Page 133 - Probability Demystified

122 CHAPTER 7 The Binomial Distribution

SOLUTION:
1
n ¼ 180 and p ¼ since there is one chance in 6 to get a three on each roll.
6
1
¼ n p ¼ 180
6
¼ 30

Hence, one would expect on average 30 threes.

EXAMPLE: Twelve cards are selected from a deck and each card is replaced
before the next one is drawn. Find the average number of diamonds.

SOLUTION:
In this case, n ¼ 12 and p ¼ 13 or 1 since there are 13 diamonds and a total of
52 4
52 cards. The mean is

¼ n p
1
¼ 12
4
¼ 3

Hence, on average, we would expect 3 diamonds in the 12 draws.

Statisticians are not only interested in the average of the outcomes of a
probability experiment but also in how the results of a probability experiment
vary from trial to trial. Suppose we roll a die 180 times and record the
number of threes obtained. We know that we would expect to get about 30
threes. Now what if the experiment was repeated again and again? In this
case, the number of threes obtained each time would not always be 30 but
would vary about the mean of 30. For example, we might get 28 threes one
time and 34 threes the next time, etc. How can this variability be explained?
Statisticians use a measure called the standard deviation. When the standard
deviation of a variable is large, the individual values of the variable are
spread out from the mean of the distribution. When the standard deviation of
a variable is small, the individual values of the variable are close to the mean.
The formula for the standard deviation for a binomial distribution is
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p
standard deviation ¼ npð1 pÞ. The symbol for the standard deviation is
the Greek letter (sigma).