Mean and Standard Deviation for the Binomial Distribution - Inferences Regarding Proportions

**MEAN AND STANDARD DEVIATION FOR THE BINOMIAL DISTRIBUTION**

In Chapter 4, we discussed the mean and variance of
a continuous variable (*μ*, σ^{2} and , *S*^{2}) for the parameters and
their respective sample estimates. It is possible to compute analogs for a
dichotomous variable. As shown in Chapter 5, the binomial distribution is used
to describe the distribution of dichotomous outcomes such as heads or tails or
“successes and failures.” The mean and variance of the binomial distribution
are functions of the parameters *n* and
*p*, where *n* refers to the number in the population and *p* to the proportion of successes. This relationship between the
parameters *n* and *p* and the binomial distribution affects the way we form test
statis-tics and confidence intervals for the proportions when using the normal
approxima-tion that we will discuss in Section 10.3. The mean of the binomial
is *np* and the variance is *np*(1 – *p*), as we will demonstrate.

Recall (see Section 5.7) that for a binomial random
variable *X* with parameters *n* and *p*, *X* can take on the
values 0, 1, 2, . . . , *n* with *P*{*X
= k*} = *C*(*n*, *k*) *p ^{k}*(1 –

The algebra becomes a little more complicated than
for the proof of *E*(*X*) = *np*
shown in Display 10.1; using techniques similar to those employed in the
foregoing proof, we can demonstrate that the variance of *X*, denoted Var(*X*),
satisfies the equation Var(*X*) = *np*(1 – *p*). Equations 10.1 and 10.2 summarize the formulas for the expected
value of *X* and the variance of *X*.

For a binomial random variable *X*

*E*(*X*) =* np *(10.1)

where *n*
is the number of Bernoulli trials and *p*
is the population success probability. For a binomial random variable *X*

Var(*X*) = *np*(1 – *p*) (10.2)

where *n*
is the number of Bernoulli trials and *p*
is the population success probability. To illustrate the use of Equations 10.1
and 10.2, let us use a simple example of a Bernouilli trial in which *n* = 3 and *p* = 0.5. An illustration would be an experiment involving three
tosses of a fair coin; a head will be called a success. Then the possi-ble
number of successes on the three tosses is 0, 1, 2, or 3. Applying Equation
10.1, we find that the mean number of successes is *np* = 3 (0.5) = 1.5; applying Equation 10.2, we find that the
variance of the number of successes is *np*(1
– *p*) = 3(0.5)(1 – 0.5) = 1.5(.5) =
0.75. Had we not obtained these two simple formulas by algebra, we could have
performed the calculations from the definitions in Chapter 5 (sum-marized in
Display 10.1 and Formula D10.1).

To apply Formula D10.1, we compute the probability
of each of the successes (outcomes 0, 1, 2, and 3), multiply each of these
probabilities by the number of suc-cesses (0, 1, 2, and 3) and then sum the
results, as shown in the next paragraph.

The probability of 0 successes is *C*(3, 0)*p*^{0}(1 – *p*)^{3};
when we replace *p* by (0.5), the term *C*(3, 0)(0.5)^{0}(1 – 0.5)^{3}
= (1)(1)(0.5)^{3} = 0.125. As there are 0 successes,

**Display 10.1.** **Proof that E(X)
= np**

To conduct this proof, we will use
the formula presented in Chapter 5, Section 5.7, in which the probability of *r* successes in *n* trials was defined as *P*(*Z = r*) = *C*(*n*,* r*)*p ^{r}*(1 –

Σ_{k}*C*(*n*, *k*)*p ^{k}*(1 –

This equation holds for any positive
integer *n* and proportion 0 < *p* < 1 when the summation is taken
over 0 ≤ *k* ≤ *n*. Assume *k* ≥ 2 for the following argument. The mean denoted by *E*(*X*)
is by definition

Remember that by applying formula
D10.1, with *n* – 1 in place of *n* in the formula

since *n* – 1 is a positive integer (recall that *n* ≥ 2, implying that *n* –
1 ≥ 1). So for *n *≥* *2,* E*(*X
*=* np*. For* n *= 1,* E*(*X*) = 0(1 –* p*) + 1(*p*) =* p = np *also. So we have* *shown for any positive integer *n*, *E*(*X*) = *np*.

we multiply 0.125 by 0 and obtain 0. Consequently,
the contribution of 0 suc-cesses to the mean is 0. Next, we calculate the
probability of 1 success by using *C*(3,
1)*p*(1 –* p*)^{2}, which is the number of ways of arranging 1 success
and 2 fail-ures in a row multiplied by the probability of a particular
arrangement that has 1 success and 2 failures. *C*(3, 1) = 3, so the resulting probability is 3*p*(1 – *p*)^{2} =
3(0.5)(0.5)^{2} = 3(0.125) = 0.375. We multiply that result by 1, since
it corresponds to 1 success and we find that 1 success contributes 0.375 to the
mean of the distribution. The probability of 2 successes is *C*(3, 2)*p*^{2}(1 – *p*),
which is the number of ways of arranging 2 successes and 1 failure in a row
multiplied by the probability of any one such arrangement. *C*(3, 2) = 3, so the probability is 3*p*^{2}(1 – *p*) =
3(0.5)^{2}(0.5) = 0.375. We then multiply 0.375 by 2, as we have 2
successes, which contribute 0.750 to the mean. To obtain the final term, we
compute the probabili-ty of 3 successes and then multiply the resulting
probability by 3. The probability of 3 successes is *C*(3, 3) = 1 (since all three places have to be successes, there is
only one possible arrangement) multiplied by *p*^{3} = (0.5)^{3} = 0.125. We then multi-ply 0.125
by 3 to obtain the contribution of this term to the mean. In this case the
contribution to the mean is 0.375.

In order to obtain the mean of this distribution,
we add the four terms together. We obtain the mean = 0 + 0.375 + 0.750 + 0.375
= 1.5. Our long computation agrees with the result from Equation 10.1. For
larger values of *n* and different
val-ues of *p*, the direct calculation
is even more tedious and complicated, but Equation 10.1 is simple and easy to
perform, a statement that also holds true for the variance calculation. Note
that in our present example, if we apply the formula for the vari-ance
(Equation 10.2), we obtain a variance of *np*(1
– *p*) = 3(0.5)(0.5) = 0.750.

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