Source data
Americans’ heights
The mean, 𝜇
The standard deviation, 𝜎
Men
69.0 inches
2.8 inches
Women
64.0 inches
2.8 inches
(You will find various different values
for these measurements at various
internet sites. These are typical.)
Excel Graph of this Normal Distribution
This was not easy to do. See xzBoth.xlsx
American men's heights
-4
-3
-2
-1
z-score
0
1
2
3
4
80.2
77.4
74.6
71.8
69
66.2
63.4
60.6
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
57.8
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
x value = Height (inches)
Notice that there are TWO x-axes, TWO horizontal axes:
x-to-z and z-to-x conversions
If you know the x-value (height in inches) and you
want to convert it to a z-score (“standard score”)
𝑥−𝜇
𝑧=
𝜎
 Subtract: the 𝑥 value, minus the mean, 𝜇.
 Then divide by the standard deviation, 𝜎.
If you know the z-score and you want to convert it
to an x-value of height, in inches:
𝑥 =𝑧∙𝜎+𝜇


Multiply 𝑧 times the standard deviation, 𝜎.
Add the mean, 𝜇.
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D.R.S.
Solution Methods
If using the printed tables of Areas of the Standard Normal Distribution
Convert all 𝑥 values to 𝑧 values. Work the problem as a 𝑧 problem.
If using TI-84 or Excel
Special functions are available to work in terms of the 𝑥 values of the mean and standard deviation.
TI-84 tools
The menu with these functions isUse 2ND DISTR (on the VARS key. Use −1𝐸99 (2ND Comma for EE, which means × 10𝑝𝑜𝑤𝑒𝑟 ) for −∞ and 1𝐸99 for ∞.
Function to use
normalcdf(low x, high x, mean, stdev)
What it tells you
Area between the two x-values
Similar to z-problem function
=normalcdf(low z, high z)
Excel functions
These functions for working with area (probability) in terms of x-values are similar to the ones we’ve used for area (probability) in terms of z-values.
Function to use
What it tells you
=NORM.DIST( x-value, mean, stdev, TRUE)
Area to the left of that x-value
=NORM.INV(probability area, mean, stdev)
What x-value has that area to its left.
(For Excel 2007 and earlier, omit the periods from the function names.)
Similar to z-problem function
=NORM.S.DIST(z-value)
=NORM.S.INV(probability area)
Examples
TI-84 solution
Problem
“What percent of American men are shorter than 5’ 6’’ (66
inches)?”
Excel solution
=NORM.DIST(66, 69, 2.8, TRUE)
gives area to the left of x = 66.
Find the area under the curve to the left of x = 66.
Answer: 0.1420, or 14.20%
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D.R.S.
TI-84 solution
Problem
“What percent of American men are 6’4’’ (76 inches) or taller?”
Excel solution
= 1-NORMDIST(76, 69, 2.8, TRUE)
Find the area under the curve to the right of x = 76.
Subtract 1 minus the area to the LEFT
of x = 76 to get the area to the RIGHT
of x = 76.
= NORMDIST(70, 69, 2.8, TRUE)
- NORMDIST(68, 69, 2.8, TRUE)
Answer: .0062, or 0.62%, less than 1%.
“What percent of American men are within an inch of the mean
height, that is, between 68 and 70 inches tall?”
Find the area under the curve between x = 68 and x = 70.
Answer: 0.2790, or 27.90%
“What is the range of heights of the middle 25% of American
men?”
=NORM.INV(0.3750, 69, 2.8)
and
=NORM.INV(0.6250, 69, 2.8)
1 - 0.2500 area in the middle = 0.7500 area in the tails,
0.7500 div by 2 = 0.3750 in each tail
The low endpoint has 0.3750 area to its left.
The right endpoint has the 0.3750 left endpoint + the 0.2500
width = 0.6250 area to its left.
Answer: 68.1 to 69.9 inches (5’ 8.1’’ to 5’ 9.9’’)
“What is the height that separates the top 10% from everybody
else?”
=NORM.INV(0.9000, 69, 2.8)
Answer: 72.6 inches, or 6’ 0.6’’, about 6 ft + ½ inch tall
If the bottom quartile of Midstate League players are under 6’ 2’’ =NORM.S.INV(0.2500)
(74 inches) tall, and the mean height is 6’ 4’’ (76 inches), what is =(74-76)/NORM.S.INV(0.2500)
the standard deviation of all the players in the league?
Or
What z-score separates the bottom 25% from the top 75%? It’s 𝑥−𝜇
0.6745. Then plug in to 𝑧 = 𝜎 or 𝑥 = 𝑧 ∙ 𝜎 + 𝜇 and solve for
unknown 𝜎. (74 – 76) / -0.6745 = 2.97 or 3.0 inches
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D.R.S.