Answer:
340 of the adults in the sample voted.
The 95% confidence interval estimate of the population percent of Americans that vote is (0.6391, 0.7209). This means that we are 95% sure that the true proportion of Americans that vote is between 0.6391 and 0.7209.
Step-by-step explanation:
In 2008, a random sample of 500 american adults was take and we found that 68% of them voted. How many of the 500 adults in the sample voted?
This is 68% of 500.
So 0.68*500 = 340.
340 of the adults in the sample voted.
Now construct a 95% confidence interval estimate of the population percent of Americans that vote and write a sentence to explain the confidence interval.
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
Z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]n = 500, p = 0.68[/tex]
95% confidence interval
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.68 - 1.96\sqrt{\frac{0.68*0.32}{500}} = 0.6391[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.68 + 1.96\sqrt{\frac{0.68*0.32}{500}} = 0.7209[/tex]
The 95% confidence interval estimate of the population percent of Americans that vote is (0.6391, 0.7209). This means that we are 95% sure that the true proportion of Americans that vote is between 0.6391 and 0.7209.
A process is normally distributed with a mean of 10.2 hits per minute and a standard deviation of 1.04 hits. If a randomly selected minute has 13.9 hits, would the process be considered in control or out of control?
Answer:
This process is out of control.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
A probability is said to be unusual if it's z-score has a pvalue of 0.05 or lower, or a pvalue of 0.95 or higher.
In this problem, we have that:
[tex]\mu = 10.2 \sigma = 1.04[/tex]
If a randomly selected minute has 13.9 hits, would the process be considered in control or out of control?
The process will be considered out of control if it's z-score(Z when X = 13.9) has a pvalue of 0.95 or higher. Otherwise(it will be positive, since 13.9 is above the mean), it will be considered in control.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{13.9 - 10.2}{1.04}[/tex]
[tex]Z = 3.56[/tex]
[tex]Z = 3.56[/tex] has a pvalue of 0.9999. So there is only a 1-0.9999 = 0.0001 = 0.01% probability of getting 13.9 hits a minute.
So this process is out of control.
If using the empirical rule, since 13.9 hits is more than two standard deviations above the mean of 10.2 hits, the process could be considered out of control. However, establishing specific control limits is necessary for a definitive answer.
Explanation:To determine whether a process is in control or out of control, we assess whether a given measurement is within the expected range of a normal distribution, often using the empirical rule or control limits. Given that the process has a mean of 10.2 hits per minute and a standard deviation of 1.04 hits, under the empirical rule, approximately 95% of the data should fall within two standard deviations of the mean (that is, between roughly 8.12 and 12.28 hits).
With 13.9 hits in a randomly selected minute, this count is significantly more than two standard deviations above the mean, suggesting that the process might be out of control. However, to make a definitive statement about control status, specific control limits must be established, often based on the particular specifications of the process being monitored.
Una familia dedica dos tercios de sus ingresos a cubrir gastos de funcionamiento, ahorra la cuarta parte del total y gadta el resto en ocio.¿Qué fraccion de los ingresos invierte en ocio?
Answer:
la familia invierte 8.33% de los ingresos totales en ocio
Step-by-step explanation:
Representando los ingresos totales por I:
- Gastos de funcionamiento = 2/3*I
- Ahorro = 1/4*I
- En ocio : lo que resta = I - 2/3*I - 1/4*I = I - 11/12*I = 1/12*I (8.33% de I)
por lo tanto la familia invierte 8.33% de los ingresos totales en ocio
In a recent poll^1 of 1000 American adults, the number saying that exercise is an important part of daily life was 753. Use strategy or other technology to find a 90% confidence interval for the proportion of American adults who think exercise is an important part of daily life. The 90% confidence interval is
Answer:
[tex]0.753 - 1.64\sqrt{\frac{0.753(1-0.753)}{1000}}=0.731[/tex]
[tex]0.753 + 1.64\sqrt{\frac{0.753(1-0.753)}{1000}}=0.775[/tex]
The 95% confidence interval would be given by (0.731;0.775)
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Solution to the problem
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2 =0.005[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
The estimated proportion for this case is:
[tex] \hat p = \frac{X}{n}= \frac{753}{1000}=0.753[/tex]
If we replace the values obtained we got:
[tex]0.753 - 1.64\sqrt{\frac{0.753(1-0.753)}{1000}}=0.731[/tex]
[tex]0.753 + 1.64\sqrt{\frac{0.753(1-0.753)}{1000}}=0.775[/tex]
The 95% confidence interval would be given by (0.731;0.775)
The length of side AB is ....
Answer:
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Step-by-step explanation:
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Shureka Washbum has scores of 74, 88, 61, and 83 on her algebra tests.
a. Use an inequality to find the scores she must make on the final exam to pass the course with an average of 76 or higher, given that the final exam counts as two
tests
b. Explain the meaning of the answer to part (a).
The solution set is {x{ }
(Type an inequality.)
Step-by-step explanation:
If x is her score on the final exam, then the average is:
(74 + 88 + 61 + 83 + 2x) / 6
(306 + 2x) / 6
51 + ⅓x
We want this to be greater than or equal to 76.
51 + ⅓x ≥ 76
⅓x ≥ 25
x ≥ 75
In order to get an average of 76 or higher, Shureka's score on the final exam must be greater than or equal to 75.
An urn has 3 blue balls 4 red balls. 3 balls are drawn without replacement. find the probability of drawing 2 blue and 1 red given at least 1 blue is drawn
Answer:
Frist case: P=12/35
Second case: P=31/35
Step-by-step explanation:
An urn has 3 blue balls 4 red balls. 3 balls are drawn without replacement.
Frist case:
We calculate the number of possible combinations
{7}_C_{3}=\frac{7!}{3! · (7-3)!}=35
We calculate the number of favorable combinations
{3}_C_{2} · {4}_C_{1} =
=\frac{3!}{2! · (3-2)!} · \frac{4!}{1! · (4-1)!}
=3 · 4 = 12
Therefore, the probability is
P=12/35
Second case:
When we count on at least one ball to be blue, we go over the probability complement.
We calculate the probability that all the balls are red, then subtract this from 1.
We calculate the number of possible combinations
{7}_C_{3}=\frac{7!}{3! · (7-3)!}=35
We calculate the number of favorable combinations
{4}_C_{3} = \frac{4!}{3! · (4-3)!=4
The probability is
P=4/35.
Therefore the probability on at least one ball to be blue
P=1-4/35
P=31/35