The tread life of a particular brand of tire is a random variable best described by a normal distribution with a mean of 60,000 miles and a standard deviation of 8,300 miles. What proportion of these tires last between 40,000 and 45,000 miles

P [ 40000 ≤ X ≤ 45000 ] =  0,02773    or  2,8 %

Step-by-step explanation:

P [ 40000 ≤ X ≤ 45000 ]  = ?

To find the probablity    P [ X ≤ 40000 ] =  ?

z(s)  = ( X - x ) / σ

z(s)= ( 40000 - 6000 )/ 8300

z(s) =  -20000/8300

z(s) ≈  - 2,41

Then

from z-table we find P [ X ≤ 40000] =  0,00820

Now again:

For  P [ X ≤ 45000 ]

z₂  = ( X - x ) / σ

z₂ =  ( 45000 - 6000 ) / 8300

z₂ = - 15000 / 8300

z₂ = - 1,8072

z₂ ≈ 1,81

From z - table

For  P [ X ≤ 45000 ] =   0,03593

Then:

P [ 40000 ≤ X ≤ 45000 ] = 0,03593 - 0,00820

P [ 40000 ≤ X ≤ 45000 ] =  0,02773    or  2,8 %

hi

step-by-step explanation:

the expression (1.26)2t reveals the approximate rate of increases in the number of plants semiannually.

hope this : )