Question

Births are approximately uniformly distributed between the 52 weeks of the year. They can be said to follow a uniform distribution from one to 53 (spread of 52 weeks).

a. X ~ _________

b. Graph the probability distribution.

c. f(x) = _________

d. ì = _________

e. ó = _________

f. Find the probability that a person is born at the exact moment week 19 starts. That is, find P(x = 19) = _________

g. P(2 < x < 31) = _________

h. Find the probability that a person is born after week 40.

i. P(12 < x|x < 28) = _________

j. Find the 70th percentile.

k. Find the minimum for the upper quarter

Short answer

As per calculations

a. $X~U(1,53)$

b.

The probability distribution function will be;

localid="1649331118711" $\begin{array}{rcl}f\left(x\right)& =& \frac{1}{b-a}\\ & =& \frac{1}{53-1}\\ & =& \frac{1}{52}\end{array}$

c.

$f\left(x\right)=\left\{\begin{array}{l}\frac{1}{52}\\ 0\end{array}\right.,1<X<53$

d. 27

e. 15.01

f. 0

g. 0.5577

h. 0.25

i. 0.5926

j. 37.4

k. 40

Step-by-step solution

Definition of variables 

Variables are the terms that may have different values. Here in this distribution variable X is uniformly distributed between the 52 weeks of the year.

Findings the values of variables 

a. Uniform of all data

$X~U(1,53)$

b. The height of probability distribution

$$\begin{gathered} f\left(x\right)=\frac{1}{b-a} \\ =\frac{1}{53-1} \\ =\frac{1}{52} \end{gathered}$$

The graph of distribution is

c. The formulation of,

$\begin{array}{rcl}f\left(x\right)& =& \left\{\begin{array}{ll}\frac{1}{b-a}& ,a<X<b\\ 0& ,1<X<53\end{array}\right.\\ f\left(x\right)& =& \left\{\begin{array}{ll}\frac{1}{52}& ,1<X<53\\ 0& ,Otherwise\end{array}\right.\end{array}$

d. The mean can be calculated as;

$\begin{array}{rcl}\mu & =& \frac{a+b}{2}\\ & =& \frac{1+53}{2}\\ & =& 27\end{array}$

Hence the mean is 27

e. The standard deviation

$$\begin{gathered} \sigma =\sqrt{\frac{{\left(b-a\right)}^2}{12}} \\ \sigma =\sqrt{\frac{{\left(53-1\right)}^2}{12}} \\ \sigma =\sqrt{225.33} \\ \sigma =15.01 \end{gathered}$$

The standard Deviation is 15.01

f. For continuous probability, the point value is zero.

Hence the value of $P(x=19)=0$

g. The required probability is

$\begin{array}{rcl}P(2& <& x<31)=Base\times Height\\ & =& (32-2)\times \frac{1}{52}\\ & =& 29\times \frac{1}{52}\\ & =& 0.5577\end{array}$

h.

The value of

localid="1649333150490" $\begin{array}{rcl}P(x& >& 40)=Base\times Height\\ & =& (53-40)\times \frac{1}{52}\\ & =& 13\times \frac{1}{52}\\ & =& 0.25\end{array}$

i. The probability of

$\begin{array}{rcl}& & P(12<x\overline{)x<28})=\frac{P(12<x<28)}{P(x<28)}\\ & =& \frac{(28-12)\times \frac{1}{52}}{(28-1)\times \frac{1}{52}}\\ & =& \frac{16}{27}\\ & =& 0.5926\end{array}$

j. For the 70 percentile

$\begin{array}{rcl}P(x& <& K)=\hspace{0.17em}Base\times Height\\ 0.70& =& (K-1)\times \frac{1}{52}\\ k-1& =& 0.70\times 52\\ K& =& 36.4+1\\ & =& 37.4\end{array}$

k. The required value can be calculated by calculating the 75th percentile of the data, which can be calculated as below;

$\begin{array}{rcl}P(x& <& K)=\hspace{0.17em}Base\times Height\\ 0.75& =& (K-1)\times \frac{1}{52}\\ k-1& =& 0.75\times 52\\ K& =& 39+1\\ & =& 40\end{array}$

The value of the minimum of the upper quarter is 40.