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dmitriy555 [2]
3 years ago
14

A continuous random variable is uniformly distributed in the interval a<X<b the Lower quartile is 5 and upper quartile is

9 find
a. the values of a and b
b. p(6<X<7)
c. the cumulative distributive function

راز​
Mathematics
1 answer:
astraxan [27]3 years ago
6 0

(a) <em>X</em> has a probabiliity density of

f_X(x) = \begin{cases}\dfrac1{b-a}&\text{if }a

If the lower quartile is 5 and the upper quartile is 9, then

\displaystyle \int_a^5 f_X(x)\,\mathrm dx = 0.25 \text{ and } \int_9^b f_X(x)\,\mathrm dx = 0.25

Computing the integrals gives the following system of equations:

\displaystyle \int_a^5\frac{\mathrm dx}{b-a} = \frac{5-a}{b-a} = 0.25 \\\\ \int_9^b \frac{\mathrm dx}{b-a} = \frac{b-9}{b-a} = 0.25

5 - <em>a</em> = 0.25 (<em>b</em> - <em>a</em>)   ==>   0.75<em>a</em> + 0.25<em>b</em> = 5   ==>   3<em>a</em> + <em>b</em> = 20

<em>b</em> - 9 = 0.25 (<em>b</em> - <em>a</em>)   ==>   0.25<em>a</em> + 0.75<em>b</em> = 9   ==>   <em>a</em> + 3<em>b</em> = 36

Eliminate <em>a</em> :

(3<em>a</em> + <em>b</em>) - 3 (<em>a</em> + 3<em>b</em>) = 20 - 3×36

-8<em>b</em> = -88

==>   <em>b</em> = 11   ==>   <em>a</em> = 3

and so P(<em>X</em> = <em>x</em>) = 1/(11 - 3) = 1/8 for all 3 < <em>x</em> < 11.

(b)

\displaystyle P(6

(c) The distribution function is then

\displaystyle F_X(x) = \int_{-\infty}^x f_X(t)\,\mathrm dt = \begin{cases}0&\text{if }x\le3 \\ \dfrac x8 &\text{if }3

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