =\frac{1}{\lambda} =\frac{1}{5}=0.2\\\\ \to Variance =\frac{1}{\lambda^2} =\frac{1}{5^2} =\frac{1}{25}=0.04\\\\[/tex]
- Let [tex]T = X_1 + X_2 +...+ X_{70}[/tex], be the total transmission time.
- According to the Central Limit Theorem, T has a Normal distribution of mean, Variance.
mean [tex]= E(T) = 70 \times 0.2=14[/tex]
Variance [tex]=V(T) =70 \times 0.04 = 2.8[/tex]
Therefore
[tex]\to P(T<12) = P (Z < \frac{12-14}{\sqrt{2.8}})[/tex]
[tex]= P(Z<-1.195) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (by sy memetry) \\\\=0.5-P(0<Z<1.19)\\\\=0.5-0.3830 \\\\=0.1170\\\\[/tex]
Therefore, the final answer is "0.1170".
Learn more about the Central Limit Theorem:
brainly.com/question/898534
