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MTH410 Quiz 5
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Gravity
chapters 7 & 8
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Suppose the number of pages per book in a library has an unknown distribution with population mean 304 and population standard deviation 30. A sample of size n=70 is randomly taken from the population, and the sum of the values is taken. Using the Central Limit Theorem for Sums, what is the standard deviation for the sample sum distribution? Round your answer to two decimal places.
The Central Limit Theorem for Sums states the standard deviation of the normal distribution of sample sums is equal to the original distribution's standard deviation multiplied by the square root of the sample size, (σX)(n−−√). The original standard deviation is 30, and the sample size is 70. So, the standard deviation of the distribution of sample sums is
(σX)(√n)=(30)(√70)≈251.00
The number of square feet per house is normally distributed with a population standard deviation of 154 square feet and an unknown population mean. If a random sample of 16 houses is taken and results in a sample mean of 1550 square feet, find a 80% confidence interval for the population mean.
Confidence intervals are written as (x¯−EBM,x¯+EBM), so we need the sample mean, x¯, and the EBM. We know x¯=1550. We can use the formula to find the error bound:
EBM=(zα2)(σ/√n)=(1.282)(154/√16)≈(1.282)(38.500)≈49.36
So, the error bound (EBM) is 49.357. So we can write this confidence interval as: (1550−49.36,1550+49.36) or (1500.64,1599.36).
Jonathan, a college business student wants to study how many pages of notes fellow classmates took during a semester for a finance class. Past data reveals that pages of notes for a finance college class has a mean of 95 , with standard deviation 25 pages. He plans to take a random sample of 30 such college students and will calculate the mean pages of notes they take to compare to the known pages of notes students take for a finance college class.
For these values, the mean and standard deviation of the sampling distribution of sample means for a sample of size 30 are: μx¯=95 and σx¯=2530√=4.6.
What is the probability that the sample mean for a sample of size 30 will be at least 99?
From above, we know the sampling distribution is normally distributed with mean 95 and standard deviation 4.6. The problem is asking for the probability the sample mean will be at least 99. So, we need to compute P(x¯≥99). To compute this, we will compute the z−score of 99.
P(x¯≥99)=P(Z ≥ 99−95/4.6)=P(Z ≥ 0.87)
Next, locate the area that corresponds to z=0.87. Recall, this is the area to the left of z=0.87. This area is found where the row 0.8 and the column 0.07 intersect.
The resulting area is P(Z≤0.87)=0.8078. Finally, use the complement rule to find the area to the right:
P(Z ≥ 0.87) = 1−P(Z<0.87) = 1−0.8078 = 0.192
The pages per book in a library are normally distributed with an unknown population mean and standard deviation. A random sample of 41 books is taken and results in a sample mean of 341 pages and sample standard deviation of 22 pages.
Find the EBM, margin of error, for a 95% confidence interval estimate for the population mean using the Student's t-distribution.
We can calculate the EBM with the formula:
EBM=(tα2)(s/√n)
The question tells us s=22 and n=41. We need to find the t-value before we can find EBM. The degrees of freedom are
df = n − 1 = 41 − 1 = 40
The confidence level is 95%, so α=1−0.95=0.05, and α2=0.025. Now, we can use the table above to find the t-value for t0.025 with 40 degrees of freedom, which is 2.021. Now we can calculate the EBM.
EBM=(tα2)(s/√n)=(2.021)(22/√41)≈(2.021)(3.436)≈6.94
The delivery times for a distributing site are normally distributed with an unknown population mean and standard deviation. If a random sample of 25 deliveries is taken to estimate the mean delivery time, what t-score should be used to find a 98% confidence interval estimate for the population mean?
The sample used was 25 deliveries, so n=25. To find the degrees of freedom: df = n − 1 = 25 − 1 = 24
The confidence level is given in the scenario: 98%. So,
α=1−CL=1−0.98=0.02
But we want to use the value for α2, which is 0.022=0.01. Using the table, we need to find the row for 24 degrees of freedom, and the column for t0.01. So, the t-score we would use to find the 98% confidence interval is 2.492
The time spent, in hours, of teenagers on social media per year are normally distributed with a population standard deviation of 442 hours and an unknown population mean. If a random sample of 24 teenagers is taken and results in a sample mean of 1330 hours, find a 99% confidence interval for the population mean.
EBM = =(2.576)(442/√24)≈(2.576)(90.223)≈232.41
So, the error bound (EBM) is 232.41. So we can write this confidence interval as: (1330−232.41,1330+232.41) or (1097.59,1562.41)
The population standard deviation for the number of corn kernels on an ear of corn is 94 kernels. If we want to be 90% confident that the sample mean is within 17 kernels of the true population mean, what is the minimum sample size that should be taken?
