Question

Let $X1,...,Xn$be a set of independent and identically distributed continuous random variables having distribution function F, and let $X\left(i\right),i=1,...,n$denote their ordered values. If X, independent of the$Xi,i=1,...,n$, also has distribution F, determine

(a) $P\left\{X>X\left(n\right)\right\}$;

(b) $P\{X>X(1)\}$;

(c) $PX\left\{\right(i)<X<X(j\left)\right\},1\dots i<j\dots n$.

Short answer

(a) Probability that exactly 3 months in 6 months will have sales greater than 100 is $P(X=3)=0.3125$.

(b) Probability that total sales in 4 months is greater than 420 is 0.0228

Step-by-step solution

Introduction

A measure of spread for a random variable distribution that determines how much the values deviate from the expected value.

Given

Mean of monthly sales $=100$

Standard deviation of monthly sales $=5$

Monthly sales are independent and follow normal distribution.

Explanation (a)

Formula used:

$P(X=k)={}^{n}C_k\times p^k\times (1-p{)}^{n-k}$ Where $n$ is the number of trials p is the probability of success. Calculation: Let $M$be the monthly sale So, $M~N\left(100,5^2\right)$Probability of monthly sale greater than 100

$P(M>100)$

$=P\left(\frac{M-{\mu }_m}{{\sigma }_{-}}>\frac{100-100}{5}\right)$

$=P(z>0)$

$=1-P(z\le 0)$

From $z$table

$P(z<0)=0.5$

$P(z>0)=1-0.5=0.5$

$=P(z>0)$

$=1-P(z\le 0)$

From$z$ table

$P(z<0)=0.5$

$P(z>0)=1-0.5=0.5$

As each month sale is independent has same probability of being greater than 100 , so the number of sale

greater than 100 (let it be denoted as $\mathrm{X}$) can be modelled by binomial distribution with parameter, $n=6$and $P=0.5$

So,$X~Bin(6,0.5)$

$P(X=3)={}^{6}C_3\times 0.5^3\times (1-0.5{)}^{6-3}$

$P(X=3)=0.3125$

Explanation (b)

Formula used:

If the events are independent and follow normal distribution with same parameters then

$\sum _{i=1}^n{X}_i~N\left(n\times \mu ,n\times {\sigma }^2\right)$

Calculation:

Let $Y$denote the total sales in 4 months

So, the distribution of $Y$will be

&$Y~N\left(4\times 100,4\times 5^2\right)$

&$Y~N(400,100)$

The required probability can be calculated as follows

$P(Y>420)$

$=P\left(\frac{Y_{-{\mu }_y}}{{\sigma }_y}>\frac{420-400}{\sqrt{100}}\right)$

$=P(z>2)$

$=1-P(z\le 2)$

From z tables

$$\begin{gathered} P(z<2)=0.97725 \\ P(z>2)=1-0.97725 \\ P(z<2)=0.02275 \end{gathered}$$