Chapter 6
The Normal Distribution
In this handout:
• The standard normal distribution
• Probability calculations with normal distributions
• The normal approximation to the binomial
It is customary to denote the standard
normal variable by Z.
Figure 6.8 (p. 231)
The standard normal curve.
Figure 6.9 (p. 232)
Equal normal tail probabilities.
From the table,
P[Z ≤ 1.37] = .9147
P[Z > 1.37]
= 1 - P[Z ≤ 1.37]
= 1 - .9147 = .0853
An upper tail normal probability
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
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
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
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
Converting a normal probability to a standard
normal probability
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
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.
Figure 6.16 (p. 241)
The binomial distributions for p = .4 and n = 5, 12, 25.
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.
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