The formula for sample size is
n=z^2σ^2/EBM^2
because the confidence level is 90%. From the problem, we know that σ=94 and EBM=17. Therefore, n=(1.645)^2(94)^2/17^2≈82.74
Use n=83 to ensure that the sample size is large enough.
Suppose we know that a confidence interval is (61,71), with an error bound of 5. Find the sample mean (x¯). Give just a number for your answer. For example, if you found x¯=12, you would enter 12.
Since we know the error bound is 5, we can subtract 5 from the upper value of the confidence interval, 71.
71−5=66
Brenda wants to estimate the percentage of people who purchase music online at least once per week. She wants to create a 95% confidence interval which has an error bound of at most 2%. How many people should be polled to create the confidence interval?
Given the information in the question, EBP=0.02 since 2%=0.02 and zα2=z0.025=1.96 because the confidence level is 95%. The values of p′ and q′ are unknown, but using a value of 0.5 for p′ will result in the largest possible product of p′q′, and thus the largest possible n. If p′=0.5, then q′=1−0.5=0.5. Therefore,
n=z^2p′q′/EBP^2=1.96^2(0.5)(0.5)/0.02^2=2401.0
Round the answer up to the next integer to be sure the sample size is large enough. The sample should include 2401 people.
The lengths of text messages have an unknown distribution with mean 26 and standard deviation 5 characters. A sample, with size n=40, is randomly drawn from the population and the mean is taken. What is the probability that the mean is more than 25.4 characters?
The Central Limit Theorem for Means states that the mean of the normal distribution of means is equal to the mean of the original distribution. The standard deviation is equal to the original standard deviation divided by the square root of the sample size. So, the mean of this mean distribution is 26 and the standard deviation is 5/√40≈0.791. To find the probability using the Standard Normal Table, we find that the z-score for the value 25.4 is −0.759, using the formula z = x−μ/σ. Using the Standard Normal Table, the area to the left of z=−0.759 is 0.2236. Remember that the area to the right of z=−0.759 will be the complement of this, 1−0.2236=0.7764. So the probability is about 78%.
The shipping times for warehouses in an industry are normally distributed with an unknown population mean and standard deviation. A random sample of 37 warehouses is taken and results in a sample mean of 42 minutes and sample standard deviation of 5 minutes.Find a 95% confidence interval estimate for the population mean using the Student's t-distribution.
we can write this confidence interval as: (42−1.67,42+1.67) or (40.33,43.67).
Suppose germination periods, in days, for grass seed are normally distributed and have a known population standard deviation of 2 days and an unknown population mean. A random sample of 22 types of grass seed is taken and gives a sample mean of 46 days. Find the error bound (EBM) of the confidence interval with a 90% confidence level.
=(1.645)(22/√2)≈(1.645)(0.426)≈0.701
So, the error bound (EBM)is 0.701.
The operating costs for each machine for one day have an unknown distribution with mean 1610 and standard deviation 136 dollars. A sample, with size n=45, was randomly drawn from the population. Using the Central Limit Theorem for Means, what is the standard deviation for the sample mean distribution?
The Central Limit Theorem for Means states the standard deviation of the normal distribution of sample means is equal to the original distribution's standard deviation divided by the square root of the sample size, σX/√n. The original standard deviation is 136, and the sample size is 45. So, the standard deviation of the distribution of sample means is:
σX/√n=136/√45≈20.27
Suppose 180 randomly selected people are surveyed to determine whether or not they prefer brand named items. Of the 180 surveyed, 30 reported preferring brand named items. What are the sample proportions for successes, p′, and failures, q′?
To form the sample proportion, take the number of successes, and divide it by n, the number of trials. In this scenario, the number of successes is the number of people who prefer brand named items, 30. The total number of people surveyed was 180. =30/180=0.167
So, p′=0.167is the sample proportion, which is the point estimate of the population proportion. Since p′+q′=1, we can solve for q′.
q′=1−p′=1−0.167=0.833
Suppose the finishing times for cyclists in a race are normally distributed. If the population standard deviation is 16 minutes, what minimum sample size is needed to be 95% confident that the sample mean is within 5 minutes of the true population mean?
because the confidence level is 95%. From the problem, we know that σ=16 and EBM=5. Therefore,
n=z^2σ^2/EBM^2=(1.96)^2(16)^2/5^2≈39.34
Use n=40 to ensure that the sample size is large enough.
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