25% of the houses in a certain area have swimming pools. if 5 houses from this area are selected at random, what is the probability that no one has a swimming pool.
1.1 at least two have swimming pools.
question2
an automatic boling machines fills cola into 2litre (200cm) bottles . aconsumer advoacate wants to test the null hypothesis that the average amount filled by the machine into a bottle is at least 2000cm. a random sample of 40 bottles coming out of the machine was selected and th e exact contents of the selected bottles was recorded. the sample mean was 1,9996cm. the population standard deviation is known from the past experience to be 1.30cm. test the null hypothesis at an @of 5%
For the first problem, you can use the binomial probability function, which states: P(x) = (nCx)(p^x)[q^(n-x)]
n = 5
x = 0 (for the first part)
p = .25
q = .75 (q = 1 - p)
For the first part: P(0) = (5C0)(.25^0)(.75^5)
I'll let you finish.
For the second part of that problem, find P(2) + P(3) + P(4) + P(5). Or you can take 1 - [P(0) + P(1)].
For the second question, you can probably use a one-sample z-test.
z = (sample mean - population mean) divided by (standard deviation divided by the square root of the sample size)
Compare the test statistic to the 5% level stated in the problem. Draw your conclusions from there.
For the first problem, let's calculate the probability that no one has a swimming pool out of the 5 selected houses. We can use the binomial probability function to do this:
P(x) = (nCx)(p^x)(q^(n-x))
Where:
n = 5 (number of houses selected)
x = 0 (no one has a swimming pool)
p = 0.25 (probability of a house having a swimming pool)
q = 0.75 (probability of a house not having a swimming pool, which is 1 - p)
Now substitute these values into the formula:
P(0) = (5C0)(0.25^0)(0.75^5)
The notation (5C0) represents "5 choose 0", which calculates the number of ways to choose 0 houses out of 5, and it equals 1.
P(0) = 1 * (0.25^0) * (0.75^5)
P(0) = 1 * 1 * (0.75^5)
Now calculate the value of (0.75^5) using a calculator or by hand:
P(0) = 1 * 1 * 0.2373
P(0) = 0.2373
So the probability that none of the 5 selected houses has a swimming pool is approximately 0.2373.
For the second part of the first problem, we can calculate the probability of at least two houses having swimming pools. This can be done by finding the complement of the probability that less than two houses have swimming pools.
P(at least two) = 1 - [P(0) + P(1)]
P(at least two) = 1 - (0.2373 + P(1))
To find P(1), we can plug the values into the binomial probability function:
P(1) = (5C1)(0.25^1)(0.75^4)
Calculate (5C1) and simplify the equation:
P(1) = 5 * 0.25 * (0.75^4)
P(1) = 0.9375 * 0.3164
P(1) = 0.2969
Now substitute this value into the equation for P(at least two):
P(at least two) = 1 - (0.2373 + 0.2969)
P(at least two) = 1 - 0.5342
P(at least two) = 0.4658
So the probability of at least two houses having swimming pools is approximately 0.4658.
For the second question, we can use a one-sample z-test to test the null hypothesis. The test statistic (z) can be calculated using the formula:
z = (sample mean - population mean) / (standard deviation / √sample size)
Given the information from the problem:
sample mean = 1,999.6 cm
population mean = 2,000 cm
standard deviation = 1.30 cm
sample size = 40
Now substitute these values into the formula:
z = (1999.6 - 2000) / (1.30 / √40)
z = (-0.4) / (1.30 / (2√10))
z = (-0.4) / (1.30 / (2 * 3.162))
z ≈ -0.4 / 0.412
z ≈ -0.97
Now we can compare the test statistic to the critical value at a significance level of 5%. For a one-tailed z-test, the critical value is approximately 1.645.
Since -0.97 is less than 1.645, it falls in the non-rejection region. Therefore, we do not have enough evidence to reject the null hypothesis.
In conclusion, based on the given information, we fail to reject the null hypothesis that the average amount filled by the machine into a bottle is at least 2000 cm.