Ch12.4 The Normal Distribution
Section 12.4 Notes Page 1
µ (Mu) to denote the mean,
σ ( sigma) to denote the standard deviation of a normal distribution.
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A small σ leads to a tall, narrow curve, because most of the data is close to the mean.
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A large σ means the data is very spread out, producing a flat, wide curve.
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The area of the shaded region under the normal curve from a to b is the probability that
an observed data value will be between a and b.
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This area have already been calculate for the normal curve with mean 0 and standard
deviation of 1 in the table 2 at the back of the textbook.(call standard normal curve) Also
the equation of the standard normal curve is
Ex1. Find the given areas under the standard normal curve. (Use table 2)
Section 12.4 Notes Page 2
a) The area between Z=0 and Z=1, the shaded region.
b) The area between Z= -2.43 and Z=0
Ex2. Find the percent of the total area for the given areas under the standard normal curve.
(Use table 2)
a) The area between 0.88 standard deviations below the mean and 2.35 standard
deviations above the mean ( that is, between Z=-0.88 and Z=2.35)
Section 12.4 Notes Page 3
b) The area between 0.58 standard deviations above the mean and 1.94 standard
deviation above the mean
c) The area to the right of 2.09 standard deviations above the mean
The Z-Score for x is the number of standard deviations that x lies from the mean.
(Positive if x is above the mean, negative if x is below the mean)
Section 12.4 Notes Page 4
Ex3. If a normal distribution has mean 60 and standard deviation5, find the given z-scores.
a) The z-score for x=65
b) The z-score for x=52.5
Ex4. Dixie Office Supplies finds that its sales force drives an average of 1200miles per month
per person, with a standard deviation of 150 miles. Assume that the number of miles driven by a
salesperson is closely approximated by a normal distribution.
a) Find the probability that a salesperson drives between 1200 and 1600 miles per month.
b) Find the probability that a salesperson drives between 1000 and 1500 miles per month.
Section 12.4 Notes Page 5
Ex5. With data from the 2005-2006 National Health and Nutritional Examination Survey, we
can use 196 (mg/dL) as an estimate of the mean total cholesterol level for all Americans and 44
(mg/dL) as an estimate of the standard deviation. Assuming total cholesterol levels to be
normally distributed, what is the probability that an American chosen at random has a
cholesterol level higher than 250?
If 200 Americans are chosen at random, how many can we expect to have total cholesterol
higher than 250?
Ex6. Lengths of pregnancies of women are normally distributed with a mean of 266 days and a
standard deviation of 16 days. What percentage of children are born from pregnancies lasting
more than 274 days?