Question

4. Replacement times for TV sets are normally distributed with a mean of 8.2 years and a standard deviation of 1.1 years. (Change the final answer to a % and keep 2 decimal places) a) Find the probability that a randomly selected TV set will have a replacement time between 9.5 and 10.5 years. (Include diagram) b) Find the probability that 35 randomly selected TV sets will have a mean replacement time less than 8.0 years. (Include diagram)

Answer 1

Solution :

Given that ,

mean =   = 8.2

standard deviation = = 1.1

P(9.5< x <10.5 ) = P[(9.5-8.2) / 1.1< (x - ) / < (10.5-8.2) /1.1 )]

= P( 1.18< Z <2.09 )

= P(Z <2.09 ) - P(Z <1.18 )

Using z table   

=0.9817-0.881

probability= 0.1007

answer=10.07%

(B)

n = 35

= 8.2

= / n = 1.1 / 35 = 0.1859

P( <8.0 )

= P[( - ) / < (8.0 -8.2) / 0.1859]

= P(z <-1.08 )

Using z table

= 0.1401

probability= 0.1401

answer=14.01%