Question

In a large hospital an average of 3 out of every 5 patients ask for water with their meal. A random sample of 10 customers is selected. Assuming a binomial distribution, find the probability that i. Exactly 6 patients ask for water with their meal ii. At most 4 patients ask for water with their meal iii. At least 3 patients ask for water with their meal iv. Find the mean of this distribution v. Find the standard deviation of this distribution.
In a mainframe computer centre, execution time of programs follows an exponential distribution. The average execution time of the programs is 5 minutes. Find the probability that the execution time of programs is: i. Less than 4 minutes ii. More than 6 minutes

Answer 1

Let $X=$ number of patients who asked for water: $X\sim Bin(n, p)$

Given $p=3/5=0.6, n=10$

i. $P(X=6)=\dbinom{10}{6}(0.6)^6(1-0.6)^{10-6}=0.250822656$

ii. $P(X\leq4)=P(X=0)+P(X=1)+P(X=2)+$

$+P(X=3)+P(X=4)=$

$=0.0001048576+0.001572864+0.010616832+$

$+0.042467328+0.111476736=0.1662386176$

iii. $P(X\geq4)=1-P(X=0)-P(X=1)-$

$-P(X=2)-P(X=3)=$

$=1-0.0001048576-0.001572864-0.010616832-$

$-0.042467328=0.9452381184$

iv. $E(X)=np=10(0.6)=6$

v. $V(X)=np(1-p)=10(0.6)(1-0.6)=2.4$

$\sigma_X=\sqrt{2.4}\approx1.55$

Let $X=$ the execution time of programs: $X\sim Exponential(\lambda)$

Given $\lambda=5$

i. $P(X<4)=1-e^{-5(4)}\approx0.999999998\approx1$

ii. $P(X>6)=e^{-5(6)}\approx10^{-13}\approx0$