MATH SOLVE

4 months ago

Q:
# Historically, there is a 40% chance of having clear sunny skies in Seattle in July. Let's assume that each day is independent from any other day (although this is not really true Under this assumption, if you will spend 18 days in Seattle in July, what's the probability of having clear sunny skies on at least 13 of those days? (a) Solve this exactly (b) Solve this by approximating using a normal distribution. 8.

Accepted Solution

A:

Answer: a) 0.0058b) 0.0026Step-by-step explanation:Given : The probability of having clear sunny skies in Seattle in July : p= 0.40The 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]