Q. 6.31 - A First Course in Probability

Question

Let $X\left(1\right) \dots X\left(2\right) \dots ··· \dots X\left(n\right)$ be the ordered values of n independent uniform $(0, 1)$random variables. Prove that for $1\dots k\dots n+1,PX\left(k\right)-X(k-1)>t=(1-t)n$ where $X\left(0\right) K 0, X(n+1) K t.$

Step-by-Step Solution

Answer

The probability that at least 110 of the 400 people never eat breakfast:

$P\{110\le M+W\}=0.0749$

The probability that the number of women who never eat breakfast is at least as large as the number of men who never eat breakfast:

$P\{M\le W\}=0.3557$

Step 1: Given information (part a)

$25.2\%$ of males never eat breakfast.

$23.6\%$ of females never eat breakfast.

$200$ men and $200$ women are chosen at random.

Let M denotes the number of men that never eat breakfast and W denotes the number of women that never eat breakfast

Note that M is a binomial random variable with$p=0.236\text{ and }n=200$

W is a binomial random variable with $p=0.236 and n=200$

Step 2: Explanation (part a)

Then compute

$$\begin{gathered} E\left[M\right]=200\left(0.252\right) \\ =50.4 \end{gathered}$$

and

$$\begin{gathered} E\left[M\right]=200\left(0.236\right) \\ =47.2 \end{gathered}$$

also

$$\begin{gathered} Var \left[M\right]=200\left(0.252\right) \\ =(1-0.252) \\ \approx 50.4 \end{gathered}$$

Also,

$$\begin{gathered} Var \left[W\right]=200\left(0.236\right) \\ =(1-0.236) \\ \approx 36.1 \end{gathered}$$

thus,

we need to compute $P\{110\le M+W\}$

Approximate $M+W$ by a normal distribution

with $\mu =50.4+47.2=97.6$

and ${\sigma }^2\approx 37.7+36.1=73.8$

Thus, $P\{110\le M+W\}=0.0749$

Step 1: Given information (part b)

$25.2\%$ of males never eat breakfast.

$23.6\%$ of females never eat breakfast.

$200$ men and $200$ women are chosen at random.

Let M denotes the number of men that never eat breakfast and W denotes the number of women that never eat breakfast

Note that M is a binomial random variable with $p=0.236\text{ and }n=200$

W is a binomial random variable with $p=0.236\text{ and }n=200$

Step 2: Explanation (part b)

Then compute 

$E \left[M\right]=200\left(0.252\right)=50.4$

and

$E \left[W\right]=200\left(0.236\right)=47.2$

also

$Var \left[M\right]=200\left(0.252\right)\left(1-0.252\right)\approx 37.7$

also

$Var \left[W\right]=200\left(0.236\right)\left(1-0.236\right)\approx 36.1$

thus

we need to approximate M by a normal random variable

with $\mu =50.4 and {\sigma }^2\approx 37.3$

W by a normal random variable with 

$\mu =47.2 and {\sigma }^2\approx 36.1$

Now, we need to compute $P\{M\le W\}=P\{M-W\le 0\}$

Approximate M - W by a normal distribution with

$\mu =50.4-47.2=3.2$

and ${\sigma }^2\approx 37.3+36.1=73.8$

thus,

$$\begin{gathered} P\{M\le W\}=P\{M-W\le 0\}=P\left\{\frac{M-W-3.2}{\sqrt{73.8}}\ge \frac{-3.2}{\sqrt{73.8}}\right\}\approx \Phi \left(\frac{-3.2}{\sqrt{73.8}}\right) \\ \approx 1-\Phi (0.37) \\ =1-0.6443 \\ =0.3557 \end{gathered}$$