Answer to Question #292282 in Statistics and Probability for Yaku

Question

4. Explain in each case why the given equation cannot serve as the probability density function of a random variable that takes on values on the interval from 0 to 5:

a) f(x) = 1/10 (x - 4);

b) f(x) = 1/50 (x + 1).

5. A doctor knows from experience that 10% of of the patients to whom

he prescribes a certain blood pressure medication will have undesirable

side effects. Calculate the probability that of the 10 randomly selected

patients:

a) none will have undesirable side effects;

b) exactly 4 patients will have undesirable side effects;

c) at most 3 will have undesirable side effects;

d) at least 3 will have undesirable side effects.

6. Refer to Exercise 5. What is the expected number of patients with

undesirable side effects. What is the standard deviation of the number

of patients with undesirable side effects.

Answer 1

$4\\a)$

We have,

$f(x) = {1\over10} (x - 4);$

For this function to serve as a probability density function over the interval 0 to 5, it must satisfy the condition,

$\displaystyle\int^5_0f(x)dx=1.0$

$\displaystyle\int^5_0f(x)dx=\displaystyle\int^5_0{1\over10}(x-4)dx={1\over10}({x^2\over2}-4x)|^5_0={1\over10}({25\over2}-20)=-{7.5\over10}=-0.75\not=1.0$

Clearly, $f(x)={1\over10}(x-4)$ does not satisfy this condition thus, it can not serve as the probability density function of a random variable that takes on values on the interval from 0 to 5.

$b)$

$f(x) = {1\over50 }(x + 1)$. Same as equation in $a$ above, it must satisfy the condition below over the interval 0 to 5 as described below.

$\displaystyle\int^5_0f(x)dx=1.0$

So,

$\displaystyle\int^5_0f(x)dx=\displaystyle\int^5_0{1\over50}(x+1)dx={1\over50}({x^2\over2}+x)|^5_0={1\over50}({25\over2}+5)=0.35\not=1.0$

Since $f(x) = {1\over50 }(x + 1)$ does not satisfy this condition, it can not serve as a probability density function over the interval 0 to 5.

$5$

Let the random variable $X$ denote the patients who will undesirable side effects after prescription. $X$ follows a Binomial distribution. That is $X\sim B(n,p)$. For this case, $n=10\\p=0.10$

Therefore, this distribution is given as,

$p(X=x)=\binom{10}{x}0.1^x(1-0.1)^{10-x},\space x=0,1,2,..10$

$a)$

The probability that none will have undesirable side effects is given as,

$p(X=0)=\binom{10}{0}0.1^00.9^{10}=0.9^{10}=0.3487(4dp)$

Therefore, the probability that none of the patients will have undesirable side effects is 0.3487

$b)$

The probability that exactly 4 patients will have undesirable side effects is given as,

$p(X=4)=\binom{10}{4}0.1^40.9^6=210\times0.0001\times0.531441=0.0112(4dp)$

Therefore, the probability that exactly 4 patients will have undesirable side effects is 0.0112.

$c)$

The probability that at most 3 patients will have undesirable side effects is,

$p(X\leqslant3)=\displaystyle\sum^3_{x=0}p(X=x)=\displaystyle\sum^3_{x=0}\binom{10}{x}0.1^x(1-0.1)^{10-x}$

This is equivalent to,

$\binom{10}{0}0.1^0(0.9)^{10}+\binom{10}{1}0.1^1(0.9)^{9}+\binom{10}{2}0.1^2(0.9)^{8}+\binom{10}{3}0.1^3(0.9)^{7}= 0.3486784+ 0.3874205+ 0.1937102+0.05739563=0.9872048$

Therefore, the probability that at most 3 patients will have undesirable side effects is 0.9872048

$d)$

The probability that at least 3 will have undesirable side effects is given as,

$p(X\geqslant3)=\displaystyle\sum^{10}_{x=3}\binom{10}{x}0.1^x(1-0.1)^{10-x}$

$p(X\geqslant3)$ can also be given as,

$p(X\geqslant3)=1-p(X\leqslant2)=1-\{p(X=0)+p(X=1)+p(x=2)\}=1-\{\binom{10}{0}0.1^0(0.9)^{10}+\binom{10}{1}0.1^1(0.9)^{9}+\binom{10}{2}0.1^2(0.9)^{8}\}=1-\{0.3486784+ 0.3874205+ 0.1937102\}=1- 0.9298092=0.0701908$

Therefore, the probability that at least 3 patients will have undesirable side effects is 0.0701908.

$6$

$a)$

From exercise 5 above, the expected number of patients with undesirable side effects is,

$E(x)=n\times p=10\times0.1=1$

The expected number of patients with undesirable side effects is therefore equal to 1.

$b)$

To find the standard deviation, we first determine the variance of the number

of patients with undesirable side effects given as,

$var(x)=n\times p\times (1-p)=10\times0.1\times 0.9=0.9$

The standard deviation is,

$sd(x)=\sqrt{var( x)}=\sqrt{0.9}=0.9486833$

Thus, the standard deviation of the number of patients with undesirable side effects is 0.948683.