1 A company knows that replacement times for the DVD players

1. A company knows that replacement times for the DVD players it produces are Normally distributed with a mean of 8.3 years and a standard deviation of 1.9 years.
Find the proportion of a randomly selected DVD players that will have a replacement time less than 3.9 years?
P(X < 3.9 years) =

4 decimal places.

If the company wants to provide a warranty so that only 2.6% of the DVD players will be replaced before the warranty expires, what is the time length of the warranty?
warranty =  years

To 1 decimal place.

B. A population of values has a normal distribution with =223.7 and =56.9. You intend to draw a random sample of size n=244.

Find the probability that a single randomly selected value is between 217.5 and 234.6.
P(217.5 < X < 234.6) =  Round to 4 decimal places.

Find the probability that the sample mean is between 217.5 and 234.6.

P(217.5 < X¯¯¯ < 234.6) =  Round to 4 decimal places.

Solution

1)
Normal Distribution
Mean ( u ) =8.3
Standard Deviation ( sd )=1.9
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P(X < 3.9) = (3.9-8.3)/1.9
= -4.4/1.9= -2.3158
= P ( Z <-2.3158) From Standard Normal Table
= 0.0103                  

P ( Z > x ) = 0.026
Value of z to the cumulative probability of 0.026 from normal table is 1.94
P( x-u/ (s.d) > x - 8.3/1.9) = 0.026
That is, ( x - 8.3/1.9) = 1.94
--> x = 1.94 * 1.9+8.3 = 11.9917 ~ 12 Years