Nonstandard Normal Distributions

Here are nonstandard density curves with a mean equal to 15 and various standard deviations.

Model Problem¹

U.S. Air Force once used ACES-II ejection seats designed for men weighing between 140 lb and 211 lb. Given that women’s weights are normally distributed with a mean 171.1 lb and a standard deviation of 46.1 lb (based on data from the National Health Survey), what percentage of women have weights that are within those limits? Were many excluded from those past specifications?

Model Solution when Using z-tables

Given: Distribution Statement: heightW ~ N(mean = 171.1lb.,sd = 46.1lb.) and heightW boundaries: 140lb and 211lb

Objective(s): a. The proportion of women’s heights within the boundaries.
b. Were many women excluded from those past specifications?

Solution Plan

1. We will sketch and label nonstandard distribution of women’s heights, (heightW) and include distribution statement as its title.
2. Next, we will calculate z-scores for the boundaries and determine the area under density cure within these limits.

3. Finally, we will provide a written statement regarding obtained results.

1. Solution

Lower boundary: x=140
Upper boundary: x=211

2. Model solution when using z-tables

Now we will convert boundaries x=140 and x=211 to respective z-scores:

x = 140 corresponds to z = -0.67
x = 211 corresponds to z = 0.84

The area under the density curve within the limits is found from z-table it is equal to 0.5564. The cumulative area under both tails is 1 – 0.5564=0.4436

3. Statements about obtained results

3a. Just about 55.64% of women met past weight requirements.

3b. About 44.00% of women were excluded from past weight specifications.

Model Solution when Using R for Calculations

For part a. we will use: pnorm(x,mean,sd) code in the following way:

pnorm(211, mean=171.1, sd=46.1)-pnorm(140,mean=171.1, sd=46.1)

## [1] 0.556662

For part b we will use: code, 1- the previous answer:

1-0.556662

## [1] 0.443338

However, we may use more fancy code, (cut/paste) as follow:

1 -(pnorm(210, mean=171.1, sd=46.1)-pnorm(140,mean=171.1, sd=46.1))

## [1] 0.4493441

Assignment 3

Extra Credit: #29 and #33 from page: 252-253. Note: Follow model solution, part; (1), (2), and (3) in order to obtain full credit.

Extracted from M.F Triola, Essential s of Statistics Sixth Edition, Pearson. Essentials of Statistics page: 252, #27↩

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1. Extracted from M.F Triola, Essential s of Statistics Sixth Edition, Pearson. Essentials of Statistics page: 252, #27↩

The Normal Distribution

Here is the Standard Normal Distribution Density Curve is also known as the Bell Curve.

How do we find the area under the Standard Normal Density Curve?

Example 1

We will use the following syntax:

pnorm(0,mean=0,sd=1)

## [1] 0.5

Example 2

How do we find the area under the Standard Normal Density Curve to the left of z = 1.25?

pnorm(1.25, mean=0,sd=1)

## [1] 0.8943502

Example 3

How do we find the area under the Standard Normal Density Curve to the right of z = 1.5?

We can use the following syntax:

1 - pnorm(1.5, mean=0, sd=1)

## [1] 0.0668072

Alternatively the code below

pnorm(1.5, mean=0, sd = 1, lower.tail = FALSE)

## [1] 0.0668072

Example 4

How do we determine the area under the Standard Normal Curve between two z-values, (eg. z = -1.25 and z = 1.25)?

We can calculate the difference of two areas as follows:

pnorm(1.25, mean = 0, sd = 1) - pnorm(-1.25, mean = 0, sd = 1)

## [1] 0.7887005

Summary:

The following code pnorm(z, mean, sd) provides the area under the Standard Normal Density Curve to the left of z-value.

Assignment 2

The bone density test scores follow standard normal distribution for children, healthy premenopausal young women, and men under 50. Assume that a randomly selected person is subjected to a bone density test. Find the probability that this person has a score:

1. z < -2.5
2. -2.5 < z < -1
3. z > -1.

Note: For each part: a, b and c include distribution statement, density curve with shaded area. Also, use a full and complete sentence as the answer to a, b, and c part.
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