Answer:
2(x+8)
Step-by-step explanation:
b. Answer the questions above for a sample of size n=64
c.find the probability that the average diameter of pistonrings from a sample size 16 is more than 11.95cm
d. For which of the above two random saples is Xbar morelikely to be within .01cm of 12cm? Explain.
Using the normal distribution and the central limit theorem, it is found that:
a) The sampling distribution is approximately normal,centered at 12 cm and with a standard deviation of 0.01 cm.
b) The sampling distribution is approximately normal,centered at 12 cm and with a standard deviation of 0.005 cm.
c) 100% probability that the average diameter of piston rings from a sample size 16 is more than 11.95 cm.
d) Due to the lower standard error, the sample of 64 is more likely to be within 0.01 cm of 12 cm.
In a normal distribution with mean and standard deviation, the z-score of a measure X is given by:
In this problem:
Item a:
Sample of 16, thus and
The sampling distribution is approximately normal,centered at 12 cm and with a standard deviation of 0.01 cm.
Item b:
Sample of 64, thus and
The sampling distribution is approximately normal,centered at 12 cm and with a standard deviation of 0.005 cm.
Item c:
This probability is 1 subtracted by the p-value of Z when X = 11.95, thus:
By the Central Limit Theorem:
has a p-value of 0.
1 - 0 = 1.
100% probability that the average diameter of piston rings from a sample size 16 is more than 11.95 cm.
Item d:
Due to the lower standard error, the sample of 64 is more likely to be within 0.01 cm of 12 cm.
A similar problem is given at brainly.com/question/24663213
Answer:
The answer is below
Step-by-step explanation:
Given that:
mean value (μ) = 12 cm and standard deviation (σ) = 0.04 cm
a) Since a random sample (n) of 16 rings is taken, therefore the mean (μx) ans standard deviation (σx) of the sample mean Xbar is given by:
The sampling distribution of Xbar is centered about 12 cm and the standard deviation of the Xbar distribution is 0.01 cm
b) Since a random sample (n) of 64 rings is taken, therefore the mean (μx) ans standard deviation (σx) of the sample mean Xbar is given by:
The sampling distribution of Xbar is centered about 12 cm and the standard deviation of the Xbar distribution is 0.005 cm
c) n = 16 and the raw score (x) = 11.95 cm
The z score equation is given by:
P(x > 11.95 cm) = P(z > -5) = 1 - P(z < -5) = 1 - 0.000001 ≅ 1 ≅ 100%
d) for n = 64, the standard deviation is 0.01 cm, therefore it is more likely to be within .01cm of 12cm
Answer:
uh get it i guess?
Step-by-step explanation:
yes
Answer:
Step-by-step explanation:
608,000
you go to thousands look next number left and round up or down and the rest after the thousands becomes zeros
608,149 rounded to the nearest thousand is 608,000. The process involved looking at the digit in the hundreds place, determining it was less than 5, and therefore replacing it and all numbers to its right with zeroes.
To round 608,149 to the nearest thousand, you need to look at the digit in the hundreds place. This is the 1 in 608,149. Because 1 is less than 5, you leave the number in the thousands place (the 8 in 608,149) as it is, and change all the numbers to its right (149) in the original number to zeros. So 608,149 rounded to the nearest thousand is 608,000.
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