Answer to Question #210389 in Statistics and Probability for Reyan Hassan

Question

In a school, 30 percent of the students do not travel by bus. From a sample

of 10 students find the probability that (a) at least 3 students travel by bus

(b) at most 4 students travel by bus (c) between 2 and 7 clients students

travel by bus (exclusive)

Answer 1

Let $X=$ the number of students who travels by bus: $X\sim Bin (n, p).$

Given $n=10, q=0.3, p=1-q=1-0.3=0.7.$

(a)

$P(X\geq3)=1-P(X=0)-P(X=1)-P(X=2)$

$=1-\dbinom{10}{0}(0.7)^{0}(0.3)^{10-0}-\dbinom{10}{1}(0.7)^{1}(0.3)^{10-1}$

$-\dbinom{10}{2}(0.7)^{2}(0.3)^{10-2}=0.9984096136$

$\approx0.9984$

(b)

$P(X\leq4)=P(X=0)+P(X=1)+P(X=2)$

$+P(X=3)+P(X=4)=\dbinom{10}{0}(0.7)^{0}(0.3)^{10-0}$

$+\dbinom{10}{1}(0.7)^{1}(0.3)^{10-1}+\dbinom{10}{2}(0.7)^{2}(0.3)^{10-2}$

$+\dbinom{10}{3}(0.7)^{3}(0.3)^{10-3}+\dbinom{10}{4}(0.7)^{4}(0.3)^{10-4}$

$=0.0473489874\approx0.0473$

(c)

$P(2<X<7)=P(X=3)+P(X=4)$

$+P(X=5)+P(X=6)=\dbinom{10}{3}(0.7)^{3}(0.3)^{10-3}$

$+\dbinom{10}{4}(0.7)^{4}(0.3)^{10-4}+\dbinom{10}{5}(0.7)^{5}(0.3)^{10-5}$

$+\dbinom{10}{6}(0.7)^{6}(0.3)^{10-6}=0.009001692$

$+0.036756909+0.1029193452$

$+0.200120949=0.3487988952\approx0.3488$