Question #335084

The U.S. Postal Service reports 95% of first-class mail within the same city is delivered within 2 days of the time of mailing. Six letters are randomly sent to different locations.

a. What is the probability that all six arrive within 2 days?

b. What is the probability that exactly five arrive within 2 days?

c. Find the mean number of letters that will arrive within 2 days.

d. Compute the variance and standard deviation of the number that will arrive within 2 days.


1
Expert's answer
2022-05-01T15:53:52-0400

We have a Bernoulli trial - exactly two possible outcomes, "success" (the letter arrives within 2 days) and "failure" (it doesn't arrive within 2 days) and the probability of success is the same every time the experiment is conducted (a letter sent), p=0.95,q=1p=10.95=0.05p=0.95, q=1-p=1-0.95=0.05.

The probability of each result

P(X=k)=(nk)pkqnk==(6k)0.95k0.056k==6!k!(6k)!0.95k0.056k.P(X=k)=\begin{pmatrix}n\\k\end{pmatrix}\cdot p^k\cdot q^{n-k}=\\ =\begin{pmatrix}6\\k\end{pmatrix}\cdot 0.95^k\cdot 0.05^{6-k}=\\ =\cfrac{6!}{k!\cdot(6-k)!}\cdot 0.95^k\cdot 0.05^{6-k}.


a. P(X=6)=6!6!0!0.9560.050=0.7351.b. P(X=5)=6!5!1!0.9550.051=0.2321.c. μ=np=60.95=5.7.d. σ2=npq=60.950.05=0.285;σ=0.285=0.534.\text{a. } P(X=6)=\cfrac{6!}{6!\cdot0!}\cdot 0.95^{6}\cdot 0.05^{0}=0.7351.\\ \text{b. } P(X=5)=\cfrac{6!}{5!\cdot1!}\cdot 0.95^{5}\cdot 0.05^{1}=0.2321.\\ \text{c. } \mu=np=6\cdot0.95=5.7.\\ \text{d. } \sigma^2=npq=6\cdot0.95\cdot0.05=0.285;\\ \sigma=\sqrt{0.285}=0.534.

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