Unformatted text preview: nts on a Normal Curve 44 Example 5.3 #1
• Refer to the above figure.
• In panel (a), st is located on the horizontal axis under
the righttail of the normal curve having mean µ =
100 and standard deviation σ = 10
• The z value corresponding to st is
st − µ st − 100
z=
=
σ
10 • In panel (b), the righttail area is 0.05, so the
In
cumulative area under the standard normal curve to z
is 1 – 0.05 = 0.95
is
45 Example 5.3 #2 Use the standard normal table to find the z value associated with a table
Use
entry of 0.95
entry But do not find 0.95; instead find values that bracket 0.95 For a table entry of 0.9495, z = 1.64
For For a table entry of 0.9505, z = 1.65
For For an area of 0.95, use the z value midway between these
For So z = 1.645
So st − 100
= 1.645
10
Solving for st gives st = 100 + (1.645 × 10) = 116.45
Rounding up, 117 tapes should be stocked so that the
probability of running out will not be more than 5 percent
46 Finding a Tolerance Interval
Finding a tolerance interval [µ ± kσ] that contains 99%
of the measurements in a normal population 47 Normal Approximation to the Binomial( 连连连连连连连连连 ) • The figure below shows several binomial
distributions
• Can see that as n gets larger and as p gets
closer to 0.5, the graph of the binomial
distribution tends to have the symmetrical,
bellshaped, form of the normal curve 48 Normal Approximation to the Binomial 49 Normal Approximation to the Binomial
• Generalize observation from last slide for large p
• Suppose x is a binomial random variable, where n is
the number of trials, each having a probability of
success p
• Then the probability of failure is 1 – p
• If n and p are such that np ≥ 5 and n(1 – p) ≥ 5, then x
is approximately normal with µ = np and σ = np (1 − p )
50 Example 5.4
Example Normal Approximation to a Binomial #1
• Suppose there is a binomial variable x with n = 50
Suppose
and p = 0.5
• Want the probability of x = 23 successes in the n =
Want
50 trials
50
• Want P(x = 23)
Want
• Approximating by using the normal curve with
µ = np = ( 50 × 0.50 ) = 25 σ = np(1 − p ) = 50 × 0.50 × 0.50 = 3.5355
51 Normal Approximation to a Binomial #2 • With continuity correction, find the normal probability
P(22.5 ≤ x ≤ 23.5)
• See figure below 52 Normal Approximation to a Binomial #3 For x = 22.5, the corresponding z value is
z= x −µ 22.5 −25
=
= − .71
0
σ
3.5355 For x = 23.5, the corresponding z value is
x −µ 23.5 −25
z=
=
= − .42
0
σ
3.5355 Then P(22.5 ≤ x ≤ 23.5) = P(0.71 ≤ z ≤ 0.42)
= 0.7611 – 0.6628
= 0.0983
Therefore the estimate of the binomial probability P(x = 23)
is 0.0983
53 54 55...
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This note was uploaded on 09/16/2012 for the course 123 123 taught by Professor Vincent during the Spring '12 term at Ill. Chicago.
 Spring '12
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