Question

The EERI team builds multi-story balsa wood structures for a seismic competition. One major scoring criteria is the weight of the building (lighter buildings score higher). Balsa wood can be purchased in three different densities -- low, medium and high. The methodology the students are using to measure density has a standard deviation of 3 lb/ft³. It is assumed that density is normally distributed.

Published Population Statistics	Low Density lb/ft3	Medium Density lb/ft3	High Density lb/ft3
Mean	8	12	16

1. The team ordered what they thought was medium density wood. They analyzed 1 piece of wood and found it to have a density of 11 lb/ft³. What is the probability of measuring a density less than or equal to 11 if it were medium density wood?

Recognizing there is variability, they measure several additional pieces of wood. After measuring 3 total samples, the new average density was 10.45 lb/ft³.

2. What is the probability of an average value more extreme than this if it were low density wood?
3. What is the probability of an average value more extreme than this if it were medium density wood?
4. Recognizing that the average value is closer to medium density than low, the team assumes it is medium. How many samples would they have to measure to be certain (less than 1% chance) that it was not low density wood, assuming the sample average stays constant?

Answer 1

Let denote the medium density of wood with a standard deviation = 3lb / ft³

Assuming that the density is normally distributed, the standard normal estimate can be defined as:

We are asked to find the probability:

........(subject to rounding error)

From standard normal table:

= 0.38591

The probability of measuring a density less than or equal to 11, if it were a medium density wood = 0.386

(*To obtain the exact probability, without rounding the Z score to two decimal places, using excel:

= 0.37070

We get the probability = 0.37

2.

Given:  , n = 3, and

  

= 1.41

To find:

Pr(Z > 1.41)

From standard normal table,

= 1 - 0.92073

= 0.0793

The probability of an average value more extreme than this if it were low-density wood = 0.0793

3.

Given:  , n = 3, and

  

= -0.89

To find:

Pr(Z > -0.89)

From standard normal table,

= 1 - 0.18673

= 0.81327

The probability of an average value more extreme than this if it were low-density wood = 0.8133

4.Assuming 99% confidence interval:

Given: Margin of Error = 1%,

By definition of Margin of error,

For alpha = 0.01, Z_1-0.01/2 = Z_0.995 = 2.58

Substituting the values,

Squaring both sides,

= 59.908

Assuming 95% confidence interval:

Given: Margin of Error = 1%,

By definition of Margin of error,

For alpha = 0.05, Z_1-0.05/2 = Z_0.975 = 1.96

Substituting the values,

Squaring both sides,

= 34.574