Download MM207 Unit 5 Chapter 5 Project Key Form 7

MM207 Unit 5 Sample Problems
Similar Examples:
1. Use the Standard Normal Distribution table to find the indicated area under the standard
normal curve.
(References: example 3 - 6 pages 244 - 247, end of section exercises 21 – 40 page
249)
a.
Between z = 0 and z = 1.85
Solution: Refer to Table 4 – Standard Normal Distribution in the text.
For z = 0 the area to the left is 0.5000
Or, using Excel, NORMSDIST(0) = 0.5000
For z = 1.85 the area to the left is 0.9678
Or, using Excel, NORMSDIST(1.85) = 0.9678
0.9678 – 0.5000 = 0.4678
b.
To the left of z = -2.13
Solution: Refer to Table 4 – Standard Normal Distribution in the text.
For z = -2.13 the area to the left is 0.0166
Or, using Excel, NORMSDIST(-2.13) = 0.0166
c.
Between z = 0.43 and z = 1.68
Solution: Refer to Table 4 – Standard Normal Distribution in the text.
For z = 0.43 the area to the left is 0.6664
For z = 1.68 the area to the left is 0.9535
0.9535 – 0.6664 = 0.2871
d.
To the right of z = 2.07
Solution: Refer to Table 4 – Standard Normal Distribution in the text.
For z = 2.07 the area to the left is 0.9808
1 – 0.9808 = 0.0192
Similar Example:
2. The thicknesses of widgets produced by a new machine are normally distributed
with a mean of 0.72 cm and a standard deviation of 0.02 cm. What percent of the
widgets will have a diameter less than 0.75 cm?
Solution: P(x < 0.75) =
x
0.75  0.72 


P z 
  P z 
  P ( z  1.5)
 
0.02 


Refer to Table 4 – Standard Normal Distribution in the text. For z = 1.5 the area to the
left is 0.9332 or 93.32%.
Or, using Excel, NORMDIST(x, mean, standard dev, cumulative) =
NORMDIST(0.75,0.72,0.02,TRUE) = 0.9332 or 93.32%.
Similar Example:
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MM207 Unit 5 Sample Problems
3. For a certain IQ test, scores are normally distributed. The mean is 100 and the
standard deviation is 15. Find the probability that an individual will have an IQ score
less than 125.
Solution: P(x < 125) =
x
125  100 


P z 
  P z 
  P ( z  1.67)
 
15



Refer to Table 4 – Standard Normal Distribution in the text. For z = 1.67 the area to
the left is 0.9525.
Or, using Excel, NORMDIST(x, mean, standard dev, cumulative) =
NORMDIST(125,100,15,TRUE) = 0.9522.
Similar Examples:
4. Answer the questions about the specified normal distribution. (References:
example 4 and 5 page 264 - 265, end of section exercises 39 – 45 pages 267
- 268)
a.
Heights of young women are normally distributed with a mean of 64 inches and a
standard deviation  of 2.7 inches. Find the height that represents the 20th
percentile.
Solution: Refer to Table 4 – Standard Normal Distribution in the text.
z for P20 is -0.84.
To convert z to x use x =  + z = 64 + (-0.84)(2.7) = 61.7 inches.
Or, using Excel, NORMINV(probability, mean, standard dev) =
NORMINV(.20,64,2.7) = 61.7 inches.
b.
Scores on the SAT test for one year were normally distributed with a mean of
1025 and standard deviation  of 205. Find the score that marks the top 10%.
Solution: For the top 0.10, the area to the left is 1.00 – 0.10 = 0.90.
Refer to Table 4 – Standard Normal Distribution in the text. z = 1.28
To convert z to x use x =
 + z = 1025 + (1.28)(205) = 1287.
Or, using Excel, NORMINV(probability, mean, standard dev) =
NORMINV(.90,1025,205) = 1288.
Similar Examples:
5. Find the probabilities.
(References: example 5 and 6 page 276 - 277, end of section exercises 25 31 pages 280 - 281)
a.
Scores on the SAT test for one year were normally distributed with a mean of
1025 and standard deviation  of 205. If the scores from 64 randomly selected
individuals are chosen, what is the probability that the mean score would be
greater than 1100?
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MM207 Unit 5 Sample Problems
Solution:




1100  1025 

P( x > 1100) = P z 
 P( z  2.93)


205


64


Refer to Table 4 – Standard Normal Distribution in the text. For z = 2.93 the area
to the left is 0.9983.
1 - 0.9983 = 0.0017
Or, using Excel, NORMDIST(x, mean, standard dev, cumulative) =
NORMDIST(1100,1025,205/SQRT(64),TRUE) = 0.9983.
1 - 0.9983 = 0.0017
b.
Heights of young women are normally distributed with a mean of 64 inches and a
standard deviation  of 2.7 inches. If the heights from 25 randomly selected
individuals are chosen, what is the probability that their mean height is less than
63 inches?
Solution:




63  64 

P( x < 63) = P z 
 P( z  1.85)

2.7 


25 

Refer to Table 4 – Standard Normal Distribution in the text. For z = -1.85 the area
to the left is 0.0322.
Or, using Excel, NORMDIST(x, mean, standard dev, cumulative) =
NORMDIST(63,64,2.7/SQRT(25),TRUE) = 0.0320.
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