Suppose the driving distance to work for residents of a certain community has a mean of 21 miles and a standard deviation of 3.6 miles. what is the probability that an individual drives between 19 and 26 miles to work? assume that the distribution is normal

Step-by-step explanation:

We start by defining the random variable X.

X : ''The driving distance to work for residents of a certain community''

X can be modeled as a normal random variable.

X ~ N (μ,σ)

Where μ is the mean and σ is the standard deviation.

For this problem :

X ~ N (21,3.6)

Where the unit for the mean and the standard deviation is miles.

We are looking for

Given a normal random variable,we can subtract it the mean and then divide by the standard deviation in order to obtain a new random variable ''Z''.

Where ''Z'' is a new normal random variable.This is called standardizing.

Z ~ N (0,1)

The mean of Z is 0 and the standard deviation is 1.

We can find the  cumulative probability distribution of Z in any table.

Φ(a) given a certain value ''a''.

For the problem :

Φ(1.3888) - Φ(-0.5555) = 0.9177 - 0.2877 = 0.63

The probability that an individual drives between 19 and 26 miles to work is 0.63

muyb ajo

step-by-step explanation:

jegosu

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2