The inside diameter of a piston ring is a random variable with mean 12 cm.
A random sample of 16 piston rings finds a mean and standard deviation of
12.02 cm and 0.04 cm, respectively. The numerical value of the standard
error of the mean, SEM or s.e.(Xbar), is
Records going back more than a century show that the amount of monsoon rainfall varies from year to year according to a distribution that is approximately Normal with mean 852 millimeters (mm) and standard deviation 82 mm.
Let's consider the next two years as a random sample of two years. What is the probability the average monsoon rainfall of these two years would be less than 700 mm? Give your numerical answer and what you typed in your calculator to obtain it.
PLEASE HELP ME IN THESE 2 QUESTIONS.
A random sample of 16 piston rings finds a mean and standard deviation of
12.02 cm and 0.04 cm, respectively. The numerical value of the standard
error of the mean, SEM or s.e.(Xbar), is
Records going back more than a century show that the amount of monsoon rainfall varies from year to year according to a distribution that is approximately Normal with mean 852 millimeters (mm) and standard deviation 82 mm.
Let's consider the next two years as a random sample of two years. What is the probability the average monsoon rainfall of these two years would be less than 700 mm? Give your numerical answer and what you typed in your calculator to obtain it.
PLEASE HELP ME IN THESE 2 QUESTIONS.
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Hd -
The numerical value of the standard error of the mean, SEM or s.e.(Xbar), is 0.04/sqrt(16)
SEM = .04/4 = 0.01 cm
What is the probability the average monsoon rainfall of these two years would be less than 700 mm?
First change the data to z-values like this:
P(X<700) = P(z < (700-852)/82) = P(z< - 1.854)
Next, look up this z-value in a Standard Normal table:
P(z< - 1.854) = 0.03187 (you may have a statistics calculator in which case just use the Normal cdf)
Finally, the problem says to assume the two years are random, so:
P(two years would be less than 700 mm) = (0.03187)(0.03187) = 0.00102
Hope that helps
The numerical value of the standard error of the mean, SEM or s.e.(Xbar), is 0.04/sqrt(16)
SEM = .04/4 = 0.01 cm
What is the probability the average monsoon rainfall of these two years would be less than 700 mm?
First change the data to z-values like this:
P(X<700) = P(z < (700-852)/82) = P(z< - 1.854)
Next, look up this z-value in a Standard Normal table:
P(z< - 1.854) = 0.03187 (you may have a statistics calculator in which case just use the Normal cdf)
Finally, the problem says to assume the two years are random, so:
P(two years would be less than 700 mm) = (0.03187)(0.03187) = 0.00102
Hope that helps