Question

(c) The binomial distribution where n = 16 and p = 0.85? What is the expected value and the standard deviation?

Answer 1

Answer:

The expected value and the standard deviation of the given binomial distribution are 13.6 and 1.42 respectively.

Explanation:

Given,

n = 16

p = 0.85

Formulae,

EV = np;

SD = root of (np)(1-p)

Therefore,

EV = (16)(0.85)

EV = 13.6

SD = root of (16)(0.85)(0.15)

SD = root of 2.04

SD = 1.42

Answer 2

SOLUTION

GIVEN

The binomial distribution where n = 16 and p = 0.85

TO DETERMINE

• The expected value

• The standard deviation

CONCEPT TO BE IMPLEMENTED

If a trial is repeated n times and p is the probability of a success and q that of failure then the probability of r successes is

$\displaystyle \sf{ \sf{P(X=r) = \: \: }\large{ {}^{n} C_r}\: {p}^{r} \: \: {q}^{n - r} } \: \: \: \: \: where \: q \: = 1 - p$

EVALUATION

Here it is given that n = 16 and p = 0.85

q = 1 - p = 1 - 0.85 = 0.15

The Binomial distribution is given by

$\displaystyle \sf{ P(X=r) = {}^{16} C_r\: {(0.85)}^{r} \times {(0.15)}^{16 - r} }$

The expected value

= E(X)

= np

$\sf = 16 \times 0.85$

$\bf = 13.6$

The standard deviation

$\sf = \sqrt{npq}$

$\sf = \sqrt{16 \times 0.85 \times 0.15}$

$\sf = \sqrt{2.04}$

$\bf \approx 1.43$

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