Question

consider a binomial experiment with n = 20 and p = 0.70.
a. compute f(12) (to 4 decimals).
b. compute f(16) (to 4 decimals).
c. compute p(x ≥ 16) (to 4 decimals).
d. compute p(x ≤ 15) (to 4 decimals).
e. compute e(x).
f. compute var(x) (to 1 decimal) and σ (to 2 decimals).

Explanation:

Step1: Recall binomial probability formula

The binomial probability formula is $f(x)=C(n,x)\times p^{x}\times(1 - p)^{n - x}$, where $C(n,x)=\frac{n!}{x!(n - x)!}$, $n$ is the number of trials, $x$ is the number of successes, and $p$ is the probability of success on a single - trial. Here $n = 20$ and $p=0.7$.

Step2: Compute $f(16)$

First, calculate the combination $C(20,16)=\frac{20!}{16!(20 - 16)!}=\frac{20!}{16!4!}=\frac{20\times19\times18\times17}{4\times3\times2\times1}=4845$. Then, $f(16)=C(20,16)\times(0.7)^{16}\times(0.3)^{4}$.
$(0.7)^{16}\approx0.003323293$ and $(0.3)^{4}=0.0081$. So $f(16)=4845\times0.003323293\times0.0081\approx0.1304$.

Step3: Compute $P(x\geq16)$

$P(x\geq16)=P(x = 16)+P(x = 17)+P(x = 18)+P(x = 19)+P(x = 20)$.
$P(x = k)=C(20,k)\times(0.7)^{k}\times(0.3)^{20 - k}$.
$P(x = 17)=C(20,17)\times(0.7)^{17}\times(0.3)^{3}$, $C(20,17)=\frac{20!}{17!(20 - 17)!}=1140$, $(0.7)^{17}\approx0.002326305$, $(0.3)^{3}=0.027$, $P(x = 17)=1140\times0.002326305\times0.027\approx0.0716$.
$P(x = 18)=C(20,18)\times(0.7)^{18}\times(0.3)^{2}$, $C(20,18)=\frac{20!}{18!(20 - 18)!}=190$, $(0.7)^{18}\approx0.001628413$, $(0.3)^{2}=0.09$, $P(x = 18)=190\times0.001628413\times0.09\approx0.0278$.
$P(x = 19)=C(20,19)\times(0.7)^{19}\times(0.3)^{1}$, $C(20,19)=20$, $(0.7)^{19}\approx0.001139889$, $(0.3)^{1}=0.3$, $P(x = 19)=20\times0.001139889\times0.3\approx0.0068$.
$P(x = 20)=C(20,20)\times(0.7)^{20}\times(0.3)^{0}$, $(0.7)^{20}\approx0.000797923$, $P(x = 20)=1\times0.000797923\times1\approx0.0008$.
$P(x\geq16)=0.1304 + 0.0716+0.0278+0.0068+0.0008=0.2374$.

Step4: Compute $P(x\leq15)$

Since $P(x\leq15)=1 - P(x\geq16)$, so $P(x\leq15)=1 - 0.2374 = 0.7626$.

Answer:

b. $0.1304$
c. $0.2374$
d. $0.7626$