Compare your answers.

  1. The heights of 1567 Douglas fir trees were measured to have a mean of 126.8 ft and a standard deviation of 43.6 ft.

      1. Assuming this data is normal, determine the percent of trees that you would expect to be shorter than 24 ft.

    Sketch a normal curve to represent the problem.

    The unknown area represents the percentage of trees that are shorter than 24 ft. Determining the z-score will help you to determine the unknown area.

    Now use technology or a z-score table to determine the area.

    The area to the left of a z-score of -2.36 is 0.0091, so approximately 0.91% of the trees will be shorter than 24 ft.

      1. Assuming the data is normal, determine the percentage of trees you expect to be taller than 241 ft.

    Sketch a normal curve to represent the problem.

    The unknown area represents the percentage of trees that are taller than 241 ft. Determining the z-score will help you to determine the unknown area.

    Now use a z-score table or technology to determine the unknown area. Using technology, the area is 0.0044. Approximately 0.44% of the trees will be taller than 241 ft.



      1. The shortest and tallest trees measured were 24 ft and 241 ft. Based on the information provided, is it reasonable to treat this data as normal? Explain.

    The data is spread from nearly three standard deviations below the mean to nearly three standard deviations above the mean since 24 ft corresponds to a z-score of -2.36 and 241 ft corresponds to a z-score of 2.62. This is consistent with data that is normally distributed. Also, the mean is approximately half way between the maximum and minimum values, suggesting the data is fairly symmetrical. This data appears to be fairly normal.

    Note that no data set is exactly normal and that care needs to be taken when interpreting data as normal. For example, the normal model of the Douglas fir trees suggests there is a small probability that a tree will have a negative height since a z-score of -3 corresponds to a tree height of -4 ft. Clearly this is not possible and illustrates a difference between the normal curve and an actual distribution.

  2. Suppose a set of data is normally distributed with a mean of 43 and a standard deviation of 12. Determine an upper limit and a lower limit that would encompass the middle 90% of the data.

    Draw a diagram to represent the problem.



    The goal is to determine values x1 and x2 such that 90% of the data lies between them. The question asks for the middle 90%, so 5% of the data should lie below x1 and 5% of the data should lie above x2.



    In other words, x1 has 5% of the data below it and x2 has 95% of data below it.




    Use technology or the z-score tables to determine the corresponding z-scores.



    The areas 0.0495 and 0.0505 both appear in the table and are equidistant from 0.05 or 5%. Using either z-score or averaging the two is reasonable.

    z1 = -1.645



    Similarly, z2 = 1.645

    These z-scores can now be used to determine the x-values.


    The middle 90% of data will fall between data values of approximately 23.3 and 62.7.

 

For further information about z-scores, see pp. 283 - 291 of Principles of Mathematics 11.

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