# Consider a binomial experiment with n=20 and p=0.70 a) Compute f(12) b)

Consider a binomial experiment with $$\displaystyle{n}={20}$$ and $$\displaystyle{p}={0.70}$$
a) Compute f(12)
b) Compute f(16)
c) Compute $$\displaystyle{P}{\left({x}\geq{16}\right)}$$
d) Compute $$\displaystyle{P}{\left({x}\le{15}\right)}$$
e) Compute E(x)
f) Compute $$\displaystyle{v}{a}{r}{\left({x}\right)}$$ and $$\displaystyle\sigma$$

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David Clayton
Step 1
Given:
$$\displaystyle{p}={0.70}$$
$$\displaystyle{n}={20}$$
Formula binomial probability:
$f(k)=(\begin{matrix} n \\ k \end{matrix})\times p^{k}\times(1-p)^{n-k}$
a) $f(12)=(\begin{matrix} 20 \\ 12 \end{matrix})\times0.70^{12}\times(1-0.70)^{20-12}=0.114397$
b $f(16)=(\begin{matrix} 20 \\ 16 \end{matrix})\times0.70^{16}\times(1-0.70)^{20-16}=0.130421$
c) Add the corresponding probabilities:
$$\displaystyle{P}{\left({X}\geq{16}\right)}={f{{\left({16}\right)}}}+{f{{\left({17}\right)}}}+{f{{\left({18}\right)}}}+{f{{\left({19}\right)}}}+{f{{\left({20}\right)}}}={0.2375}$$
d) Use the complement rule for probabilities:
$$\displaystyle{P}{\left({X}\le{15}\right)}={1}-{P}{\left({X}\geq{16}\right)}={1}-{0.2375}={0.7625}$$
Step 2
e) The mean of binomial distribution is the product of the sample size n and the probability p:
$$\displaystyle\mu={n}{p}={20}\times{0.20}={14}$$
f) The standard deviation of a binomial distribution is the square root of the product of the sample size n and the probabilities p and q. The variance is the square of the standard deviation.
$$\displaystyle\sigma^{{{2}}}={n}{p}{q}={n}{p}{\left({1}-{p}\right)}={20}{\left({0.70}\right)}{\left({1}-{0.70}\right)}={4.2}$$
$$\displaystyle\sigma=\sqrt{{{n}{p}{q}}}=\sqrt{{{n}{p}{\left({1}-{p}\right)}}}=\sqrt{{{20}{\left({0.70}\right)}{\left({1}-{0.70}\right)}}}\approx{2.0494}$$
###### Not exactly what you’re looking for?
levurdondishav4
Step 1
Wherever a random variable X can be modeled as a binomial random variable we write :
$$\displaystyle{X}\sim{B}{i}{\left({n},\ {p}\right)}$$
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 :
$$\displaystyle{P}{\left({X}={x}\right)}={f{{\left({x}\right)}}}={\left({n}{C}{x}\right)}\times{\left({p}^{{{x}}}\right)}\times{\left({1}-{p}\right)}^{{{n}-{x}}}$$
Where $$\displaystyle{P}{\left({X}={x}\right)}$$ is the probability of the variable X to assume the value x.
Where $$\displaystyle{n}{C}{x}$$ is the combinatorial number define as :
$$\displaystyle{n}{C}{x}={\frac{{{n}!}}{{{x}!{\left({n}-{x}\right)}!}}}$$
In our question
$$\displaystyle{X}\sim{B}{i}{\left({20},\ {0.70}\right)}$$
Now let's calculate the probabilities :
$$\displaystyle{f{{\left({12}\right)}}}={P}{\left({X}={12}\right)}={\left({20}{C}{12}\right)}\times{\left({0.70}\right)}^{{{12}}}\times{\left({1}-{70}\right)}^{{{20}-{12}}}={0.1144}$$
1) $$\displaystyle{f{{\left({16}\right)}}}={P}{\left({X}={16}\right)}={\left({20}{C}{16}\right)}\times{\left({0.70}\right)}^{{{16}}}\times{\left({1}-{0.70}\right)}^{{{20}-{16}}}={0.1304}$$
$$\displaystyle{P}{\left({X}\geq{16}\right)}\Rightarrow$$
2) $$\displaystyle{P}{\left({X}\geq{16}\right)}={P}{\left({X}={16}\right)}+{P}{\left({X}={17}\right)}+{P}{\left({X}={18}\right)}+{P}{\left({X}={19}\right)}+{P}{\left({X}={20}\right)}$$
3) $$\displaystyle{P}{\left({X}={17}\right)}={\left({20}{C}{17}\right)}\times{\left({0.70}\right)}^{{{17}}}\times{\left({1}-{0.70}\right)}^{{{20}-{17}}}={0.0716}$$
4) $$\displaystyle{P}{\left({X}={18}\right)}={\left({20}{C}{18}\right)}\times{\left({0.70}\right)}^{{{18}}}\times{\left({1}-{0.70}\right)}^{{{20}-{18}}}={0.0278}$$
5) $$\displaystyle{P}{\left({X}={19}\right)}={\left({20}{C}{19}\right)}\times{\left({0.70}\right)}^{{{19}}}\times{\left({1}-{0.70}\right)}^{{{20}-{19}}}={0.0068}$$
6) $$\displaystyle{P}{\left({X}={20}\right)}={\left({20}{C}{20}\right)}\times{\left({0.70}\right)}^{{{20}}}\times{\left({1}-{0.70}\right)}^{{{20}-{20}}}={0.0008}$$
Step 2
Using (1), (3), (4). (5) and (6) in (2):
$$\displaystyle{P}{\left({X}\geq{16}\right)}={0.1304}+{0.0716}+{0.0278}+{0.0068}+{0.0008}={0.2374}$$
Now:
$$\displaystyle{P}{\left({X}\le{15}\right)}$$
$$\displaystyle{P}{\left({X}\le{15}\right)}={1}-{P}{\left({X}\geq{16}\right)}$$
$$\displaystyle{P}{\left({X}\le{15}\right)}={1}-{0.2374}={0.7626}$$
Finally,
$$\displaystyle{E}{\left({X}\right)}=\mu{\left({X}\right)}$$
$$\displaystyle{E}{\left({X}\right)}$$ is the mean of the variable X
In this case, X is a binomial random variable and its mean can be calculated as
$$\displaystyle{E}{\left({X}\right)}={\left({n}\right)}\times{\left({p}\right)}$$
In the question :
$$\displaystyle{E}{\left({X}\right)}={\left({20}\right)}\times{\left({0.70}\right)}={14}$$
nick1337

Step 1
NSK
Given that X is binomial $$n=20;\ p=0.70$$
a) Compute f(12) (to 4 decimals).
$$=0.1144$$
b) Compute f(16) (to 4 decimals).
$$=0.1304$$
c) Compute $$P(x \geq16)$$ (to 4 decimals)
$$=0.7624$$
d) Compute $$P(x \le15)$$ (to 4 decimals).
$$=0.5836$$
e) Compute $$E(x)=np=14$$
NSK
f) Compute Var(x) (to 1 decimal) and (to 2 decimals).
$$Var(x)=npq=4.2$$