Answer:
The correct option is;
C. 0.3337
Step-by-step explanation:
We note that for a binomial probability distribution, we have;
[tex]p(r\geq \gamma)=\sum_{\gamma = r}^{ n}\dbinom{n}{r}\cdot \left (p\right )^{\gamma }\cdot \left (1-p\right )^{n - \gamma}[/tex]
Which gives;
[tex]p(x \geq 90)=\sum_{r = 90}^{ 100}\dbinom{100}{90}\cdot \left (0.88\right )^{\gamma }\cdot \left (1-0.88\right )^{100 - \gamma}[/tex]
[tex]\dbinom{100}{90}\cdot \left (0.88\right )^{90 }\cdot \left (1-0.88\right )^{10} = 0.108033[/tex]
[tex]\dbinom{100}{91}\cdot \left (0.88\right )^{91 }\cdot \left (1-0.88\right )^{9} = 0.08706[/tex]
[tex]\dbinom{100}{92}\cdot \left (0.88\right )^{92 }\cdot \left (1-0.88\right )^{8} = 0.062456[/tex]
[tex]\dbinom{100}{93}\cdot \left (0.88\right )^{93 }\cdot \left (1-0.88\right )^{7} = 0.039399[/tex]
[tex]\dbinom{100}{94}\cdot \left (0.88\right )^{94 }\cdot \left (1-0.88\right )^{6} = 0.021516[/tex]
[tex]\dbinom{100}{95}\cdot \left (0.88\right )^{95 }\cdot \left (1-0.88\right )^{5} = 0.09965[/tex]
[tex]\dbinom{100}{96}\cdot \left (0.88\right )^{96 }\cdot \left (1-0.88\right )^{4} = 0.003806[/tex]
[tex]\dbinom{100}{97}\cdot \left (0.88\right )^{97 }\cdot \left (1-0.88\right )^{3} = 0.001151[/tex]
[tex]\dbinom{100}{98}\cdot \left (0.88\right )^{98 }\cdot \left (1-0.88\right )^{2} = 0.000258[/tex]
[tex]\dbinom{100}{99}\cdot \left (0.88\right )^{99 }\cdot \left (1-0.88\right ) = 0.0000383[/tex]
[tex]\dbinom{100}{100}\cdot \left (0.88\right )^{100 }\cdot \left (1-0.88\right )^{0} = 0.00000281[/tex]
P(r≥90) is therefore 0.1083 + 0.08706 + 0.062456 + 0.039399 + 0.021516 + 0.09965 + 0.003806 + 0.001151 + 0.000258 + 0.0000383 + 0.00000281 = 0.333685 ≈ 0.3337.
