Question

The life of Sunshine CD players is normally distributed with a mean of 4.2 years and a standard deviation of 1.3 years. A CD player is guaranteed for 3 years.

1. We are interested in the length of time a CD player lasts.

a. Find the probability that a CD player will break down during the guarantee period. P(0<x<____) = _______

b. Sketch and shade a normal curve to indicate the probability we are interested in. Be sure to label and scale the horizontal axis.

2. Some CD players last up to 6 years.

a. Find the probability that a CD player will last between 2.75 and 6 years.

b. Sketch and shade a normal curve as in question 1.b.

3. An executive wants to know the 90th percentile for CD player life.

a. Find the 90th percentile for the length of time a CD player lasts.

b. Sketch and shade a normal curve as in question 1.b.

4. The engineers want to know what is typical for these CD players.

a. Find the lower and upper bounds for CD player life for the middle 85% of CD players produced

b. Sketch and shade a normal curve as in question 1.b.

Answer 1

The life of Sunshine CD players is normally distributed with a mean ( µ ) = 4.2 years and a standard deviation (σ) = 1.3 years.

A CD player is guaranteed for 3 years

1) P( 0 < x < 3 ) = P( x ≤ 3 ) - P( x ≤ 0)

=   

= P( z ≤ -0.92 ) - P( z ≤ -3.23 )

= 0.1788 - 0.0006 ----- ( From z score table )

= 0.1782

#2) P( 2.75 ≤ x ≤ 6 ) = P( x ≤ 6 ) - P ( x ≤ 2.75 )

  

= P( z ≤ 1.38 ) - P( z ≤ -1.12 )

= 0.9162 - 0.1314 ----- ( From z score table )

= 0.7848

#3)  the 90th percentile for CD player life.

We have to find x such that area to the left is 0.90 , so first we need to find the z score such that area to the left is 0.90

So such z score is 1.28

x = (1.28*1.3) + 4.2

x = 5.864

#4) the lower and upper bounds for CD player life for the middle 85% of CD players produced

We have to find x₁ and x₂ such that area bounded between them is 0.85

The total area under the curve is 1 , then middle area between x₁ and x₂ is 0.85 therefore area to the left of x₁ is 0.075 and area to the right of x₂ is 0.075 [ since (1-0.85)/2 = 0.075 ]

z score corresponding to x₁ = -1.44

Therefore x₁ = (-1.44*1.3) + 4.2  

x₁ = 2.328

z score corresponding to x₂ = 1.44

Therefore x₂ = (1.44*1.3) + 4.2  

x₂ = 6.072

So  lower and upper bounds for CD player life for the middle 85% of CD players produced are ( 2.328 , 6.072 )