# Math Help - [SOLVED] Statistics and Probability:

1. ## [SOLVED] Statistics and Probability:

Out of 20 questions I have managed to complete all but these four, so any assistance would be awesome.

Not sure if I am in brain freeze mode or what:

6.22: According to the national Marine manufacturers Association, 50% of the population of Vermont were boating participants during the most recent year. For the randomly selected sample of 20 Vermont residents, with the discrete random variable x= the number in the sample who were boating participants that year, determine the following:
a) E (x)
b) P (x ≤ 8)
c) P (x=10)
d) P (x=12)
e) P (7 ≤ x ≤ 13)

6.30: It has been estimated that one in five Americans suffers from allergies. The president of Hargrove University plans to randomly select 10 students from the undergraduate population of 1400 to attend a dinner at his home. Assuming that Hargrove students are typical in terms of susceptibility to allergies and that the college presidents home happens to contain just about every common allergin to which afflicted persons react, what is the probability that at least 8 of the students will be able to stay for the duration of the event?

7.33: It has been reported that households in the west spend an annual average of $6050.00 for groceries. Assume a normal distribution with a standard deviation of$1500.
a) what is the probability that randomly selected western household spends more than $6350 for groceries? b) How much money would a western household have to spend on groceries per year in order to be at the 99th percentile (ie only 1% of western households would spend more on groceries)? 7.54: A random variable is Poisson distributed with λ=.50 arrivals per minute. For the corresponding exponential distribution, and the x = minutes until the next arrival, indentify the mean of x and determine the following: a) P (x ≤0.5) b) P (x ≤1.5) c) P (x ≥ 2.5) d) P (x ≥ 3.0) 2. Originally Posted by loriemomof2 [snip] 7.54: A random variable is Poisson distributed with λ=.50 arrivals per minute. For the corresponding exponential distribution, and the x = minutes until the next arrival, indentify the mean of x and determine the following: a) P (x ≤0.5) b) P (x ≤1.5) c) P (x ≥ 2.5) d) P (x ≥ 3.0) If the number of customers arriving per minute has a Poisson distribution with mean 0.5, then the intervals between arrivals has an exponential distribution with mean time between arrivals of $\frac{1}{0.5} = 2$ minutes. 3. Originally Posted by loriemomof2 [snip] 7.33: It has been reported that households in the west spend an annual average of$6050.00 for groceries. Assume a normal distribution with a standard deviation of $1500. a) what is the probability that randomly selected western household spends more than$6350 for groceries?
b) How much money would a western household have to spend on groceries per year in order to be at the 99th percentile (ie only 1% of western households would spend more on groceries)?
[snip]
a) $z = \frac{6350 - 6050}{1500} = 0.2$.

Calculate Pr(Z > 0.2) = 1 - Pr(Z < 0.2). Using tables, I assume ....?

b) Use tables (in reverse) to get the value of z* such that Pr(Z < z*) = 0.99. Substitute this value into $z^* = \frac{x - 6050}{1500}$ and solve for $x$.

4. Originally Posted by loriemomof2
[snip]

6.30: It has been estimated that one in five Americans suffers from allergies. The president of Hargrove University plans to randomly select 10 students from the undergraduate population of 1400 to attend a dinner at his home. Assuming that Hargrove students are typical in terms of susceptibility to allergies and that the college presidents home happens to contain just about every common allergin to which afflicted persons react, what is the probability that at least 8 of the students will be able to stay for the duration of the event?
[snip]
Technically X ~ Hypergeometric(N = 1400, n = 10, D = 280). But since N is large compared to n you can make the approximation

X ~ Binomial(n = 10, p = 1/5).

Calculate 1 - Pr(X < 8).

5. Originally Posted by loriemomof2
Out of 20 questions I have managed to complete all but these four, so any assistance would be awesome.

Not sure if I am in brain freeze mode or what:

6.22: According to the national Marine manufacturers Association, 50% of the population of Vermont were boating participants during the most recent year. For the randomly selected sample of 20 Vermont residents, with the discrete random variable x= the number in the sample who were boating participants that year, determine the following:
a) E (x)
b) P (x ≤ 8)
c) P (x=10)
d) P (x=12)
e) P (7 ≤ x ≤ 13)

[snip]
X ~ Binomial(n = 20, p = 1/2)

(assuming that the population of Vermont is much larger than 20, of course )

6. ## Thanks for the response:

Originally Posted by mr fantastic
If the number of customers arriving per minute has a Poisson distribution with mean 0.5, then the intervals between arrivals has an exponential distribution with mean time between arrivals of $\frac{1}{0.5} = 2$ minutes.
Thanks for the response I had gotten this far however was not sure how to complete the problem to get the answers to the list below.

a) P (x ≤0.5)
b) P (x ≤1.5)
c) P (x ≥ 2.5)
d) P (x ≥ 3.0)

7. Originally Posted by mr fantastic
a) $z = \frac{6350 - 6050}{1500} = 0.2$.

Calculate Pr(Z > 0.2) = 1 - Pr(Z < 0.2). Using tables, I assume ....?

b) Use tables (in reverse) to get the value of z* such that Pr(Z < z*) = 0.99. Substitute this value into $z^* = \frac{x - 6050}{1500}$ and solve for $x$.
Just got my assignment back today and according to the instructor the answers are:
a) .4207
b) \$9545

8. Originally Posted by mr fantastic
Technically X ~ Hypergeometric(N = 1400, n = 10, D = 280). But since N is large compared to n you can make the approximation

X ~ Binomial(n = 10, p = 1/5).

Calculate 1 - Pr(X < 8).
Fortunately we received a message from the instructor saying we only had to submit 4 of the 8 problems as the more I look at this whole thing the more confused I get will be glad when this class is finally over....