Chapter 6 The Normal Distribution

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. Figure 6.9 (p. 232)

Equal normal tail probabilities.

It is customary to denote the standard normal variable by Z.

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

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

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.

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