Page 29 - Quality control of pharmaceuticals (07-PA 704)

Adding these two areas we find that the probability that a cow´s milk
production will lie in the interval from 60 pounds to 90 pounds is 0.7176.

Example
Annual incomes for career service employees at a university are
normally distributed with a mean of $6200 and standard deviation of
$900. Find the probability that an employee chosen at random will
have an annual income less than $5000; an income greater than
$7000.

z = (5000 - 6200) / 900 = -1200/900= -1.33
Referring to Table 1, the area between µ = 6200 and a value 1.33 standard

deviations to the left of the mean is 0.4082.
Hence the probability of observing an annual income less than $5000 is

0.5 - 0.4082 = 0.0918.
Similarly, to compute the probability of observing a salary above $7000,

we determine the area between 6200 and 7000.
z =(7000 - 6200) / 900 = 0.89
Referring to Table 1, the area corresponding to z = 0.89 is 0.3133.
Hence the desired probability is, 0.5 - 0.3133 = 0.1867
Example 4:
The amount of aspirin in the analgesic tablets from a particular

manufacturer is known to follow a normal distribution with µ = 250