3.5 Applying the Normal Distribution (Z-Scores)

The Graph:

Review of the Normal Distribution

- it is symmetrical; the mean, median and mode are equal and fall at the line of symmetry- it is shaped like a bell, peaking in the middle and sloping down toward the sides. It approaches zero at the extremes- approximately 68% of the data is within one standard deviation of the mean- approximately 95% of the data is within two standard deviations of the mean- approximately 99.7% of the data is within three standard deviations of the mean

- the area under any normal curve equals 1.

Properties:

Note: the percent of data between two values equals the area under the curve between those two values.

The Notation:

X~N(10, 22)

This normal distribution would have a mean of 10, a standard deviation of 2 and a variance of 4

Example:

How can you use the normal curve to accurately determine the percent of data that lies above or below a given value? How can the normal curve be used to compare two different data from two different data sets?

You must use the standard normal distribution

0 1-3 -2 -1 2 3

The Standard Normal Distribution

The standard normal distribution is a special normal distribution with a mean of 0 and a standard deviation of 1.

Each element of a normal distribution can be translated to the same place on a standard normal distribution by determining the number of standard deviations a given score lies away from the mean (z-score)

z-score: the number of standard deviations a given piece of data is above or below the mean.

a) Show the position of x = 10, on a normal curve for the distribution

Example 1

b) For the distribution , determine the number of standard deviations x=10 is above or below the mean.

Note: a positive z-score indicates that the value lies above the mean and a negative z-score indicates that the value lies below the mean.

c) Translate the value x=10 from the normal distribution to the same place on the standard normal distribution using the z-score.

Note: This graph shows how the value x = 10 is 1.2 standard deviation below the mean. This value is in the same position on this graph as it was for the normal distribution in part a).

Example 2For the distribution X~N (14, 42), determine the number of standard deviations each piece of data above or below the mean.

a) x = 11 b) x = 21.5

This piece of data is 0.75 standard deviations below the mean.

This piece of data is 1.875 standard deviations above the mean

Caley scored 84% in her Data Management course, while Lauren, who attends a different Data Management class, scored 83%. If Caley's class average is 74% with a standard deviation of 8, and Lauren's class average is 70% with a standard deviation of 9.8, use z-scores to determine who has the better mark.

Note: z-scores are used to standardize the data so that they can be accurately compared.

Example 3

Caley Lauren

Conclusion: Lauren's result is 1.327 standard deviations above the mean while Caley's is 1.25 standard deviations above the mean. Lauren's result is slightly better.

Z-Score Table

The z-score table on pages 398 & 399 is used to find the proportion (percentage) of data that has an equal or lesser z-score than a given value.

How to use the z-score table:

If Patrick's error total in keyboarding class is 2.43 standard deviations below the mean, then his z-score is -2.43. To find the proportion of data that has a lower (or equal) z-score, find the value -2.43 on the z-score table.

The z-score table states that only 0.75% of a normal distribution has a lower z-score. Therefore 99.25% has a higher z-score.

This proportion (percentage) of data with a lower z-score is equal to the percentile the value can be categorized by.

Percentile: the kth percentile is the least data value that is greater than k% of the population.

Percentile

Example:

- for example, the 84th percentile is the value where 84% of the data is below that value.

Don't forget about quartiles:

- Q1 is the 25th percentile- Q2 is the 50th percentile- Q3 is the 75th percentile

0 1 2 3-1-2-3

Example 4

Perch in a lake have a mean length of 20cm and a standard deviation of 5cm. Find the percent of the population that is less than or equal to the following lengths (the percentile).

a) 22cm

Of the fish in this lake, 65.54% are 22cm long or less. This fish is in the 66th percentile.

normalcdf(-1E99, 0.4) or normalcdf(-1E99, 22, 20, 5)

0 1 2 3-1-2-3

b) 16 cm

0 1 2 3-1-2-3

Of the fish in this lake, 21.19% are 16cm long or less. This fish is in the 21st percentile.

normalcdf(-1E99, -0.8) or normalcdf(-1E99, 16, 20, 5)

c) 28 cm

Of the fish in this lake, 94.52% are 28cm long or less. This fish is in the 95th percentile.

normalcdf(-1E99, 1.6) or normalcdf(-1E99, 28, 20, 5)

0 1 2 3-1-2-3

d) 4 cm

Note: If z < -2.99, the percentile = 0%

If z > 2.99, the percentile = 100%

Of the fish in this lake, 0% are 4cm long or less.

0 1 2 3-1-2-3

Using TI-83 for Normal Distributions

Go back and try Example 4 using the ti-83 calculator

0 1-3 -2 -1 2 3

Example 5Using the normal distribution X~N(7, 2.22), find the percent of data that is within the given intervals:

a) 3 < x < 6

Note: by subtracting the percentages obtained from the z-score table, you obtain the percentage of the data that fills the interval.

normalcdf(-1.82, -0.45) ornormalcdf(3, 6, 7, 2.2)

b) 7 < x < 15

0 1-3 -2 -1 2 3

normalcdf(0, 3.64) ornormalcdf(7, 15, 7, 2.2)

Complete worksheet