Transcript Chapter 6 Slides

INTRODUCTORY STATISTICS
Chapter 6 THE NORMAL DISTRIBUTION
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FIGURE 6.1
If you ask enough people about their shoe size, you will find that your graphed data is
shaped like a bell curve and can be described as normally distributed.
(credit: Ömer Ünlü)
A NORMAL DISTRIBUTION
NORMAL DISTRIBUTION AND THE 68-95-99.7 RULE
The mean, µ, is in the center. In general,
•
68% of data falls within one standard deviation
•
95% of data falls within two standard deviations
•
99.7% of data falls within three standard deviations
Z-SCORES
Z-scores are used to evaluate how far something is from the mean
with normal distributions often.
Recall: a value’s z-score =
𝑣𝑎𝑙𝑢𝑒−𝜇
𝜎
Example:
If heights are normally distributed with a mean of 67 inches and a
standard deviation of 4, calculate the z-scores for the following
heights:
a) 62 in.
b) 70 in.
c) 75 in.
Which height would be considered the most unusual?
EXAMPLE:
Scores on a standardized test are normally distributed with a mean of
54 and a standard deviation of 7. If X = a student’s score, then we
would write 𝑋~𝑁(54,7). Calculate the z-scores for the following
results:
a) 35
b) 77
c) 90
d) You received an 62 on the particular test. Your parents are upset.
Justify mathematically to them why your score isn’t so bad.
PRACTICE:
Olivia’s school has an average GPA of 3.0 with a 0.7 standard
deviation. She has a GPA of 3.2. Sam’s school has an average GPA
of 2.7 with a 0.3 standard deviation. Sam has a GPA of 2.9. Who has
the best GPA and why?
SECTION 6.3: USING THE NORMAL DISTRIBUTION
Distribution is written as X~𝑁(𝑚𝑒𝑎𝑛, 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛)
TOTAL AREA ADDS UP TO 1
Probability to the right is always 1 - P(X<x)
Ex. P(X>65) = 1- P(X<65)
USING A CALCULATOR TO COMPUTE PROBABILITY
Go to 2nd Dist and choose normalcdf(
Enter normalcdf(lower value, uppervalue, mean, standard deviation)
Ex. For an exam, scores are normally distributed with a mean of 71 and
standard deviation of 5. Calculate:
a)
The probability of scoring less than 60.
b)
Normalcdf(0,60,71,5)=0.139
The probability of scoring less than 80.
Normalcdf(0,80,71,5)=0.964
c)
The probability of scoring more than 80.
1-0.964=0.036 or Normalcdf(80,1EE99,71,5)=0.036
d)
The probability of scoring more than 90
Normalcdf(90,1EE99,71,5)=7.237x10-5 (which is practically 0)
USING A CALCULATOR TO FIND PERCENTILES
The invNorm( function allows you to find a percentile value.
Enter invNorm(area to the left, mean, standard deviation)
Ex. In this example, find the 90th percentile if the mean is 63 and the standard deviation is 9. (𝑋~𝑁(63,9))
invNorm(0.9,63,9)=74.534
EXAMPLE
Heights for a school are normally distributed with a mean of 65 inches
and a standard deviation of 7inches.
a) Write the distribution
b) Find the probability that a student is over 70 inches. Sketch the
graph.
c) Find the probability that a student is under 63 inches. Sketch the
graph.
d) Find the 80th percentile of heights.
EXAMPLE: THE MEAN IS 2 AND THE STANDARD
DEVIATION IS 0.3
Find P(1.8<X<2.75)
STADIUM CAPACITIES