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A continuous random variable,
X, has probability density function given by:

f(x)={[k(x-3),3 <= x < 7],[k(11-x),7 <= x <= 11],[0," otherwise "]:}
a Show that
k=(16)/(.16).
[3]
c Find the value of
a if
P(3 <= X <= a)=0.75.
[3]

A continuous random variable, $X$ , has probability density function given by: $\newline$ $\mathrm{f}(x)=\left\{\begin{array}{ll}k(x-3) & 3 \leqslant x<7 \\ k(11-x) & 7 \leqslant x \leqslant 11 \\ 0 & \text { otherwise }\end{array}\right.$ $\newline$ a Show that $k=\frac{16}{.16}$ . $\newline$ [ $3$ ] $\newline$ c Find the value of $a$ if $\mathrm{P}(3 \leqslant X \leqslant a)=0.75$ . $\newline$ [ $3$ ]

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Find probabilities using the normal distribution I

Full solution

Q. A continuous random variable,

X

, has probability density function given by:

\newline

\mathrm{f}(x)=\left\{\begin{array}{ll}k(x-3) & 3 \leqslant x<7 \\ k(11-x) & 7 \leqslant x \leqslant 11 \\ 0 & \text { otherwise }\end{array}\right.

\newline

a Show that

k=\frac{16}{.16}

.

\newline

[

3

]

\newline

c Find the value of

a

if

\mathrm{P}(3 \leqslant X \leqslant a)=0.75

.

\newline

[

3

]

Calculate Area $1$ : To find $k$ , we need to ensure the total area under the probability density function equals $1$ . $\newline$ Calculate the area under the first part of the function from $3$ to $7$ : $\newline$ Area $1 = \int_{3}^{7} k(x-3) \, dx$ . $\newline$ Area $1 = \frac{k}{2} \cdot (x-3)^2$ from $3$ to $7$ . $\newline$ Area $1 = \frac{k}{2} \cdot (7-3)^2 - \frac{k}{2} \cdot (3-3)^2$ . $\newline$ Area $1 = \frac{k}{2} \cdot 16 - \frac{k}{2} \cdot 0$ . $\newline$ Area $1$ $0$ .
Calculate Area $2$ : Calculate the area under the second part of the function from $7$ to $11$ : $\newline$ Area $2$ = $\int_{7}^{11} k(11-x) \, dx$ . $\newline$ Area $2$ = $\frac{k}{2} \cdot (11-x)^2$ from $7$ to $11$ . $\newline$ Area $2$ = $\frac{k}{2} \cdot (11-7)^2 - \frac{k}{2} \cdot (11-11)^2$ . $\newline$ Area $2$ = $\frac{k}{2} \cdot 16 - \frac{k}{2} \cdot 0$ . $\newline$ Area $2$ = $8k$ .
Set Total Area: Add both areas and set the sum equal to $1$ for the total probability: $\newline$ Total Area = Area $1$ + Area $2$ . $\newline$ $1 = 8k + 8k$ . $\newline$ $1 = 16k$ . $\newline$ Solve for k: $\newline$ $k = \frac{1}{16}$ .
Find $k$ : Convert $k$ to decimal form: $\newline$ $k = \frac{1}{16} = 0.0625$ . $\newline$ But the question states $k$ should be $\frac{16}{0.16}$ , which is $100$ . $\newline$ There's a mistake in the calculation.

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