1. Finding Probability Using the Normal Curve
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HAWKES LEARNING SYSTEMS
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Section 6.3
Finding Probability Using the Normal Curve

2. Probability less than some value P(X<-10)
HAWKES LEARNING SYSTEMS
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Continuous Random Variables
6.3 Finding the Probability Using the Normal Curve
Four basic types of probability problems:
Probability less than some value P(X<-10)
Probability greater than some value P(X>10)
Probability between two values P(-10<X<10)
Probability less than one value and greater than another value P(X<-10 and >10)

3. HAWKES LEARNING SYSTEMS
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Continuous Random Variables
6.3 Finding the Probability Using the Normal Curve
Determine the probability:
Deviation IQ scores, sometimes called Wechsler IQ scores, are scores with a mean of 100 and a standard deviation of 15. What percentage of the general population have IQ’s lower than 92?
Solution:
m = 100, s = 15, x = 92
P(z < -0.53)
= = 29.81%

4. HAWKES LEARNING SYSTEMS
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Continuous Random Variables
6.3 Finding the Probability Using the Normal Curve
Determine the probability:
Deviation IQ scores, sometimes called Wechsler IQ scores, are scores with a mean of 100 and a standard deviation of 15. What percentage of the general population have IQ’s larger than 130?
Solution:
m = 100, s = 15, x = 130
P(z > 2.00)
= = 2.28%

5. HAWKES LEARNING SYSTEMS
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Continuous Random Variables
6.3 Finding the Probability Using the Normal Curve
Determine the probability:
Deviation IQ scores, sometimes called Wechsler IQ scores, are scores with a mean of 100 and a standard deviation of 15. What percentage of the general population have IQ’s between 90 and 110?
Solution:
m = 100, s = 15, x1 = 90 and x2 = 110
P(-0.67 < z < 0.67)
= = 49.72%

6. m = 100, s = 15, x1 = 80 and x2 = 120 P(z < -1.33 or z > 1.33)
HAWKES LEARNING SYSTEMS
math courseware specialists
Continuous Random Variables
6.3 Finding the Probability Using the Normal Curve
Determine the probability:
Deviation IQ scores, sometimes called Wechsler IQ scores, are scores with a mean of 100 and a standard deviation of 15. What percentage of the general population have IQ’s less than 80 and greater than 120?
Solution:
m = 100, s = 15, x1 = 80 and x2 = 120
P(z < or z > 1.33)
= = 18.35%

7. HAWKES LEARNING SYSTEMS
math courseware specialists
Continuous Random Variables
6.3 Finding the Probability Using the Normal Curve
Determine the probability:
In a recent year, the ACT scores for high school students with a 3.50 to 4.00 GPA were normally distributed, with a mean of 24.2 and a standard deviation of A student who took the ACT during this time is selected. Find the probability that the student’s ACT score is less than 20.
Solution:
m = 24.2, s = 4.2, x = 20
P(z < -1.00)
= = 15.87%

8. HAWKES LEARNING SYSTEMS
math courseware specialists
Continuous Random Variables
6.3 Finding the Probability Using the Normal Curve
Determine the probability:
In a recent year, the ACT scores for high school students with a 3.50 to 4.00 GPA were normally distributed, with a mean of 24.2 and a standard deviation of A student who took the ACT during this time is selected. Find the probability that the student’s ACT score is greater than 31.
Solution:
m = 24.2, s = 4.2, x = 31
P(z > 1.62)
= = 5.26%

9. m = 24.2, s = 4.2, x1 = 25 and x2 = 32 P(0.19 < z < 1.86)
HAWKES LEARNING SYSTEMS
math courseware specialists
Continuous Random Variables
6.3 Finding the Probability Using the Normal Curve
Determine the probability:
In a recent year, the ACT scores for high school students with a 3.50 to 4.00 GPA were normally distributed, with a mean of 24.2 and a standard deviation of A student who took the ACT during this time is selected. Find the probability that the student’s ACT score is between 25 and 32.
Solution:
m = 24.2, s = 4.2, x1 = 25 and x2 = 32
P(0.19 < z < 1.86)
= = 39.33%

10. Std Deviation = 6
Z = (147 – 144) / 6 = 0.5
P (x > 147) = 1 – P(0.5) = 1 –
P (x > 147) =

12. Z = (630 – 600) / 20 = 1.5
P (x <= 630) = P(Z) =

14. P (1169 < x < 1400) = P(z2) – P(z1)
P (1169 < x < 1400) = P(1) – P(-1.31)
P (1169 < x < 1400) = –
P (1169 < x < 1400) =

16. P (61 < x < 159) = P(z2) – P(z1)
P (61 < x < 159) = P(1.3929) – P( )
P (61 < x < 159) = –
P (61 < x < 159) =