A trimming operation at a local manufacturer produces rods whose length conforms to unifrom distribution with a minimum of 18cm and maximum of 33cm. What is the probability that a randomly selected rod will be at least 25.8 cm long?

The probability that randomly selected road will be at least 25.8 cm long will be 48%.

Step-by-step explanation:

Given: uniform distribution with min and max values of 18 and 33 respectively.

To find : probability density for upper cumulative frequency i.e 25.8 at least means x25.8  upto 33 cm i.e the maximum limit of function.

Solution:

we have by definition , of uniform distribution

we get , probability density function defines as :

F(x,a,b)=  

            =1/(33-18)=1/15=0.0667.

this is probability density function.

here the x=25.8 , a=18 and b=33

for lower cumulative frequency it defines as ;

P(x,a,b)= =25.8-18/33-18=0.52

for upper cumulative frequency it defines as ;

Q(x,a,b)=b-x/b-a=33-25.8/33-18=0.48

here at least 25.8 cm probability means it should be greater than a value(18cm) hence it is provided by the upper cumulative frequency

i.e. Q(x,a,b)=0.48

The probability that randomly selected road will be at least 25.8 cm long will be 48%.

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so: