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A recent survey found that 70% of Indiana rivers State College students plan to go on a vacation after graduation suppose 5% of Indiana rivers State College students are randomly selected and let X be the number of students that plans to go on a vacation after graduation out of the sample of 5 used a binomial probability formula or the binomial probability table to construct the probability distribution of X also used a binomial probability formula or the binomial probability table to construct the cumulative probability distribution

User Nikovn
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Given:

Probability of Indiana rivers State College students plan to go on a vacation after graduation is(P) 70%=0.7.

Sample size =n=5

x= the random sample of number of students that plans to go on a vacation after graduation out of the sample of 5

The formula for binomial distribution is,


\begin{gathered} P(X=x)=^nC_x(p)^x(q)^(n-x) \\ \text{Here, p=0.7} \\ q=1-p=1-0.7=0.3 \\ ^nC_x=(n!)/(x!(n-x)!) \end{gathered}

The probability that no more than 4 students out of random sample of 5 students plan to go on vaccination after graduation is given by,


\begin{gathered} P(X\leq4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4) \\ =^5C_0(0.7)^0(0.3)^(5-0)+^5C_1(0.7)^1(0.3)^(5-1)+^5C_2(0.7)^2(0.3)^(5-2)+^5C_3(0.7)^3(0.3)^(5-3)+^5C_4(0.7)^4(0.3)^(5-4) \\ =0.0024+0.0284+0.1323+0.3087+0.3602 \\ =0.832 \end{gathered}

Hence, the probability P(X≤4)=0.832.

User LimeRed
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