1. Chapter 6 The Normal Distribution
In this handout:
The standard normal distribution
Probability calculations with normal distributions
The normal approximation to the binomial

2. Figure 6.8 (p. 231) The standard normal curve.
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.

3. An upper tail normal probability
From the table,
P[Z ≤ 1.37] = .9147
P[Z > 1.37]
= 1 - P[Z ≤ 1.37]
= = .0853
An upper tail normal probability

4. Figure 6.11 (p. 233) Normal probability of an interval.
P[-.155 < Z < 1.6]
= P[Z < 1.6] - P[Z < -.155] = = .5068

5. 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] = = .0466

6. 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] = = 0.975
From the table, = P[Z < 1.96]. Thus, z = 1.96

7. 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 < ] = Thus z = 1.645

8. Converting a normal probability to a standard normal probability

9. 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]
= = .6678

10. 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 (p. 241) The binomial distributions for p = .4 and n = 5, 12, 25.

11. 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.

12. Example:
Suppose n=15 and p=.4. Compute P[X = 7].
Mean = np = 15 * .4 = 6
Variance = np(1-p) = 6 * .6 = sd = 1.897
P[6.5 < X < 7.5]
= P[(6.5 -6)/1.897 < Z < ( )/1.897]
= P[.264 < Z < .791] = = .1814