Answer:
There is a 90.46% probability that it was repaired by a competent shop, given that it was repaired correctly.
Step-by-step explanation:
We have these following probabilities:
An 87% probability that an air conditioner repair shop is competent.
A 13% probability that an air conditioner repair shop is incompetent.
An 85% probability that an compotent shop can repair the air.
A 60% probability than an incompetent shop can repair the air.
This can be formulated as the following problem:
What is the probability of B happening, knowing that A has happened?
It can be calculated by the following formula
[tex]P = \frac{P(B).P(A/B)}{P(A)}[/tex]
Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.
In this problem we have that:
Probability that it was repaired by a competent shop, given that it was repaired correctly.
P(B) is the probability that it was repaired by a competent shop. 87% of the shops are competent, so [tex]P(B) = 0.87[/tex]
P(A/B) is the probability that it was repaired correctly, given that it was repaired by a competent shop. There is an 85% probability that an compotent shop can repair the air. So [tex]P(A/B) = 0.85[/tex]
P(A) is the probability that an air was repaired correctly.
This is 85% of 87% and 60% of 13%. So
[tex]P(A) = 0.85*0.87 + 0.60*0.13 = 0.8175[/tex]
Finally
[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.87*0.85}{0.8175} = 0.9046[/tex]
There is a 90.46% probability that it was repaired by a competent shop, given that it was repaired correctly.
what percentage of eligibible american vote? 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? Now construct a 95% confidence interval estimate of the population percent of Americans that vote and write a sentence to explain the confidence interval.
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.
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)
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
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
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.
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.
The mean cost of a meal for two in a mid-range restaurant in Tokyo is $40 (Numbeo.com website, December 14, 2014). How do prices for comparable meals in Hong Kong compare? The DATAfile HongKongMeals contains the costs for a sample of 42 recent meals for two in Hong Kong mid-range restaurants.
Data:
Meal Price ($)
22.78
33.89
22.77
18.04
23.29
35.28
42.38
36.88
38.55
41.68
25.73
34.19
31.75
25.24
26.32
19.57
36.57
32.97
36.83
30.17
37.29
25.37
24.71
28.79
32.83
43.00
35.23
34.76
33.06
27.73
31.89
38.47
39.42
40.72
43.92
36.51
45.25
33.51
29.17
30.54
26.74
37.93
a. With 95% confidence, what is the margin of error (to 2 decimals)?
b. What is the 95% confidence interval estimate of the population mean (to 2 decimals)?
c. How do prices for meals for two in mid-range restaurants in Hong Kong compare to prices for comparable meals in Tokyo restaurants?
a. To find the margin of error, we first calculate the standard error of the mean using the formula [tex]\(SE = \frac{s}{\sqrt{n}}\)[/tex], where s is the sample standard deviation and n is the sample size. Then, we use the formula for the margin of error [tex]\(ME = Z \times SE\)[/tex], where Z is the z-score corresponding to the desired level of confidence.
b. Once we have the margin of error, we can construct the confidence interval estimate of the population mean by adding and subtracting the margin of error from the sample mean.
c. To compare the prices for meals for two in mid-range restaurants in Hong Kong to those in Tokyo, we can use the confidence interval estimate of the population mean. If the confidence interval includes the mean price for Tokyo ($40), it suggests that there may not be a significant difference in prices between the two cities. However, if the confidence interval does not include $40, it suggests that there may be a significant difference in prices between the two cities.
Explanation:
a. Margin of error:
1. Calculate the sample mean [tex](\( \bar{x} \))[/tex] and sample standard deviation s from the given data.
2. Determine the sample size n.
3. Find the standard error of the mean SE using the formula [tex]\( SE = \frac{s}{\sqrt{n}} \)[/tex].
4. Look up the z-score corresponding to the desired level of confidence (e.g., 95%) from the standard normal distribution table.
5. Multiply the z-score by the standard error to find the margin of error [tex](\( ME \))[/tex].
b. Confidence interval estimate:
1. Calculate the margin of error ME.
2. Subtract the margin of error from the sample mean to find the lower bound of the confidence interval.
3. Add the margin of error to the sample mean to find the upper bound of the confidence interval.
c. Price comparison:
1. Check if the confidence interval estimate of the population mean includes the mean price for Tokyo ($40).
2. If the confidence interval includes $40, it suggests that there may not be a significant difference in prices between the two cities.
3. If the confidence interval does not include $40, it suggests that there may be a significant difference in prices between the two cities.