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4 years ago

A uniform distribution is defined over the interval from 6 to 10.

Mathematics

1 answer:

zhuklara [117]4 years ago

8 0

Answer:

Refer below

Step-by-step explanation:

a) a and b are the lower and higher values of the interval for which uniform distribution is defined.

Here a= 6 and b =10

b) Mean of the uniform distribution= (a+b)/2 = (6+10)/2 =8

Or int x (1/4) dx = x^2/8 = 8

c) Variance of the uniform distribution = (b^2-a^2)/12 = (100-64)/12

= 36/12 =3

Std dev = sq rt of 3 = 1.732

d) To find total area

PDF of the distribution = 1/(b-a) = 1/4, 6<x<10

Area = \int 6 to 10 of 1/4 dx

= x/4

Subtitute limits

= (10-6)/4 =1

So total area = 1

d)P(X>7) = int 7 to 10 of 1/4 dx = 3/4

e) P(7<x<9) = Int 7 to 9 of 1/4 dx = 2/4 = 1/2

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I need helping answering 15. I need to create a table and graph the function.

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ANSWER and EXPLANATION

We want to order the functions from widest to narrowest:

$\begin{gathered} y=3x^2 \\ y=2x^2 \\ y=4x^2 \end{gathered}$

To do this, we have to plot the graphs of the functions by using a table of values.

Let us find the values of the functions for values of x = -2, 0, 2

For the first function:

$\begin{gathered} y=3x^2 \\ \Rightarrow x=-2: \\ y=3(-2)^2=3*4 \\ y=12 \\ \Rightarrow x=0: \\ y=3(0)^2=3*0 \\ y=0 \\ \Rightarrow x=2: \\ y=3(2)^2=3*4 \\ y=12 \end{gathered}$

Hence, its table is:

For the second function:

$\begin{gathered} y=2x^2 \\ \Rightarrow x=-2: \\ y=2(-2)^2=2*4 \\ y=8 \\ \Rightarrow x=0: \\ y=2(0)^2=2*0 \\ y=0 \\ \Rightarrow x=2: \\ y=2(2)^2=2*4 \\ y=8 \end{gathered}$

Hence, its table is:

For the third function:

$\begin{gathered} y=4x^2 \\ \Rightarrow x=-2: \\ y=4\left(-2\right)^2=4*4 \\ y=16 \\ \Rightarrow x=0: \\ y=4(0)^2=4*0 \\ y=0 \\ \Rightarrow x=2: \\ y=4(2)^2=4*4 \\ y=16 \end{gathered}$

Hence, its table is:

Now, let us plot the graphs of the functions:

Therefore, from the graph, we see that the order of the functions from widest to narrowest is: