Answer: a) 0.0058
b) 0.0026
Step-by-step explanation:
Given : The probability of having clear sunny skies in Seattle in July : p= 0.40
The number of days spent in Seattle in July: n= 18
a) Using, Binomial probability formula : [tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex]
The probability of having clear sunny skies on at least 13 of those days:-
[tex]P(x\geq13)=P(13)+P(14)+P(15)+P(16)+P(17)+P(18)\\\\=^{18}C_{13}(0.4)^{13}(0.6)^5+^{18}C_{14}(0.4)^{14}(0.6)^4+^{18}C_{15}(0.4)^{15}(0.6)^3+^{18}C_{16}(0.4)^{16}(0.6)^2+^{18}C_{17}(0.4)^{17}(0.6)^1+^{18}C_{18}(0.4)^{18}(0.6)^0[/tex]
[tex]=\dfrac{18!}{13!5!}(0.4)^{13}(0.6)^5+\dfrac{18!}{14!4!}(0.4)^{14}(0.6)^4+\dfrac{18!}{15!3!}(0.4)^{15}(0.6)^3+\dfrac{18!}{16!2!}(0.4)^{16}(0.6)^2+(18)(0.4)^{17}(0.6)^1+(1)(0.4)^{18}[/tex]
[tex]=0.00447111249474+0.00106455059399+0.000189253438931+0.0000236566798664+0.00000185542587187+0.000000068719476736[/tex]
[tex]=0.00575049735288\approx0.0058[/tex]
b) On converting binomial to normal distribution, we have
[tex]\text{Mean=}\mu=np= 18\times0.40=7.2\\\\\text{Standard deviation}=\sigma=\sqrt{np(1-p)}\\\\=\sqrt{18(0.40)(1-0.40)}=2.07846096908\approx2.08[/tex]
Let x be the number of days having clear sunny skies in Seattle in July.
Then, using [tex]z=\dfrac{x-\mu}{\sigma}[/tex] we have
[tex]z=\dfrac{13-7.2}{2.08}=2.78846153846\approx2.79[/tex]
P-value = [tex]P(x\geq13)=P(z\geq2.79)=1-P(z<2.79)[/tex]
[tex]=1-0.9973645=0.0026355\approx0.0026[/tex]
