Welcome to Chapter 6!

The standard Deviation (SD) as a Ruler and

The Normal Model

Why should we learn this chapter?

  • To describe quantitative data with a normal model
  • To compare “apples and oranges”with the same scale
    (Compare different normal distributions using Z score)
  • To make confident conclusions about real-life data

Applications in the real world:

The midterm scores for a random SSU class can be described by a Normal model with a mean of 70 and standard deviation of 10(out of 100 points).

  1. You can find the percent in any interval:
  1. What % of students has a score between 60 and 90?
  2. What % of students is below 60?
  3. What % of students is above 90?
  1. You can find the score needed for any percentile:
  1. If a student wants to be better than 95% of the class, what score does (s)he need?
  2. What score is at the90th percentile of the class?
  3. What score is worse than60% of the class?

What are the key points to learn?

The normal distribution model

The 68-95-99.7 rule

Check whether a distribution is normal

The effect of shifting a quantitative variable

The effect of rescaling a quantitative variable

To calculate Z-score

To make conclusions with Z-score

How do we accomplish them?

Picture! Picture! Picture!
Draw and label the normal distribution model.

The 68-95-99.7 Rule only applies to normal data.

68-95-99.7 are the percents within 1-2-3 standard

deviations from the mean.

Check whether a quantitative variable is normal with the
Normal Probability Plot. If the plot fits a line, it is normal.

See the effect of shifting a quantitative variable:
When we add or subtract a constant to all values in a data set, only its Center is shifted, but its Shape and Spread are not changed.

See the effect of rescaling a quantitative variable:
When we multiply or divide a constant to a data set, its Shape, Center, and Spread all rescaled.

Compare “apples and oranges” (different normal distributions) by putting them on the same scale: Z score.

What is the Z-score of any data point?
Z-score measures its distance from the mean; it is the number of standard deviations away from the mean.

Calculate Z-score:
Z = (data value - mean)/standard deviation

Make confident decisions about normal model:

Applications in the real world:

The midterm scores for a random SSU class can be described by a Normal model with a mean of 70 and standard deviation of 10(out of 100 points).

  1. Find the percent in any interval for any normal data:
  1. What % of students has a score between 60 and 90?

Since the scores ~ Normal (70, 10), the answer is:
normalcdf (60, 90, 70, 10)=

  1. What % of students is below 60?

normalcdf (___,60, 70, 10)=

  1. What % of students is above 90?

normalcdf (90, ____, 70, 10)=

  1. Find the percentile for any normal data:
  1. If a student wants to be better than 95% of the class, what score does (s)he need?

invNorm(0.95, 70, 10) = the 95th percentile.

  1. What score is at the 90th percentile of the class?

invNorm(____, 70, 10) =

  1. What score is worse than60% of the class?

invNorm(____, 70, 10) = the _____ percentile.

Note:
The data given and numbers you enter are in green.

The TI calculator commands and outputs are in purple.

For any variable X ~ Normal (mean, standard deviation), to find the % of the distribution in any interval from low to high, denoted as (low, high),

Use the DIST (2nd, VARS, 2) function in TI:
normalcdf (low, high, mean, standard deviation) = %

If the high point is infinity, we use as many 9’s as needed to represent infinity. For almost all cases, we use 9999 for +infinity and – 9999 for –infinity.

Therefore, the % of X in (low, high) = normalcdf (low, high,mean, standard deviation)

And, the % of X < high= normalcdf (- 9999, high, mean, standard deviation)

And, the % of X > low = normalcdf (low, 9999, mean, standard deviation)

Given X ~ Normal (70, 10), what % of X is in (60, 90)? What % of X is <60? What % of X is >90?

2. To find the p percentile (the data value that is above p% of the distribution or p% of the distribution is below it),

Use the DIST (2nd, VARS, 3) function in TI:

invNorm(p, mean, standard deviation) = the p percentile.

Be sure to enter p as a decimal.

If a student wants to be above 95% of the class, what score does (s)he need?
invNorm(0.95, 70, 10) = the 95th percentile.

Practice what we learned: Either one or both of the following

  1. Textbook HW – Team exercise
  2. Pop Quiz

Page 1 Ai-Chu Wu, Ph. D. 9/29/2018