Answer:
The probability is [tex]P( X \le 4 ) = 0.0054[/tex]
Step-by-step explanation:
From the question we are told that
The percentage that are on time is p = 0.76
The sample size is n = 11
Generally the percentage that are not on time is
[tex]q = 1- p[/tex]
[tex]q = 1- 0.76[/tex]
[tex]q = 0.24[/tex]
The probability that no more than 4 of them were on time is mathematically represented as
[tex]P( X \le 4 ) = P(1 ) + P(2) + P(3) + P(4)[/tex]
=> [tex]P( X \le 4 ) = \left n } \atop {}} \right.C_1 p^{1} q^{n- 1} + \left n } \atop {}} \right.C_2p^{2} q^{n- 2} + \left n } \atop {}} \right.C_3 p^{3} q^{n- 3} + \left n } \atop {}} \right.C_4 p^{4} q^{n- 4}[/tex]
[tex]P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{11- 1} + \left 11 } \atop {}} \right.C_2p^{2} q^{11- 2} + \left 11 } \atop {}} \right.C_3 p^{3} q^{11- 3} + \left 11 } \atop {}} \right.C_4 p^{4} q^{11- 4}[/tex]
[tex]P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{10} + \left 11 } \atop {}} \right.C_2p^{2} q^{9} + \left 11 } \atop {}} \right.C_3 p^{3} q^{8} + \left 11 } \atop {}} \right.C_4 p^{4} q^{7}[/tex]
[tex]= \frac{11! }{ 10! 1!} (0.76)^{1} (0.24)^{10} + \frac{11!}{9! 2!} (0.76)^2 (0.24)^{9} + \frac{11!}{8! 3!} (0.76)^{3} (0.24)^{8} + \frac{11!}{7!4!} (0.76)^{4} (0.24)^{7}[/tex]
[tex]P( X \le 4 ) = 0.0054[/tex]
