Question

A particular telephone number is used to receive both voice

calls and fax messages.

Suppose that $25\%$ of the incoming calls involve fax messages, and consider a sample of $25$ incoming calls.

What is the probability that

$a$. At most $6$ of the calls involve a fax message?

$b$. Exactly $6$ of the calls involve a fax message?

$c$. At least $6$ of the calls involve a fax message?

$d$. More than $6$ of the calls involve a fax message?

$e$. What is the expected number of calls among the $25$ that involve a fax message?

$f$. What is the standard deviation of the number among the $25$ calls that involve a fax message?

$g$. What is the probability that the number of calls among the $25$ that involve a fax transmission exceeds the expected number by more than $2$ standard deviations?

Answer 1

Explanation for the each part:

$\left(a\right)$ Using Binomial distribution with given values as $n=25$ and $p=0.25,q=0.75$ so we have, probability as

$P(X\le 6)=\sum _{x=}^6{}^{25}C_x{(0.25)}^x{(0.75)}^{25-x}$

on solving above sum, we get

$P(X\le 6)=0.5611$

$\left(b\right)$ Here, we can use direct value as

$$\begin{gathered} P(X=6)={}^{25}C_6{(0.25)}^6{(0.75)}^{19} \\ P(X=6)=0.182 \end{gathered}$$

$\left(c\right)$ Here,

$$\begin{gathered} P(X\ge 6)=1-P(X<6) \\ P(X\ge 6)=1-0.5611-0.182 \\ P(X\ge 6)=0.6217 \end{gathered}$$

$\left(d\right)$ Here,

$$\begin{gathered} P(X\ge 6)=1-P(X\le 6) \\ P(X\ge 6)=1-0.5611 \\ P(X\ge 6)=0.4389 \end{gathered}$$

$\left(e\right)$ Here, expected number can be calculated as

$$\begin{gathered} E\left(X\right)=np \\ E\left(X\right)=25\times 0.25 \\ E\left(X\right)=6.25 \end{gathered}$$

$\left(f\right)$ Using standard deviation formula as

$$\begin{gathered} S.D=\sqrt{npq} \\ S.D=\sqrt{25\times 0.25\times 0.75} \\ S.D=2.16 \end{gathered}$$

$\left(g\right)$ Here, $P(X\le 10)=\sum _{x=0}^{x=10}{}^{25}C_x{(0.25)}^x{(0.75)}^{25-x}P(X\le 10)=0.97$