Question

Consider a binomial experiment with n = 20 and p = .70. if you calculate the binomial probabilities manually, make sure to carry at least 4 decimal digits in your calculations. compute f(12) (to 4 decimals). 0.1143 compute f(16) (to 4 decimals). 0.1304 compute p(x 16) (to 4 decimals). 0.2375 compute p(x 15) (to 4 decimals). 0.7624 compute e(x). 14

Answer 1

Answer:

Here are the results for the first 2 parts.

Step-by-step explanation:

f(x)   =    nCr p^r q^(n-r)

So f(12) = 20C12 (0.70^12 q^(0.30)^(20-12)

=  20C8 0.7^12 0.30^8

= 0.1144 .

f(16)  =  0.1304.

Answer 2

Answer:

Step-by-step explanation:

The question is incomplete. p(x 16) is actually $P(X\geq 16)$ ; p(x 15) is actually $P(X\leq 15)$ and e(x) is $E(X)$

Wherever a random variable X can be modeled as a binomial random variable we write :

X ~ Bi (n,p)

Where ''n'' is the number of Bernoulli experiments taking place (whose variable is called binomial random variable).

And where ''p'' is the success probability.

In a Bernoulli experiment we define which event will be a ''success''

In order to calculate the probabilities for the variable X we can use the following equation :

$P(X=x)=f(x)=(nCx).(p^{x}).(1-p)^{n-x}$

Where ''$P(X=x)$'' is the probability of the variable X to assume the value x.

Where ''$nCx$'' is the combinatorial number define as :

$nCx=\frac{n!}{x!(n-x)!}$

In our question

X ~ Bi (20,0.70)

Now let's calculate the probabilities :

$f(12)=P(X=12)=(20C12).(0.70)^{12}.(1-0.70)^{20-12}=0.1144$

$f(16)=P(X=16)=(20C16).(0.70)^{16}.(1-0.70)^{20-16}=0.1304$ (I)

$P(X\geq 16)$ ⇒

$P(X\geq 16)=P(X=16)+P(X=17)+P(X=18)+P(X=19)+P(X=20)$ (II)

$P(X=17)=(20C17).(0.70)^{17}.(1-0.70)^{20-17}=0.0716$ (III)

$P(X=18)=(20C18).(0.70)^{18}.(1-0.70)^{20-18}=0.0278$ (IV)

$P(X=19)=(20C19).(0.70)^{19}.(1-0.70)^{20-19}=0.0068$ (V)

$P(X=20)=(20C20).(0.70)^{20}.(1-0.70)^{20-20}=0.0008$ (VI)

Using (I), (III), (IV), (V) and (VI) in (II) :

$P(X\geq 16)=0.1304+0.0716+0.0278+0.0068+0.0008=0.2374$

Now :  

$P(X\leq 15)$

$P(X\leq 15)=1-P(X\geq 16)$

$P(X\leq 15)=1-0.2374=0.7626$

Finally,

$E(X)=$ μ (X)

$E(X)$ is the mean of the variable X

In this case, X is a binomial random variable and its mean can be calculated as

$E(X)=(n).(p)$

In the question :

$E(X)=(20).(0.70)=14$