the sides of a triangle are in the ratio 3:4:5 what is the length of each side if the perimeter of the triangle is 90cm
1 answer:
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Answer:
1) X[bar]= $262.60
2) S= $83.55
3) [$57.47; $152.55]
Step-by-step explanation:
Hello!
The objective is to study the average daily cost for business travel in the United States. To test this a sample of the estimated living expenses to 10 international cities was taken. The study variable is X: estimated living costs (including a single room at 4 stars hotel, beverages, breakfast, taxi fares, and incidental costs) on an international city.
Sample data:
n=10
242.87, 260.93, 194.19, 260.86, 355.36, 211.00, 284.08, 139.16, 436.72, 240.87
∑Xi= 2626.04
∑Xi²= 752439.39
1)
The sample mean is calculated by adding all the observed costs and divide it by the sample size:
X[bar]= ∑Xi/n = 2626.04/10= $262.60
2)
To calculate the standard deviation of the sample you have to calculate the sample variance first using the formula:
S²=
S²=
S²= 6981.1979
S= √S² = √6981.1979 = $83.55
3)
To calculate a Confidence Interval for the population standard deviation, you have to first calculate the Confidence Interval for the population variance and then calculate it's square root to obtain the interval for the standard deviation.
This happens because to study a population parameter you need a statistic that follows the following conditions: 1) have known distribution, 2) include the population parameter and it's point estimator, 3) the only incognito of the statistic equation must be the parameter under study. There is no statistic that meets these conditions, but the chi-square statistic meets these conditions and allows you to study the population variance. And this is only possible if the study variable has a normal distribution.
Assuming that the daily living costs have a normal distribution, the formula of the Confidence Interval for the population variance is:
[$²3302.88; $²23270.66]
So with a confidence level of 95% you'd expect that the confidence interval
[$²3302.88; $²23270.66] will contain the population variance of the estimated living costs (including single room at 4 stars hotel, beverages, breakfast, taxi fares and incidental costs) on an international city.
Now you calculate the square root of the interval:
[√$²3302.88; √$²23270.66]
[$57.47; $152.55]
And at the same level, you'd expect that the interval [$57.47; $152.55] contains the population standard deviation of the estimated living costs (including a single room at 4 stars hotel, beverages, breakfast, taxi fares, and incidental costs) on an international city.
I hope it helps!
Answer:
D. 68%
Step-by-step explanation:
The following statistics are given;
sample mean = 47%
s = 5% ; the sample standard deviation
n = 625 ; the sample size
The confidence interval for a population mean is given as;
sample mean ± z-score*
Substituting the above values we have;
47 ± z-score*
47 ± z-score*0.2
The confidence interval has been given as;
lower limit = 46.8%
upper limit = 47.2%
We can use any of these two values with the above expression to solve for the z-score. Using the lower limit we have the following equation;
47 - z-score*0.2 = 46.8
-z-score*0.2 = 46.8 - 47
-z-score*0.2 = -0.2
z-score = 1
The area of the standard normal curve between -1 and +1 will be the confidence level for the given confidence interval.
Pr( -1<Z<1 ) = 0.68 = 68%
From the Empirical rule