chapter 6 the normal distribution
DESCRIPTION
Chapter 6 The Normal Distribution. In this handout: The standard normal distribution Probability calculations with normal distributions The normal approximation to the binomial. Figure 6.8 (p. 231) The standard normal curve. It is customary to denote the standard normal variable by Z. - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 6 The Normal Distribution
In this handout:
• The standard normal distribution
• Probability calculations with normal distributions
• The normal approximation to the binomial
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Figure 6.8 (p. 231)The standard normal curve. Figure 6.9 (p. 232)
Equal normal tail probabilities.
It is customary to denote the standard normal variable by Z.
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An upper tail normal probability
From the table,
P[Z ≤ 1.37] = .9147
P[Z > 1.37]
= 1 - P[Z ≤ 1.37]
= 1 - .9147 = .0853
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Figure 6.11 (p. 233)Normal probability of an interval.
P[-.155 < Z < 1.6]
= P[Z < 1.6] - P[Z < -.155] = .9452 - .4384 = .5068
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Figure 6.12 (p. 233)Normal probabilities for Example 3.
P[Z < -1.9 or Z > 2.1]
= P[Z < -1.9] + P[Z > 2.1] = .0287 + .0179 = .0466
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Determining an upper percentile of the standard normal distribution
• Problem: Locate the value of z that satisfies P[Z > z] = 0.025
• Solution: P[Z < z] = 1- P[Z > z] = 1- 0.025 = 0.975
From the table, 0.975 = P[Z < 1.96]. Thus, z = 1.96
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Determining z for given equal tail areas
• Problem: Obtain the value of z for which P[-z < Z < z] = 0.9
• Solution: From the symmetry of the curve, P[Z < -z] = P[Z > z] = .05
From the table, P[Z < -1.645] = 0.05. Thus z = 1.645
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Converting a normal probability to a standard normal probability
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Converting a normal probability to a standard normal probability (example)
• Problem: Given X is N(60,4),
find P[55 < X < 63]
• Solution: The standardized variable is Z = (X – 60)/4
x=55 gives z=(55-60)/4 = -1.25
x=63 gives z=(63-60)/4 = .75
Thus,
P[55<X<63] = P[-1.25 < Z < .75]
= P[Z < .75] - P[Z < -1.25]
= .7734 - .1056 = .6678
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Figure 6.16 (p. 241)The binomial distributions for p = .4 and n = 5, 12, 25.
When the success probability p of is not too near 0 or 1
and the number of trials is large,
the normal distribution serves
as a good approximation
to the binomial probabilities.
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How to approximate the binomial probability by a normal?
The normal probability assigned to a single value x is zero.
However, the probability assigned to the interval x-0.5 to x+0.5 is the appropriate comparison (see figure).
The addition and subtraction of 0.5 is called the continuity correction.
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Example:
Suppose n=15 and p=.4. Compute P[X = 7].
Mean = np = 15 * .4 = 6
Variance = np(1-p) = 6 * .6 = 3.6 sd = 1.897
P[6.5 < X < 7.5]
= P[(6.5 -6)/1.897 < Z < (7.5 - 6)/1.897]
= P[.264 < Z < .791] = .7855 - .6041 = .1814