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KonstantinChe [14]
3 years ago
10

Historically, there is a 40% chance of having clear sunny skies in Seattle in July. Let's assume that each day is independent fr

om 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.
Mathematics
1 answer:
Igoryamba3 years ago
6 0

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 : P(x)=^nC_xp^x(1-p)^{n-x}

The probability of having clear sunny skies on at least 13 of those days:-

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

=\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}

=0.00447111249474+0.00106455059399+0.000189253438931+0.0000236566798664+0.00000185542587187+0.000000068719476736

=0.00575049735288\approx0.0058

b) On converting binomial to normal distribution, we have

\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

Let x be the number of days having clear sunny skies in Seattle in July.

Then, using z=\dfrac{x-\mu}{\sigma} we have

z=\dfrac{13-7.2}{2.08}=2.78846153846\approx2.79

P-value = P(x\geq13)=P(z\geq2.79)=1-P(z

=1-0.9973645=0.0026355\approx0.0026

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We have to write the transition matrix M for the population.

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-  For smokers who smoke a pack or less per day, there is a 10% probability of quitting and a 10% probability of increasing to more than a pack per day. <em>Then, the probability of staying in the same state is 80%.</em>

- For smokers who smoke more than a pack per day, there is an 8% probability of quitting and a 10% probability of dropping to a pack or less per day. <em>Then, the probability of staying in the same state is 82%.</em>

<em />

The transition matrix becomes:

\begin{vmatrix} &NS&P1&PM\\NS&  0.90&0.08&0.02 \\  P1&0.10&0.80 &0.10 \\  PM& 0.08 &0.10&0.82 \end{vmatrix}

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\left[\begin{array}{ccc}5000&2500&2500\end{array}\right] * \left[\begin{array}{ccc}0.90&0.08&0.02\\0.10&0.80 &0.10\\0.08 &0.10&0.82\end{array}\right] =\left[\begin{array}{ccc}4950&2650&2400\end{array}\right]

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To calculate the the state for the second month, we us the state of the first of the month and multiply it one time by the transition matrix:

\left[\begin{array}{ccc}4950&2650&2400\end{array}\right] * \left[\begin{array}{ccc}0.90&0.08&0.02\\0.10&0.80 &0.10\\0.08 &0.10&0.82\end{array}\right] =\left[\begin{array}{ccc}4912&2756&2332\end{array}\right]

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If we repeat this multiplication 12 times from the actual state (or 10 times from the two-months state), we will get the state a year from now:

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In a year, we will have 4,793 non-smokers, 3,005 smokers of one pack and 2,202 smokers of more than one pack.

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