91Ó°ÊÓ

The Uniform Distribution, sometimes also known as a rectangular distribution, is a distribution that has constant probability.

It is symmetrical distribution on interval $(a,b)$with probability density function $f\left(x\right)=\frac{1}{(b-a)},x∈(a,b)$and zero everywhere else.

Let random variable $X$has uniform distribution on $(a,b)$. Then $E\left(X\right)=\frac{a+b}{2}\mathrm{and}Var\left(X\right)=\frac{{(b-a)}^2}{12}$.

Let $A\mathrm{and}B$are two events such that $P\left(B\right)>0$.

The Conditional probability $P(A\overline{)B)}$is :

$P(A\overline{)B)}=\frac{P(Aâˆ\copyright B)}{P\left(B\right)}$and zero elsewhere.

(a) According to the given question, the age of a first grader on September 1 at Garden Elementary School is uniformly distributed from $5.8\mathrm{and}6.8$years.

So, the variable$X$represents the age of a first grader.

(b) According to the question, the age of a first grader, is uniformly distributed over the interval $(5.8,6.8)$.

Since$X$represents the age of a first grader, hence$X~U(5.8,6.8)$

(c) The probability density function for $X$is defined by

$f\left(x\right)=\left\{\begin{array}{l}1,x∈(5.8,6.8)\\ 0,elsewhere\end{array}\right.$

The graph of the probability density function is :

$f\left(x\right)=1,5.8<x<6.8$

(d) Since the probability density function is $f\left(x\right)=\frac{1}{(b-a)},x∈(a,b)$

And localid="1648181714679" $X~U(5.8,6.8)$, so :

$f\left(x\right)=\frac{1}{(6.8-5.8)}$

localid="1648181822391" $⇒f\left(x\right)=\left\{\begin{array}{l}1,x∈(5.8,6.8)\\ 0,\mathrm{elsewhere}\end{array}\right.$

(e) The mean of uniform distribution over the interval $(5.8,6.8)$is given by :

role="math" localid="1648182095415" $$\begin{gathered} \mathrm{μ}=\frac{a+b}{2} \\ =\frac{5.8+6.8}{2} \\ ⇒\mathrm{μ}=6.3 \end{gathered}$$

(f) We first find the Variance, which is given by:

$$\begin{gathered} Var\left(X\right)=\frac{{(b-a)}^2}{12} \\ =\frac{{(6.8-5.8)}^2}{12} \\ =\frac{1}{12} \\ =0.084 \end{gathered}$$

Now, Standard Deviation, $\mathrm{σ}=\sqrt{Var\left(X\right)}$

$$\begin{gathered} ⇒\mathrm{σ}=\sqrt{0.084} \\ =0.289 \end{gathered}$$

(g)

$$\begin{gathered} P(X>6.5)={∫}_{6.5}^{∞}f\left(X\right).dx \\ ={∫}_{6.5}^{6.8}1.dx \\ =X_{6.5}^{6.8} \\ =6.8-6.5 \\ =0.3 \end{gathered}$$

(h)

(i) The${70}^{th}$percentile means the probability of $0.7$for $X∈(5.8,6.8)$

That is, $P(X≤x)=\left\{\begin{array}{l}1,x>6.8\\ 0,x<5.8\end{array}\right.$

Now,

$$\begin{gathered} 0.7=P(X≤x) \\ =F\left(x\right) \\ =\frac{x-5.8}{6.8-5.8} \\ ⇒0.7=x-5.8 \\ ⇒x=6.5 \end{gathered}$$

Thus, the ${70}^{th}$percentile for the age of first graders on September 1 at Garden Elementary School is$6.5$