Answer to Question #163289 in Statistics and Probability for Dush

Question

i. Suppose in a district grade 1 through grade 5, 60 percent of the population favor a particular school. A simple random sample of 350 is surveyed.

a. b. c. d. e.

What is the probability that at least 140 favor the school? What is the probability that at most 170 favor the school? What is the probability that more than 160 favor the school? What is the probability that fewer than 145 favor the school? What is the probability that exactly 180 favor the school?

ii. The annual

of downtown Memphis follows a Poisson distribution with mean 6.5.

number of earthquakes registering at least 2.5 on the Richter Scale and having an epicenter within 40 miles

a. What is the probability that at least 9 such earthquakes will strike next year?

b. What is the probability that at least 25 such earthquakes will strike during the next 5 years?

Answer 1

(i) As 60% of population favour a particular school,

Probability of success, $p=\dfrac{60}{100}=0.6$

$q=1-p=1-0.6=0.4$

Using Binomal Distribution,

$P(X=x)=^nC_rp^rq^{n-r}$

(a) P( at least 140 favor the school)=$P(X\ge140)=1-P(X<140)$

$=1-[^{350}C_{139}(0.6)^{139}(0.4^{211})+.....+^{350}C_1(0.6)(0.4)^{349}]$

(b) P( at most 170 favor the school)=$P(X\le 170)=P(X\le170)$

$= ^{350}C_{1}(0.6)^{1}(0.4)^{349}+.....+^{350}C_{170}(0.6)^{170}(0.4)^{180}$

(c) P(more than 160 favor the school)=$P(X>160)$

$=1-[^{350}C_{1}(0.6)^{1}(0.4)^{349}+.....+^{350}C_{160}(0.6)^{160}(0.4)^{190}]$

(d) P(Fewer than 145 favor the school)=$P(X<145)$

$=^{350}C_{1}(0.6)^{1}(0.4)^{349}+.....+^{350}C_{144}(0.6)^{144}(0.4)^{206}]$

(e) P(exactly 180 favor the school)=$P(X=180)$

$=^{350}C_{180}(0.6)^{180}(0.4)^{170}$

(ii) We can use the Poisson distribution to calculate the exact probability. Using the Poisson table with $\lambda=6.5,$ we get:

(a) P( at least 9 earthquake in an year)

$P(Y\ge 9)=1-P(Y\le 8)=1-0.792=0.208$

Now, let's use the normal approximation to the Poisson to calculate an approximate probability. we have to make a continuity correction. Doing so, we get:

$P(Y\ge 9)=P(Y\ge 8.5)$

$P(Y\ge 9)=P(Y\ge 8.5)=P(Z>\dfrac{8.5-6.5}{\sqrt{6.5}})=P(Z>0.78)=0.218$

So, in summary, we used the Poisson distribution to determine the probability that $Y$ is at least $9$ is exactly $0.208$ , and we used the normal distribution to determine the probability that $Y$ is at least $9$ is approximately $0.218$

(b) P( at least 25 earthquake in 5 years)=P(at least 5 earthquakes in 1 year)

$P(Y\ge 5)=1-P(Y\le 4)=1-0.4352=0.4351$

Now, let's use the normal approximation to the Poisson to calculate an approximate probability. we have to make a continuity correction. Doing so, we get:

$P(Y\ge 5)=P(Y\ge 4.5)$

$P(Y\ge 5)=P(Y\ge 4.5)=P(Z>\dfrac{6.5-4.5}{\sqrt{6.5}})=P(Z>0.98)=0.1269$

So, in summary, we used the Poisson distribution to determine the probability that $Y$ is at least $25$ is exactly $0.1269$ , and we used the normal distribution to determine the probability that $Y$ is at least $25$ is approximately $0.1269$.