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

kloseyq 2021-12-21 Answered
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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Expert Answer

David Clayton
Answered 2021-12-22 Author has 4057 answers
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.
\(\displaystyle\sigma^{{{2}}}={n}{p}{q}={n}{p}{\left({1}-{p}\right)}={20}{\left({0.70}\right)}{\left({1}-{0.70}\right)}={4.2}\)
\(\displaystyle\sigma=\sqrt{{{n}{p}{q}}}=\sqrt{{{n}{p}{\left({1}-{p}\right)}}}=\sqrt{{{20}{\left({0.70}\right)}{\left({1}-{0.70}\right)}}}\approx{2.0494}\)
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levurdondishav4
Answered 2021-12-23 Author has 5828 answers
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 :
\(\displaystyle{E}{\left({X}\right)}={\left({20}\right)}\times{\left({0.70}\right)}={14}\)
0
nick1337
Answered 2021-12-27 Author has 10160 answers

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).
\(Var(x)=npq=4.2\)

0

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