Transcript The Standard Normal Distribution

The Standard Normal
Distribution
Standard Normal Density Curve
• The standard normal distribution
has a mean = 0 and a standard
deviation = 1
• Values on the x-axis are the
number of standard deviations
from the mean, called z-scores
• Interactive Applet:
www.whfreeman.com/sta
• What percent of the
observations lie within 1
standard deviation?
• What if we wanted to know the
percent of observations that lie
within 1.5 standard deviations?
Using z-score table
• The area to the left of the
corresponding z- score is the
percentage of observations, P,
that fall below the z-score.
• 1. Use the table to find the
percentage of values below
z = -1.2.
• 11.51%
• 2. What is P(z > -1.2)?
• 100 – 11.51 = 88.49%
• 3. What is P(-1.2 < z < 1.2)?
• 88.49 – 11.51 = 76.98%
Using z-scores with a calculator
• Use the normalcdf command
to compute probabilities using
the normal curve.
• Press 2nd, DIST,
normalcdf(lower boundary,
upper boundary). Use -10 or
10 for ± infinity.
• For example, to find P(z < 1) =
normalcdf(-10, 1)
• 1. Use the calculator to find the
percentage of values below
z = -1.2.
• 2. What is P(z > -1.2)?
• 3. What is P(-1.2 < z < 1.2)?
Normalcdf(-10,-1.2)= .1151 = 11.51%
Normalcdf(-1.2, 10)= .8849 = 88.49%
Normalcdf(-1.2,1.2)= .7699 = 76.99%
(z- scores)
Z=
A positive z- score indicates the number of standard
deviations above the mean
A negative z- score indicates the number of standard
deviations below the mean.
Using the normal curve to find percents
• Find percent below (or the percentile of) a certain value
– Use the formula to find the z-score and use table or use calculator:
normalcdf with lower boundary of -10, upper is z
• Find percent above a certain value
– Use the formula to find the z-score and subtract table value from
100% or use calculator: normalcdf (lower boundary is z, upper is 10)
• Find percent between two certain values
– Use the formula to find two z-scores, one for each value
– Use the table to find two percents and subtract them from each other
….or….
– Use calculator: normalcdf(lower is 1st z, upper is 2nd z)
SAT Example
•
•
The approximately normal distribution of SAT scores
for the 2009 incoming class at the University of Texas
had a mean 1815 and a standard deviation 252.
What is the z- score for a University of Texas student
who got 1901 on the SAT?
– Z = 1900 – 1815 = 0.34
252
What percentage did better than this student?
– P(z > .34) = Normalcdf(.34, 10) = .366 or 36.6%
What percentile was a student
at if he made a 2200?
z = (2200-1815)/252 = 1.53
normalcdf(-10,1.53) = 94%
Heights Example
•
The heights of 18 to 24 year old males in the US are
approximately normal with mean 70.1 inches and standard
deviation 2.7 inches. The heights of 18 to 24 year old females
have a mean of 64.8 inches and a standard deviation of 2.5
inches.
•
Estimate the percentage of US males
between 18 and 24 who are 6ft tall or taller.
–
–
•
Z = 72 – 70.1= .7037
2.7
P(z > .7037) = normalcdf(.7037, 10) = .2408 = 24.08%
Estimate the percentage of US females between 18 and 24 who
are between 5 feet and 5’5”.
–
Z = 60 – 64.8 = -1.92
z = 65 – 64.8 = .08
2.5
2.5
P(-1.92 < z < .08) = normalcdf(-1.92,.08) = .5044 = 50.44%
Comparison Example
•
Your friend Mark scored a 43 on Mrs. Johnson's
chemistry test. The mean score in her class was a 38
and the standard deviation was 4. Marshall scored a
67 on his AP Calculus test and the mean in that class
was a 65, with a standard deviation of 7. Whose score
was better, relative to their classes?
–
–
Mark: z = (43 – 38)/4 = 1.25
P(z < 1.25) = normalcdf(-10, 1.25) = .894 = 89.4th percentile
–
–
Marshall: z = (67 – 65)/7 = .286
P(z < .286) = normalcdf(-10, .286) = .613 = 61.3rd percentile
–
Mark did better because he was
further above the mean than Marshall
and was in the 89th percentile
compared to Marshall who was only
in the 61st percentile.
Born to Run…
• A study of elite distance runners found a mean body
weight of 139.1 pounds with a standard deviation of 10.6
pounds. Assuming the distribution of weights is
approximately normal, make a sketch of the weight distribution.
• What percent of runners would have a body
weight between 107.3 and 149.7 lb? 84%
• 16% of runners would have a body weight
107.3 117.9 128.5 139.1 149.7 160.3 170.9
less than how many pounds? 128.5 pounds
• What percent of runners would have a body weight less than 130
pounds? Z = (130-139.1)/10.6 = .858; normalcdf(-10, -.858) = 19.5%
• What percent of runners would have a body weight greater than 160
pounds? Z = (160-139.1)/10.6 = 1.97; normalcdf(1.97,10) = 2.4%