Computing Probabilities for Normal Random VariablesThe cumulative distribution function (CDF) for the standard normal random
variable can be computed with the
where If the probability for only one number is required then the
where The
where
ExampleThis example is from an exercise in Newbold [1995, Exercise 19, Chapter 5]. Suppose that anticipated consumer demand for a product next month can be represented by a normal random variable with mean 1200 units and standard deviation 100 units. Denote the random variable by X.
For some value x, the CDF is P(X<x)=F(x).
The random variable Z=(X Questions and Solutions(a) What is the probability that sales will exceed 1000 units ? We need to find: P(X > 1000) = 1 - P(X < 1000) = 1 - F(1000) Note that: P(X < 1000) = P(Z < (1000-1200/100) ) = P(Z < -2) (b) What is the probability that sales will be between 1100 and 1300 units ? We need to find: P(1100 < X < 1300) = P(X < 1300) - P(X < 1100) = F(1300) - F(1100) Note that: P(X < 1100) = P(Z < (1100-1200/100) ) = P(Z < -1) and P(X < 1300) = P(Z < (1300-1200/100) ) = P(Z < 1) (c) The probability is 0.10 that sales will be more than how many units ? We need to find a value b such that: P(X > b) = .10 The SHAZAM commands (filename:
The SHAZAM output can be viewed. The answers to the questions are: (a) (b) (c) [SHAZAM Guide home] SHAZAM output|_SAMPLE 1 1 |_* Part (a) |_GEN1 ANSWER = 1 - NCDF((1000-1200)/100) |_PRINT ANSWER ANSWER .9772499 |_* Part (b) |_GEN1 ANSWER = NCDF((1300-1200)/100) - NCDF((1100-1200)/100) |_PRINT ANSWER ANSWER .6826895 |_* Part (c) |_* Compute the variance of X |_GEN1 SIG2=100**2 |_GEN1 PROB=.1 |_DISTRIB PROB / TYPE=NORMAL MEAN=1200 VAR=SIG2 INVERSE CRITICAL=ANSWER NORMAL DISTRIBUTION - MEAN= 1200.0 VARIANCE= 10000. PROBABILITY CRITICAL VALUE PDF PROB ROW 1 .10000 1328.2 .17550E-02 |_PRINT ANSWER ANSWER 1328.155 |_STOP [SHAZAM Guide home] |