Suppose that the value of a stock varies each day from $10.22 to $40.58 with a uniform distribution. Given that the stock is greater than $20, find the probability that the stock is more than $34. Give your answer accurate to four decimal places.

0.3197 = 31.97% probability that the stock is more than $34.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X lower than x is given by the following formula.

For this problem, we have that:

Suppose that the value of a stock varies each day from $10.22 to $40.58 with a uniform distribution.

This means that

Given that the stock is greater than $20

This means that we can use

Find the probability that the stock is more than $34.

Either the stock is 34 or less, or it is more than 34. The sum of the probabilities of these events is decimal 1. So

We want P(X > 34).

0.3197 = 31.97% probability that the stock is more than $34.

the first dot plot best represents the data

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let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!