PLEASE HELP!! The batteries from a certain manufacturer have a mean lifetime of 860 hours, with a standard deviation of 70 hours. Assuming that the lifetimes are normally
distributed, complete the following statements.
(a) Approximately of the batteries have lifetimes between 790 hours and
930 hours.
Example
(b) Approximately 95% of the batteries have lifetimes between hours and
hours.

A) 99.74%

B) 633.6 hours and 986.4 hours

Step-by-step explanation:

Solution :

Given that ,

mean =  u =810

standard deviation = sigma  = 90    

P(540< x <1080 ) = P[(540-810) /90 < (x - u   ) /  sigma < (1080-810) /90 )]

= P( -3< Z < 3)

= P(Z <3 ) - P(Z < -3)

Using z table    

=0.9987-0.0013

= 0.9974

answer=99.74%

(B)

middle 95% of score is

P(-z < Z < z) = 0.95

P(Z < z) - P(Z < -z) = 0.95

2 P(Z < z) - 1 = 0.95

2 P(Z < z) = 1 + 0. 95= 1.95

P(Z < z) = 1.95/ 2 = 0.975

P(Z <1.96 ) = 0.975

z  ±1.96    

Using z-score formula  

x= z * sigma  +  u

x= -1.96* 90+810

x= 633.6

Using z-score formula  

x= z * sigma  +  u

x=1.96* 90+810

x= 986.4

633.6 hours and 986.4 hours

7

step-by-step explanation:

i dont have an answer

step-by-step explanation: