a school newspaper reporter decides to randomly survey 19 students to see if they will attend tet (vietnamese new year) festivities this year. based on past years, she knows that 25% of students attend tet festivities. we are interested in the number of students who will attend the festivities.
part (a)
part (b)
part (c)
part (d)
part (e)
find the probability that at most 6 students will attend. (round your answer to four decimal places )
part (f)
find the probability that more than 5 students will attend. (round your answer to four decimal places )
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Question

Accepted Answer

# Explanation:
## Step1: Identify the distribution
This is a binomial distribution problem. The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n = 19$ (number of trials, i.e., number of students surveyed), $p=0.25$ (probability of success, i.e., probability that a student will attend Tet festivities), and $C(n,k)=\frac{n!}{k!(n - k)!}$

## Step2: Calculate $P(X\leq6)$ for part (e)
\[
\begin{align*}
P(X\leq6)&=\sum_{k = 0}^{6}C(19,k)\times(0.25)^{k}\times(0.75)^{19 - k}\
C(19,0)&=\frac{19!}{0!(19-0)!}=1\
P(X = 0)&=1\times(0.25)^{0}\times(0.75)^{19}\approx0.0014\
C(19,1)&=\frac{19!}{1!(19 - 1)!}=19\
P(X = 1)&=19\times(0.25)^{1}\times(0.75)^{18}\approx0.0107\
C(19,2)&=\frac{19!}{2!(19-2)!}=\frac{19\times18}{2\times1}=171\
P(X = 2)&=171\times(0.25)^{2}\times(0.75)^{17}\approx0.0393\
C(19,3)&=\frac{19!}{3!(19 - 3)!}=\frac{19\times18\times17}{3\times2\times1}=969\
P(X = 3)&=969\times(0.25)^{3}\times(0.75)^{16}\approx0.0865\
C(19,4)&=\frac{19!}{4!(19-4)!}=\frac{19\times18\times17\times16}{4\times3\times2\times1}=3876\
P(X = 4)&=3876\times(0.25)^{4}\times(0.75)^{15}\approx0.1491\
C(19,5)&=\frac{19!}{5!(19 - 5)!}=\frac{19\times18\times17\times16\times15}{5\times4\times3\times2\times1}=11628\
P(X = 5)&=11628\times(0.25)^{5}\times(0.75)^{14}\approx0.1914\
C(19,6)&=\frac{19!}{6!(19-6)!}=\frac{19\times18\times17\times16\times15\times14}{6\times5\times4\times3\times2\times1}=27132\
P(X = 6)&=27132\times(0.25)^{6}\times(0.75)^{13}\approx0.1974\
P(X\leq6)&=0.0014 + 0.0107+0.0393 + 0.0865+0.1491+0.1914+0.1974\
P(X\leq6)&\approx0.6758
\end{align*}
\]

## Step3: Calculate $P(X>5)$ for part (f)
Since $P(X>5)=1 - P(X\leq5)$
$$
\begin{align*}
P(X\leq5)&=\sum_{k = 0}^{5}C(19,k)\times(0.25)^{k}\times(0.75)^{19 - k}\
P(X\leq5)&=0.0014 + 0.0107+0.0393 + 0.0865+0.1491+0.1914\
P(X\leq5)&\approx0.4784\
P(X>5)&=1-0.4784\
P(X>5)&\approx0.5216
\end{align*}
$$

# Answer:
For part (e): $0.6758$
For part (f): $0.5216$

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QUESTION IMAGE

Question

Explanation:

Step1: Identify the distribution

Step2: Calculate \(P(X\leq6)\) for part (e)

Step3: Calculate \(P(X>5)\) for part (f)

Answer: